CONTENTS
3.1
Introduction
3.2
Projectile
3.3
Assumptions of projectile motion
3.4
Principle of physical independence of motions
3.5
Types of projectile motion
3.6
Oblique projectile
3.7
Horizontal projectile
3.8
Projectile on an inclined plane
3.9
Variables of circular motion
3.10
Centripetal acceleration
3.11
Centripetal force
3.12
Centrifugal force
3.13
Work done by the centripetal force
3.14
Skidding of vehicle on a level road
3.15
Skidding of object on a rotating platform
3.16
Bending of a cyclist
3.17
Banking of a road
3.18
Overturning of vehicle
3.19
Motion of charged particle in magnetic field
3.20
Reaction of road on car
3.21
Non-uniform circular motion
3.22
Equations of circular motion
3.23
Motion in vertical circle
3.24
Conical pendulum
Sample Problems
Practice Problems (Basic and Advance Level)
Answer Sheet of Practice Problems
Chapter
3
The lion and ring-master both perfectly
understand the concept of projectile motion
and apply it in their circus to cross the fire-
wheel which placed between the two table
and at a particular height from the ground.
134 Motion in Two Dimension
The motion of an object is called two dimensional, if two of the three co-ordinates are
required to specify the position of the object in space changes w.r.t time.
In such a motion, the object moves in a plane. For example, a billiard ball moving over the
billiard table, an insect crawling over the floor of a room, earth revolving around the sun etc.
Two special cases of motion in two dimension are 1. Projectile motion 2. Circular
motion
PROJECTILE MOTION
3.1 Introduction.
A hunter aims his gun and fires a bullet directly towards a monkey sitting on a distant tree.
If the monkey remains in his position, he will be safe but at the instant the bullet leaves the
barrel of gun, if the monkey drops from the tree, the bullet will hit the monkey because the
bullet will not follow the linear path.
The path of motion of a bullet will be parabolic and this motion of bullet is defined as
projectile motion.
If the force acting on a particle is oblique with initial velocity then the motion of particle is
called projectile motion.
3.2 Projectile.
Motion in Two Dimension 135
A body which is in flight through the atmosphere but is not being propelled by any fuel is
called projectile.
Example: (i) A bomb released from an aeroplane in level flight
(ii) A bullet fired from a gun
(iii) An arrow released from bow
(iv) A Javelin thrown by an athlete
3.3 Assumptions of Projectile Motion.
(1) There is no resistance due to air.
(2) The effect due to curvature of earth is negligible.
(3) The effect due to rotation of earth is negligible.
(4) For all points of the trajectory, the acceleration due to gravity ‘g’ is constant in
magnitude and direction.
3.4 Principles of Physical Independence of Motions.
(1) The motion of a projectile is a two-dimensional motion. So, it can be discussed in two
parts. Horizontal motion and vertical motion. These two motions take place independent of each
other. This is called the principle of physical independence of motions.
(2) The velocity of the particle can be resolved into two mutually perpendicular
components. Horizontal component and vertical component.
(3) The horizontal component remains unchanged throughout the flight. The force of
gravity continuously affects the vertical component.
(4) The horizontal motion is a uniform motion and the vertical motion is a uniformly
accelerated retarded motion.
3.5 Types of Projectile Motion.
(1) Oblique projectile motion (2) Horizontal projectile motion (3) Projectile motion
on an inclined plane
3.6 Oblique Projectile.
Y
Y
X
X
Y
136 Motion in Two Dimension
In projectile motion, horizontal component of velocity (u cos
), acceleration (g) and
mechanical energy remains constant while, speed, velocity, vertical component of velocity (u
sin
), momentum, kinetic energy and potential energy all changes. Velocity, and KE are
maximum at the point of projection while minimum (but not zero) at highest point.
(1) Equation of trajectory : A projectile thrown with velocity u at an angle
with the
horizontal. The velocity u can be resolved into two rectangular components.
v cos
component along X–axis and u sin
component along Y–axis.
For horizontal motion x = u cos
t
cosu
x
t
…. (i)
For vertical motion
2
2
1
)sin( gttuy
…. (ii)
From equation (i) and (ii)
22
2
cos
2
1
cos
sin u
x
g
u
x
uy
22
2
cos
2
1
tan u
gx
xy
This equation shows that the trajectory of projectile is parabolic because it is similar to
equation of parabola
y = ax – bx2
Note : Equation of oblique projectile also can be written as
R
x
xy 1tan
(where R = horizontal range =
g
u
2sin
2
)
Sample problems based on trajectory
Problem 1. The trajectory of a projectile is represented by
2/3 2
gxxy
. The angle of projection is
(a) 30o (b) 45o (c) 60o (d) None of these
Solution : (c) By comparing the coefficient of x in given equation with standard equation
22
2
cos2
tan u
gx
xy
3tan
60
Problem 2. The path followed by a body projected along y-axis is given as by
2
)2/1(3 xxy
, if g = 10
m/s, then the initial velocity of projectile will be – (x and y are in m)
(a)
sm/103
(b)
sm/102
(c)
sm/310
(d)
sm/210
Solution : (b) By comparing the coefficient of x2 in given equation with standard equation
.
cos2
tan 22
2
u
gx
xy
X
Y
O
u
u sin
u cos
y
x
P
Motion in Two Dimension 137
2
1
cos2 22
u
g
Substituting
= 60o we get
sec/102mu
.
Problem 3. The equation of projectile is
4
5
16 2
x
xy
. The horizontal range is
(a) 16 m (b) 8 m (c) 3.2 m (d) 12.8 m
Solution : (d) Standard equation of projectile motion
R
x
xy 1tan
Given equation :
4
5
16 2
x
xy
or
5/64
116 x
xy
By comparing above equations
5
64
R
=12.8 m.
(2) Displacement of projectile
)r(
: Let the particle acquires a position P having the
coordinates (x, y) just after time t from the instant of projection. The corresponding position
vector of the particle at time t is
r
as shown in the figure.
jyixr ˆˆ
….(i)
The horizontal distance covered during time t is given as
tuxtvx x
cos
….(ii)
The vertical velocity of the particle at time t is given as
,)( 0gtvv yy
….(iii)
Now the vertical displacement y is given as
2
2/1sin gttuy
….(iv)
Putting the values of x and y from equation (ii) and equation (iv) in equation (i) we obtain
the position vector at any time t as
jgttuitur ˆ
2
1
)sin(
ˆ
)cos( 2
2
22 2
1
)sin()cos(
gttutur
u
gt
u
gt
tur
sin
2
12
and
)/(tan 1xy
)cos(
2/1sin
tan 2
1
tu
gtut
or
cos2
sin2
tan 1u
gtu
Note : The angle of elevation
of the highest point of the
projectile and the angle of projection
are related
to each other as
tan
2
1
tan
v
x
v
y
v
P (x,
y)
x
O
y
r
X
Y
vi
u
X
Y
O
H
R
138 Motion in Two Dimension
Sample problems based on displacement
Problem 4. A body of mass 2 kg has an initial velocity of 3 m/s along OE and it is subjected to a force of
4 Newton’s in OF direction perpendicular to OE. The distance of the body from O after 4
seconds will be
(a) 12 m (b) 28 m (c) 20 m (d) 48 m
Solution : (c) Body moves horizontally with constant initial velocity 3 m/s upto 4 seconds
mutx1243
and in perpendicular direction it moves under the effect of
constant force with zero initial velocity upto 4 seconds.
2
)(
2
1tauty
2
2
1
0t
m
F
2
4
2
4
2
1
m16
So its distance from O is given by
2222 )16()12( yxd
md 20
Problem 5. A body starts from the origin with an acceleration of 6 m/s2 along the x-axis and 8 m/s2
along the y-axis. Its distance from the origin after 4 seconds will be [MP PMT 1999]
(a) 56 m (b) 64 m (c) 80 m (d) 128 m
Solution : (c) Displacement along X- axis :
2
2
1tatux xx
m48)4(6
2
12
Displacement along Y- axis :
mtatuy yy 64)4(8
2
1
2
122
Total distance from the origin
myx 80)64()48(2222
(3) Instantaneous velocity v : In projectile motion, vertical component of velocity changes
but horizontal component of velocity remains always constant.
Example : When a man jumps over the hurdle leaving behind its skateboard then vertical
component of his velocity is changing, but not the horizontal component, which matches with
the skateboard velocity.
As a result, the skateboard stays underneath him, allowing him to land on it.
x
O
y
E
F
u
d
Motion in Two Dimension 139
Let vi be the instantaneous velocity of projectile at time t direction of this velocity is along
the tangent to the trajectory at point P.
jvivv yxi ˆ
22 yxi vvv
222 )sin(cos gtuu
sin2
222 gtutguvi
Direction of instantaneous velocity
cos
sin
tan u
gtu
v
v
x
y
or
sectantan 1u
gt
(4) Change in velocity : Initial velocity (at projection point)
juiuuiˆ
sin
ˆ
cos
Final velocity (at highest point)
jiuu fˆ
0
ˆ
cos
(i) Change in velocity (Between projection point and highest point)
juuuu if ˆ
sin
When body reaches the ground after completing its motion then final velocity
juiuu fˆ
sin
ˆ
cos
(ii) Change in velocity (Between complete projectile motion)
iuuuu if ˆ
sin2
Sample problems based on velocity
Problem 6. In a projectile motion, velocity at maximum height is [AIEEE 2002]
(a)
2
cos
u
(b)
cosu
(c)
2
sin
u
(d) None of these
Solution : (b) In a projectile motion at maximum height body possess only horizontal component of
velocity i.e. u cos
.
Problem 7. A body is thrown at angle 30o to the horizontal with the velocity of 30 m/s. After 1 sec, its
velocity will be (in m/s) (g = 10 m/s2)
(a)
710
(b)
10700
(c)
7100
(d)
Solution : (a) From the formula of instantaneous velocity
sin2
222 tgutguv
o
v30sin1103021)10()30(222
sm /710
Problem 8. A projectile is fired at 30o to the horizontal. The vertical component of its velocity is 80 ms–
1. Its time of flight is T. What will be the velocity of the projectile at t = T/2
(a) 80 ms–1 (b)
380
ms–1 (c) (80/
3
) ms–1 (d) 40 ms–1
Solution : (b) At half of the time of flight, the position of the projectile will be at the highest point of the
parabola and at that position particle possess horizontal component of velocity only.
140 Motion in Two Dimension
Given uvertical
80sin
u
smu o/160
30sin
80
./38030cos160cos smuu o
horizontal
Problem 9. A particle is projected from point O with velocity u in a direction making an angle
with
the horizontal. At any instant its position is at point P at right angles to the initial direction
of projection. Its velocity at point P is
(a) u tan
(b) u cot
(c) u cosec
(d) u sec
Solution : (b) Horizontal velocity at point
cos'' uO
Horizontal velocity at point
sin'' vP
In projectile motion horizontal component of velocity
remains constant throughout the motion
cossin uv
cotuv
Problem 10. A particle P is projected with velocity u1 at an angle of 30o with the horizontal. Another
particle Q is thrown vertically upwards with velocity u2 from a point vertically below the
highest point of path of P. The necessary condition for the two particles to collide at the
highest point is
(a)
21 uu
(b)
21 2uu
(c)
2
2
1u
u
(d)
21 4uu
Solution : (b) Both particle collide at the highest point it means the vertical distance travelled by both the
particle will be equal, i.e. the vertical component of velocity of both particle will be equal
21 30sin uu
2
1
2u
u
21 2uu
Problem 11. Two seconds after projection a projectile is travelling in a direction inclined at 30o to the
horizontal after one more sec, it is travelling horizontally, the magnitude and direction of
its velocity are [RPET 1999]
(a)
o
m60sec,/202
(b)
o
m60sec,/320
(c)
o
m30sec,/406
(d)
o
m30sec,/640
u2
u1
30o
P
Q
90o
u
v
O
P
v sin
90o –
u sin
u cos
90o
u
v
O
P
Motion in Two Dimension 141
Solution : (b) Let in 2 sec body reaches upto point A and after one more sec upto point B.
Total time of ascent for a body is given 3 sec i.e.
3
sin g
u
t
30310sin
u
…..(i)
Horizontal component of velocity remains always constant
30coscos vu
…..(ii)
For vertical upward motion between point O and A
2sin30sin guv o
tguv Using
203030sin
o
v
30 sinAs
u
./20 smv
Substituting this value in equation (ii)
o
u30cos20cos
310
…..(iii)
From equation (i) and (iii)
320u
and
60
Problem 12. A body is projected up a smooth inclined plane (length =
m220
) with velocity u from the
point M as shown in the figure. The angle of inclination is 45o and the top is connected to a
well of diameter 40 m. If the body just manages to cross the well, what is the value of v
(a)
1
40
ms
(b)
1
240
ms
(c)
1
20
ms
(d)
1
220
ms
Solution : (d) At point N angle of projection of the body will be 45°. Let velocity of projection at this point
is v.
If the body just manages to cross the well then
wellofDiameterRange
40
2sin
2
g
v
45As
400
2v
smv /20
But we have to calculate the velocity (u) of the body at point M.
For motion along the inclined plane (from M to N)
Final velocity (v) = 20 m/s,
acceleration (a) = – g sin
= – g sin 45o, distance of inclined plane (s) =
220
m
220.
2
2)20(22 g
u
[Using v2 = u2 + 2as]
4002022 u
u
./220 sm
Problem 13. A projectile is fired with velocity u making angle
with the horizontal. What is the change
in velocity when it is at the highest point
40 m
45o
M
40 m
45o
M
R
N
u
v
u
O
B
v
A
30o
u cos
142 Motion in Two Dimension
(a) u cos
(b) u (c) u sin
(d) (u cos
– u)
Solution : (c) Since horizontal component of velocity remain always constant therefore only vertical
component of velocity changes.
Initially vertical component
sinu
Finally it becomes zero. So change in velocity
sinu
(5) Change in momentum : Simply by the multiplication of mass in the above expression of
velocity (Article-4).
(i) Change in momentum (Between projection point and highest point)
jmuppp if ˆ
sin
(ii) Change in momentum (For the complete projectile motion)
jmuppp if ˆ
sin2
(6) Angular momentum : Angular momentum of projectile at highest point of trajectory
about the point of projection is given by
mvrL
g
u
Hr 2
sin
Here 22
g
um
g
u
umL 2
sincos
2
sin
cos 23
22
Sample problems based on momentum and angular momentum
Problem 14. A body of mass 0.5 kg is projected under gravity with a speed of 98 m/s at an angle of 30o
with the horizontal. The change in momentum (in magnitude) of the body is [MP PET 1997]
(a) 24.5 N–s (b) 49.0 N–s (c) 98.0 N–s (d) 50.0 N–s
Solution : (b) Change in momentum between complete projectile motion = 2mu sin
30sin985.02
=
49 N–s.
Problem 15. A particle of mass 100 g is fired with a velocity 20 m sec–1 making an angle of 30o with the
horizontal. When it rises to the highest point of its path then the change in its momentum
is
(a)
1
sec3
mkg
(b) 1/2 kg m sec–1 (c)
1
sec2
mkg
(d) 1 kg m sec–1
Solution : (d) Horizontal momentum remains always constant
So change in vertical momentum (
p
) = Final vertical momentum – Initial vertical
momentum
sin0 mu
o
P30sin201.0||
secmkg /1
.
X
O
u
Y
P = mv
r
Motion in Two Dimension 143
Problem 16. Two equal masses (m) are projected at the same angle (
) from two points separated by
their range with equal velocities (v). The momentum at the point of their collision is
(a) Zero (b) 2 mv cos
(c) – 2 mv cos
(d) None of these
Solution : (a) Both masses will collide at the highest point of their trajectory with equal and opposite
momentum. So net momentum of the system will be zero.
Problem 17. A particle of mass m is projected with velocity v making an angle of 45o with the horizontal.
The magnitude of the angular momentum of the particle about the point of projection when
the particle is at its maximum height is (where g = acceleration due to gravity) [MP PMT 1994; UPSEAT 2000; MP PET 2001]
(a) Zero (b) mv3/
)24( g
(c) mv3/
)2( g
(d) mv2/2g
Solution : (b)
g
um
L2
sincos 23
=
)24(
3
g
mv
[As
= 45o]
Problem 18. A body is projected from the ground with some angle to the horizontal. What happens to
the angular momentum about the initial position in this motion [AIIMS 2000]
(a) Decreases (b) Increases
(c) Remains same (d) First increases and then decreases
Solution : (b)
Problem 19. In case of a projectile, where is the angular momentum minimum
(a) At the starting point
(b) At the highest point
(c) On return to the ground
(d) At some location other than those mentioned above
Solution : (a)
(7) Time of flight : The total time taken by the projectile to go up and come down to the
same level from which it was projected is called time of flight.
For vertical upward motion 0 = u sin
– gt t = (u sin
/g)
Now as time taken to go up is equal to the time taken to come down so
Time of flight
g
u
tT
sin2
2
v
v
mv cos
mv cos
144 Motion in Two Dimension
(i) Time of flight can also be expressed as :
g
u
Ty
.2
(where uy is the vertical component of
initial velocity).
(ii) For complementary angles of projection
and 90o –
(a) Ratio of time of flight =
gu
gu
T
T
/)90sin(2
/sin2
2
1
=
tan
tan
2
1
T
T
(b) Multiplication of time of flight =
g
u
g
u
TT
cos2sin2
21
g
R
TT 2
21
(iii) If t1 is the time taken by projectile to rise upto point p and t2 is the time taken in
falling from point p to ground level then
g
u
tt
sin2
21
time of flight
or
2
)(
sin 21 ttg
u
and height of the point p is given by
2
11 2
1
sin tgtuh
2
11
21 2
1
2
)( tgt
tt
gh
by solving
221ttg
h
(iv) If B and C are at the same level on trajectory and the time difference between these
two points is t1, similarly A and D are also at the same level and the time difference between
these two positions is t2 then
g
h
tt 8
2
1
2
2
Sample problems based on time of flight
Problem 20. For a given velocity, a projectile has the same range R for two angles of projection if t1 and
t2 are the times of flight in the two cases then [KCET 2003]
(a)
2
21 Rtt
(b)
Rtt
21
(c)
R
tt 1
21
(d)
2
21
1
R
tt
Solution : (b) As we know for complementary angles
g
R
tt 2
21
Rtt
21
.
Problem 21. A body is thrown with a velocity of 9.8 m/s making an angle of 30o with the horizontal. It
will hit the ground after a time [JIPMER 2001, 2002; KCET (Engg.) 2001]
(a) 1.5 s (b) 1 s (c) 3 s (d) 2 s
X
Y
O
h
t2
t1
A
B
C
D
h
O
Y
X
t
1
t
2
P
Motion in Two Dimension 145
Solution : (b)
8.9
30sin8.92sin2 o
g
u
T
= 1sec
Problem 22. Two particles are separated at a horizontal distance x as shown in figure. They are
projected at the same time as shown in figure with different initial speed. The time after
which the horizontal distance between the particles become zero is [CBSE PMT 1999]
(a)
xu 2/
(b)
ux/
(c)
xu/2
(d)
xu/
Solution : (b) Let
1
x
and
2
x
are the horizontal distances travelled by particle A and B respectively in time
t.
t
u
x 30cos.
3
1
…..(i) and
tux o 60cos
2
……(ii)
uttut
u
xx oo 60cos30cos.
3
21
utx
uxt /
Problem 23. A particle is projected from a point O with a velocity u in a direction making an angle
upward with the horizontal. After some time at point P it is moving at right angle with its
initial direction of projection. The time of flight from O to P is
(a)
g
u
sin
(b)
g
u
cosec
(c)
g
u
tan
(d)
g
u
sec
Solution : (b) When body projected with initial velocity
u
by making angle
with the horizontal. Then
after time t, (at point P) it’s direction is perpendicular to
u
.
Magnitude of velocity at point P is given by
.cot
uv
(from sample problem no. 9)
For vertical motion : Initial velocity (at point O)
sinu
Final velocity (at point P)
cosv
coscotu
Time of flight (from point O to P) = t
Applying first equation of motion
tguv
tguu
sincoscot
g
uu
t
coscotsin
22 cossin
sin g
u
g
u
cosec
Problem 24. A ball is projected upwards from the top of tower with a velocity 50 ms–1 making angle 30o
with the horizontal. The height of the tower is 70 m. After how many seconds from the
instant of throwing will the ball reach the ground
60o
30o
u
x
3/u
A
B
90o
u
P
O
v
(90 –
)
v cos
u sin
u cos
146 Motion in Two Dimension
(a) 2.33 sec (b) 5.33 sec (c) 6.33 sec (d) 9.33 sec
Solution : (c) Formula for calculation of time to reach the body on the ground from the tower of height ‘h’
(If it is thrown vertically up with velocity u) is given by
2
2
11 u
gh
g
u
t
So we can resolve the given velocity in vertical direction
and can apply the above formula.
Initial vertical component of velocity
30sin50sin
u
./25 sm
2
)25(
708.92
11
8.9
25
t
= 6.33 sec.
Problem 25. If for a given angle of projection, the horizontal range is doubled, the time of flight
becomes
(a) 4 times (b) 2 times (c)
2
times (d)
2/1
times
Solution : (c)
g
u
R
2sin
2
and
g
u
T
sin2
2
uR
and
uT
(If
and g are constant).
In the given condition to make range double, velocity must be increased upto
2
times that
of previous value. So automatically time of flight will becomes
2
times.
Problem 26. A particle is thrown with velocity u at an angle from the horizontal. Another particle is
thrown with the same velocity at an angle from the vertical. The ratio of times of flight of
two particles will be
(a) tan 2
: 1 (b) cot 2
: 1 (c) tan
: 1 (d) cot
: 1
Solution : (c) For first particles angle of projection from the horizontal is
. So
g
u
T
sin2
1
For second particle angle of projection from the vertical is
. it mean from the horizontal is
).90(
g
u
T)90(sin2
2
g
u
cos2
. So ratio of time of flight
tan
2
1
T
T
.
Problem 27. The friction of the air causes vertical retardation equal to one tenth of the acceleration
due to gravity (Take g = 10 ms–2). The time of flight will be decreased by
(a) 0% (b) 1% (c) 9% (d) 11%
Solution : (c)
g
u
T
sin2
10
11
10
1
2
2
1
g
g
g
g
g
T
T
Fractional decrease in time of flight
11
1
1
21
T
TT
30o
u
70 m
u
sin30o
Motion in Two Dimension 147
Percentage decrease = 9%
(8) Horizontal range : It is the horizontal distance travelled by a body during the time of
flight.
So by using second equation of motion
TuR
cos
)/sin2(cos guu
g
u
2sin
2
g
u
R
2sin
2
(i) Range of projectile can also be expressed as :
R = u cos
× T =
g
2u
sincos2
sin2
cos xy
u
g
uu
g
u
u
g
2u xy
u
R
(where ux and uy are the horizontal and vertical component of initial
velocity)
(ii) If angle of projection is changed from
to
= (90 –
) then range remains unchanged.
R
g
u
g
u
g
u
Ro
2sin
)]90(2sin[
'2sin
'2
2
2
So a projectile has same range at angles of projection
and (90 –
), though time of
flight, maximum height and trajectories are different.
These angles
and 90o –
are called complementary angles of projection and for
complementary angles of projection ratio of range
1
/)]90(2[sin
/2sin
2
2
2
1
gu
gu
R
R
o
1
2
1
R
R
(iii) For angle of projection
1 = (45 –
) and
2 = (45 +
), range will be same and equal to
u2 cos 2
/g.
1 and
2 are also the complementary angles.
X
O
u
Y
Horizontal range
60o
30o
Blast
u
45o
Y
H
148 Motion in Two Dimension
(iv) Maximum range : For range to be maximum
0
d
dR
0
2sin
2
g
u
d
d
cos 2
= 0 i.e. 2
= 90o
= 45o and Rmax = (u2/g)
i.e., a projectile will have maximum range when it is projected at an angle of 45o to the
horizontal and the maximum range will be (u2/g).
When the range is maximum, the height H reached by the projectile
442
45sin
2
sin max
22222 R
g
u
g
u
g
u
H
i.e., if a person can throw a projectile to a maximum distance Rmax, The maximum height to which
it will rise is
4
max
R
.
(v) Relation between horizontal range and maximum height :
g
u
R
2sin
2
and
g
u
H2
sin 22
cot4
2/sin
/2sin
22
2 gu
gu
H
R
cot4HR
(vi) If in case of projectile motion range R is n times the maximum height H
i.e. R = nH
g
u
n
g
u
2
sin2sin 222
]/4[tan n
or
]/4[tan 1n
The angle of projection is given by
]/4[tan 1n
Note : If R = H then
)4(tan 1
or
o
76
.
If R = 4H then
)1(tan 1
or
o
45
.
Sample problem based on horizontal range
Problem 28. A boy playing on the roof of a 10m high building throws a ball with a speed of 10 m/s at
an angle of 30o with the horizontal. How far from the throwing point will the ball be at
the height of 10 m from the ground (g = 10 m/s2, sin 30o =
2
1
,
2
3
30cos
o
) [AIEEE 2003]
(a) 8.66 m (b) 5.20 m (c) 4.33 m (d) 2.60 m
Solution : (a) Simply we have to calculate the range of projectile
g
u
R
2sin
2
10
)302sin()10(2
35R
meter66.8
Problem 29. Which of the following sets of factors will affect the horizontal distance covered by an
athlete in a long–jump event [AMU (Engg.) 2001]
30o
10
m
u
10
m
Motion in Two Dimension 149
(a) Speed before he jumps and his weight
(b) The direction in which he leaps and the initial speed
(c) The force with which he pushes the ground and his speed
(d) The direction in which he leaps and the weight
Solution : (b) Because range
g
ion)of project Angle(2sin)projectionof Velocity( 2
Problem 30. For a projectile, the ratio of maximum height reached to the square of flight time is (g = 10
ms–2)
[EAMCET (Med.) 2000]
(a) 5 : 4 (b) 5 : 2 (c) 5 : 1 (d) 10 : 1
Solution : (a)
g
u
H2
sin22
and
g
u
T
sin2
222
22
2/sin4
2/sin
gu
gu
T
H
4
5
8
10
8 g
Problem 31. A cricketer can throw a ball to a maximum horizontal distance of 100 m. The speed with
which he throws the ball is (to the nearest integer) [Kerala (Med.) 2002]
(a) 30 ms–1 (b) 42 ms–1 (c) 32 ms–1 (d) 35 ms–1
Solution : (c)
g
u
R2
max
= 100 (when
45
)
1000u
./62.31 sm
Problem 32. If two bodies are projected at 30o and 60o respectively, with the same velocity, then [CBSE PMT 2000; JIPMER 2002]
(a) Their ranges are same (b) Their heights are
same
(c) Their times of flight are same (d) All of these
Solution : (a) Because these are complementary angles.
Problem 33. Figure shows four paths for a kicked football. Ignoring the effects of air on the flight, rank
the paths according to initial horizontal velocity component, highest first [AMU (Med.) 2000]
(a) 1, 2, 3, 4 (b) 2, 3, 4, 1 (c) 3, 4, 1, 2 (d) 4, 3, 2, 1
Solution : (d) Range
horizontal component of velocity. Graph 4 shows maximum range, so football
possess maximum horizontal velocity in this case.
Problem 34. Four bodies P, Q, R and S are projected with equal velocities having angles of projection
15o, 30o, 45o and 60o with the horizontal respectively. The body having shortest range is [EAMCET (Engg.) 2000]
0
1
2
3
4
y
x
150 Motion in Two Dimension
(a) P (b) Q (c) R (d) S
Solution : (a) Range of projectile will be minimum for that angle which is farthest from 45°.
Problem 35. A particle covers 50 m distance when projected with an initial speed. On the same surface
it will cover a distance, when projected with double the initial speed [RPMT 2000]
(a) 100 m (b) 150 m (c) 200 m (d) 250 m
Solution : (c)
g
u
R
2sin
2
2
uR
so
2
1
2
1
2
u
u
R
R
2
2
u
u
12 4RR
= 4 50 = 200 m
Problem 36. A bullet is fired from a canon with velocity 500 m/s. If the angle of projection is 15o and g =
10 m/s2. Then the range is [CPMT 1997]
(a) 25 × 103 m (b) 12.5 × 103 m (c) 50 × 102 m (d) 25 × 102 m
Solution : (b)
g
u
RRange
2sin
)( 2
10
152sin500 2
m12500
m
3
105.12
Problem 37. A projectile thrown with a speed v at an angle
has a range R on the surface of earth. For
same v and
, its range on the surface of moon will be
(a) R/6 (b) 6 R (c) R/36 (d) 36 R
Solution : (b)
g
u
R
2sin
2
gR /1
Moon
Earth
Earth
Moon g
g
R
R
= 6
EarthMoon gg 6
1
RRR EarthMoon 66
Problem 38. A projectile is thrown into space so as to have maximum horizontal range R. Taking the
point of projection as origin, the co-ordinates of the point where the speed of the particle is
minimum are
(a) (R, R) (b)
2
,R
R
(c)
4
,
2
RR
(d)
4
,R
R
Solution : (c) For maximum horizontal Range
45
From
cot4HR
= 4H [As
= 45o, for maximum
range.]
Speed of the particle will be minimum at the highest
point of parabola.
So the co-ordinate of the highest point will be (R/2,
R/4)
Problem 39. The speed of a projectile at the highest point becomes
2
1
times its initial speed. The
horizontal range of the projectile will be
45o
Y
X
R/2
R/4
Motion in Two Dimension 151
(a)
g
u2
(b)
g
u
2
2
(c)
g
u
3
2
(d)
g
u
4
2
Solution : (a) Velocity at the highest point is given by
2
cos u
u
(given)
= 45o
Horizontal range
g
u
g
u
g
u
Ro222 )452sin(2sin
Problem 40. A large number of bullets are fired in all directions with same speed u. What is the
maximum area on the ground on which these bullets will spread
(a)
g
u2
(b)
2
4
g
u
(c)
2
4
2
g
u
(d)
2
2
2
g
u
Solution : (b) The maximum area will be equal to area of the circle with radius equal to the maximum
range of projectile
Maximum area
2
max
2Rr
2
2
g
u
2
4
g
u
[As
guRr /
2
max
for
= 45o]
Problem 41. A projectile is projected with initial velocity
./)
ˆ
8
ˆ
6( secmji
If g = 10 ms–2, then horizontal
range is
(a) 4.8 metre (b) 9.6 metre (c) 19.2 metre (d) 14.0 metre
Solution : (b) Initial velocity
smJi /
ˆ
8
ˆ
6
(given)
Magnitude of velocity of projection
22 yx uuu
22 86
= 10 m/s
Angle of projection
6
8
tan
x
y
u
u
3
4
5
4
sin
and
5
3
cos
Now horizontal range
g
u
R
2sin
2
g
u
cossin2
2
10 5
3
5
4
2)10(2
meter6.9
Problem 42. A projectile thrown with an initial speed u and angle of projection 15o to the horizontal has
a range R. If the same projectile is thrown at an angle of 45o to the horizontal with speed
2u, its range will be
(a) 12 R (b) 3 R (c) 8 R (d) 4 R
Solution : (c)
g
u
R
2sin
2
2sin
2
uR
1
2
2
1
2
1
22sin
2sin
u
u
R
R
1
2
12 8
30sin
90sin2 R
u
u
RR o
o
Problem 43. The velocity at the maximum height of a projectile is half of its initial velocity of projection
u. Its range on the horizontal plane is [MP PET 1993]
152 Motion in Two Dimension
(a)
gu 2/3 2
(b)
gu 3/
2
(c)
gu 2/3 2
(d)
gu /3 2
Solution : (a) If the velocity of projection is u then at the highest point body posses only
cosu
2
cos u
u
(given)
60
Now
g
u
R)602sin(
2
g
u2
2
3
Problem 44. A projectile is thrown from a point in a horizontal place such that its horizontal and
vertical velocity component are 9.8 m/s and 19.6 m/s respectively. Its horizontal range is
(a) 4.9 m (b) 9.8 m (c) 19.6 m (d) 39.2 m
Solution : (d) We know
g
uu
Ryx
2
8.9
6.198.92
m2.39
Where
x
u
horizontal component of initial velocity,
y
u
vertical component of initial
velocity.
Problem 45. A particle is projected with a velocity v such that its range on the horizontal plane is twice
the greatest height attained by it. The range of the projectile is (where g is acceleration due
to gravity) [BHU 1984]
(a)
g
v
5
42
(b)
2
5
4
v
g
(c)
g
v2
(d)
g
v
5
42
Solution : (a) We know
cot4HR
cot42 HH
2
1
cot
;
5
2
sin
;
5
1
cos
given 2 As HR
g
u
cos.sin.2.
Range 2
g
u5
1
.
5
2
22
g
u
5
42
Problem 46. The range R of projectile is same when its maximum heights are h1 and h2. What is the
relation between R and h1 and h2 [EAMCET (Med.) 2000]
(a)
21hhR
(b)
21
2hhR
(c)
21
2hhR
(d)
21
4hhR
Solution : (d) For equal ranges body should be projected with angle
or
)90(
o
from the horizontal.
And for these angles :
g
u
h2
sin22
1
and
g
u
h2
cos22
2
by multiplication of both height :
2
222
21 4
cossin
g
u
hh
2
22sin
16
1
g
u
2
21
16 Rhh
21
4hhR
Problem 47. A grasshopper can jump maximum distance 1.6 m. It spends negligible time on the ground.
How far can it go in 10 seconds
Motion in Two Dimension 153
(a)
m25
(b)
m210
(c)
m220
(d)
m240
Solution : (c) Horizontal distance travelled by grasshopper will be maximum for
45
m
g
u
R6.1
2
max
./4 smu
Horizontal component of velocity of grasshopper
45cos4cos
u
sm /22
Total distance covered by it in10 sec.
tuS
cos
m2201022
Problem 48. A projectile is thrown with an initial velocity of
,
ˆˆ jbiav
if the range of projectile is
double the maximum height reached by it then
(a) a = 2b (b) b = a (c) b = 2a (d) b = 4a
Solution : (c) Angle of projection
a
b
v
v
x
y11 tantan
a
b
tan
…(i)
From formula
HHR 2cot4
2
1
cot
2tan
…(ii) [As R = 2H given]
From equation (i) and (ii) b = 2a
1.6 m
152 Motion in Two Dimension
(9) Maximum height : It is the maximum height from the point of projection, a projectile
can reach.
So, by using
asuv 2
22
gHu2)sin(0 2
g
u
H2
sin 22
(i) Maximum height can also be expressed as
g
u
Hy
2
2
(where
y
u
is the vertical component of initial velocity).
(ii)
g
u
H2
2
max
(when sin2
= max = 1 i.e.,
= 90o)
i.e., for maximum height body should be projected vertically upward. So it falls back to the
point of projection after reaching the maximum height.
(iii) For complementary angles of projection
and 90o –
Ratio of maximum height =
gu
gu
H
H
o2)90(sin
2/sin
22
22
2
1
2
2
cos
sin
2
tan
2
2
1tan
H
H
Sample problem based on maximum height
Problem 49. A cricketer can throw a ball to a maximum horizontal distance of 100 m. With the same
effort, he throws the ball vertically upwards. The maximum height attained by the ball is [UPSEAT 2002]
(a) 100 m (b) 80 m (c) 60 m (d) 50 m
Solution : (d)
m
g
u
R100
2
max
(when
45
)
100010100
2u
g
u
H2
2
max
.50
102
1000 metre
(when
90
)
Problem 50. A ball thrown by one player reaches the other in 2 sec. the maximum height attained by the
ball above the point of projection will be about [Pb. PMT 2002]
(a) 10 m (b) 7.5 m (c) 5 m (d) 2.5 m
Solution : (c)
sec
g
u
T2
sin2
(given)
10sin
u
X
O
u
Y
Max. height
Motion in Two Dimension 153
Now
.5
102
)10(
2
sin 222 m
g
u
H
Problem 51. Two stones are projected with the same magnitude of velocity, but making different angles
with horizontal. The angle of projection of one is
/3 and its maximum height is Y, the
maximum height attained by the other stone with as
/6 angle of projection is [J & K CET 2000]
(a) Y (b) 2 Y (c) 3 Y (d)
3
Y
Solution : (d) When two stones are projected with same velocity then for complementary angles
and
(90o –
)
Ratio of maximum heights :
2
2
1tan
H
H
3
3
tan 2
331
2Y
H
H
Problem 52. If the initial velocity of a projectile be doubled. Keeping the angle of projection same, the
maximum height reached by it will
(a) Remain the same (b) Be doubled (c) Be quadrupled (d) Be halved
Solution : (c)
g
u
H2
2sin
2
2
uH
[As
constant]
If initial velocity of a projectile be doubled then H will becomes 4 times.
Problem 53. Pankaj and Sudhir are playing with two different balls of masses m and 2m respectively. If
Pankaj throws his ball vertically up and Sudhir at an angle
, both of them stay in our view
for the same period. The height attained by the two balls are in the ratio
(a) 2 : 1 (b) 1 : 1 (c) 1 : cos
(d) 1 : sec
Solution : (b) Time of flight for the ball thrown by Pankaj
g
u
T1
12
Time of flight for the ball thrown by Sudhir
g
u
To)90sin(2 2
2
g
u
cos2 2
According to problem
21 TT
g
u
g
u
cos22 21
cos
21 uu
Height of the ball thrown by Pankaj
g
u
H2
2
1
1
Height of the ball thrown by Sudhir
g
u
Ho
2
)90(sin22
2
2
g
u
2
cos22
2
gu
gu
H
H
2/cos
2/
22
2
2
1
2
1
= 1 [As
cos
21 uu
]
Problem 54. A boy aims a gun at a bird from a point, at a horizontal distance of 100 m. If the gun can
impart a velocity of 500 ms–1 to the bullet. At what height above the bird must he aim his
gun in order to hit it (take g = 10 ms–2)
m
u1
2m
u2
90o –
Short Trick :
Maximum height H T2
2
2
1
2
1
T
T
H
H
1
2
1
H
H
(As T1 = T2)
154 Motion in Two Dimension
[CPMT 1996]
(a) 20 cm (b) 10 cm (c) 50 cm (d) 100 cm
Solution : (a) Time taken by bullet to travel a horizontal distance of 100 m is given by
sect 5
1
500
100
In this time the bullet also moves downward due to gravity its vertical displacement
2
2
1tgh
2
5
1
10
2
1
m5/1
= 20 cm
So bullet should be fired aiming 20 cm above the bird to hit it.
Problem 55. The maximum horizontal range of a projectile is 400 m. The maximum height attained by it
will be
(a) 100 m (b) 200 m (c) 400 m (d) 800 m
Solution : (a)
mR 400
max
[when
45
]
So from the Relation
cot4HR
45cot4400 H
.100 mH
Problem 56. Two bodies are projected with the same velocity. If one is projected at an angle of 30o and
the other at an angle of 60o to the horizontal, the ratio of the maximum heights reached is
[EAMCET (Med.) 1995; Pb. PMT 2000; AIIMS 2001]
(a) 3 : 1 (b) 1 : 3 (c) 1 : 2 (d) 2 : 1
Solution : (b)
3
1
60sin
30sin
sin
sin
2
2
2
21
2
2
1 o
o
H
H
Problem 57. If time of flight of a projectile is 10 seconds. Range is 500 m. The maximum height attained
by it will be
[RPMT 1997; RPET 1998]
(a) 125 m (b) 50 m (c) 100 m (d) 150 m
Solution : (a)
sec10
sin2 g
u
T
50sin
u
so
g
u
H2
sin22
m125
102
)50(2
.
Problem 58. A man can throw a stone 80 m. The maximum height to which he can raise the stone is
(a) 10 m (b) 15 m (c) 30 m (d) 40 m
Solution : (d) The problem is different from problem no. (54). In that problem for a given angle of
projection range was given and we had find maximum height for that angle.
But in this problem angle of projection can vary,
m
g
u
R80
2
max
[for
45
]
But height can be maximum when body projected vertically up
g
u
Ho
2
90sin22
max
g
u
g
u22
2
1
2
= 40 m
Motion in Two Dimension 155
Problem 59. A ball is thrown at different angles with the same speed u and from the same points and it
has same range in both the cases. If y1 and y2 be the heights attained in the two cases, then
21 yy
(a)
g
u2
(b)
g
u2
2
(c)
g
u
2
2
(d)
g
u
4
2
Solution : (c) Same ranges can be obtained for complementary angles i.e.
and 90o –
g
u
y2
sin22
1
and
g
u
y2
cos22
2
g
u
g
u
yy 2
cos
2
sin 2222
21
g
u
2
2
(10) Projectile passing through two different points on same height at time t1 and t2 : If
the particle passes two points situated at equal height y at
1
tt
and
,
2
tt
then
(i) Height (y):
2
11 2
1
sin gttuy
.....(i)
and
2
22 2
1
sin gttuy
.....(ii)
Comparing equation (i) with equation (ii)
2
sin 21 ttg
u
Substituting this value in equation (i)
2
11
21 2
1
2gtt
tt
gy
221tgt
y
(ii) Time (t1 and t2):
2
2
1
sin gttuy
0
2sin2
2 g
y
t
g
u
t
2
sin
2
11
sin
u
gy
g
u
t
2
1sin
2
11
sin
u
gy
g
u
t
and
2
2sin
2
11
sin
u
gy
g
u
t
(11) Motion of a projectile as observed from another projectile : Suppose two balls A and
B are projected simultaneously from the origin, with initial velocities u1 and u2 at angle
1 and
2, respectively with the horizontal.
The instantaneous positions of the two balls are given by
Ball A : x1 = (u1 cos
1)t
2
111 2
1
)sin( gttuy
u2
1
u1
X
Y
O
2
B
y1 –
y2
X
O
y
y
t = t2
u
t = t1
Y
156 Motion in Two Dimension
Ball B : x2 = (u2 cos
2)t
2
222 2
1
)sin( gttuy
The position of the ball A with respect to ball B is given by
tuuxxx )coscos( 221121
tuuyyy )sinsin( 221121
Now
2211
2211 coscos
sinsin
uu
uu
x
y
constant
Thus motion of a projectile relative to another projectile is a straight line.
(12) Energy of projectile : When a projectile moves upward its kinetic energy decreases,
potential energy increases but the total energy always remain constant.
If a body is projected with initial kinetic energy K(=1/2 mu2), with angle of projection
with the horizontal then at the highest point of trajectory
(i) Kinetic energy
222 cos
2
1
)cos(
2
1muum
2
cos' KK
(ii) Potential energy
mgH
g
u
mg 2
sin 22
22 sin
2
1mu
g
u
H2
sin
As 22
(iii) Total energy = Kinetic energy + Potential energy
2222 sin
2
1
cos
2
1mumu
=
2
2
1mu
= Energy at the point of projection.
This is in accordance with the law of conservation of energy.
Sample problems based on energy
Problem 60. A projectile is projected with a kinetic energy K. Its range is R. It will have the minimum
kinetic energy, after covering a horizontal distance equal to [UPSEAT 2002]
(a) 0.25 R (b) 0.5 R (c) 0.75 R (d) R
Solution : (b) Projectile possess minimum kinetic energy at the highest point of the trajectory i.e. at a
horizontal distance
.2/R
Problem 61. A projectile is fired at 30o with momentum p. Neglecting friction, the change in kinetic
energy when it returns to the ground will be
(a) Zero (b) 30% (c) 60% (d) 100%
Solution : (a) According to law of conservation of energy, projectile acquire same kinetic energy when it
comes at same level.
Problem 62. A particle is projected making angle 45o with horizontal having kinetic energy K. The
kinetic energy at highest point will be [CBSE PMT 2000, 01; AIEEE 2002]
u
K
Y
X
K
= Kcos2
u cos
Motion in Two Dimension 157
(a)
2
K
(b)
2
K
(c) 2K (d) K
Solution : (b) Kinetic energy at the highest point
2
cos' KK
o
K45cos2
2/K
Problem 63. Two balls of same mass are projected one vertically upwards and the other at angle 60o
with the vertical. The ratio of their potential energy at the highest point is
(a) 3 : 2 (b) 2 : 1 (c) 4 : 1 (d) 4 : 3
Solution : (c) Potential energy at the highest point is given by
22 sin
2
1muPE
For first ball
90
2
12
1
)( muPE
For second ball
30)6090(oo
from the horizontal
30sin
2
1
)( 22
2muPE
2
8
1mu
1:4
)(
II
I
PE
PE
Problem 64. In the above problem, the kinetic energy at the highest point for the second ball is K. What
is the kinetic energy for the first ball
(a) 4 K (b) 3 K (c) 2 K (d) Zero
Solution : (d) KE at the highest point
22 cos
2
1muKE
For first ball
= 90o KE = 0
Problem 65. A ball is thrown at an angle
with the horizontal. Its initial kinetic energy is 100 J and it
becomes 30 J at the highest point. The angle of projection is
(a) 45o (b) 30o (c) cos–1 (3/10) (d)
)10/3(cos 1
Solution : (d) KE at highest point
2
cos' KK
2
cos10030
10
3
cos 2
10
3
cos 1
3.7 Horizontal Projectile.
A body be projected horizontally from a certain height ‘y’ vertically above the ground with
initial velocity u. If friction is considered to be absent, then there is no other horizontal force
which can affect the horizontal motion. The horizontal velocity therefore remains constant and
so the object covers equal distance in horizontal direction in equal intervals of time.
(1) Trajectory of horizontal projectile : The horizontal displacement x is governed by the
equation
x = ut
u
x
t
…. (i)
X
P(x, y)
Y
O
u
y
x
158 Motion in Two Dimension
The vertical displacement y is governed by
2
1
y
gt2 …. (ii)
(since initial vertical velocity is zero)
By substituting the value of t in equation (ii)
2
2
2
1
u
xg
y
Sample problems based on trajectory
Problem 66. An aeroplane is flying at a constant horizontal velocity of 600 km/hr at an elevation of 6
km towards a point directly above the target on the earth’s surface. At an appropriate time,
the pilot releases a ball so that it strikes the target at the earth. The ball will appear to be
falling [MP PET 1993]
(a) On a parabolic path as seen by pilot in the plane
(b) Vertically along a straight path as seen by an observer on the ground near the target
(c) On a parabolic path as seen by an observer on the ground near the target
(d) On a zig-zag path as seen by pilot in the plane
Solution : (c)
The path of the ball appears parabolic to a observer near the target because it is at rest. But
to a Pilot the path appears straight line because the horizontal velocity of aeroplane and the
ball are equal, so the relative horizontal displacement is zero.
Problem 67. The barrel of a gun and the target are at the same height. As soon as the gun is fired, the
target is also released. In which of the following cases, the bullet will not strike the target
(a) Range of projectile is less than the initial distance between the gun and the target
(b) Range of projectile is more than the initial distance between the gun and the target
(c) Range of projectile is equal to the initial distance between the gun and target
(d) Bullet will always strike the target
Solution : (a) Condition for hitting of bullet with target initial distance between the gun and target
Range of projectile.
Problem 68. A ball rolls off top of a staircase with a horizontal velocity u m/s. If the steps are h metre
high and b mere wide, the ball will just hit the edge of nth step if n equals to
Y
X
Motion in Two Dimension 159
(a)
2
2
gb
hu
(b)
2
28
gb
u
(c)
2
2
2
gb
hu
(d)
2
2
2
hb
gu
Solution : (c) By using equation of trajectory
2
2
2u
gx
y
for given condition
2
2
2
)(
u
nbg
nh
2
2
2
gb
hu
n
(2) Displacement of Projectile
)(r
: After time t, horizontal displacement
utx
and vertical
displacement
2
2
1gty
.
So, the position vector
jgtiutrˆ
2
1
ˆ2
Therefore
2
2
1
u
gt
utr
and
u
gt
2
tan 1
u
gy
2
tan 1
g
y
t2
as
(3) Instantaneous velocity : Throughout the motion, the horizontal component of the
velocity is vx = u.
The vertical component of velocity increases with time and is given by
vy = 0 + g t = g t (From v = u + g t)
So,
jvivv yx ˆˆ
=
jtgiuv ˆˆ
i.e.
2
2gtuv
2
1
u
gt
u
Again
jgyiuv ˆ
2
ˆ
i.e.
gyuv 2
2
Direction of instantaneous velocity :
x
y
v
v
tan
x
y
v
v
1
tan
u
gy2
tan 1
or
u
gt
1
tan
Where
is the angle of instantaneous velocity from the horizontal.
X
v
vx
vy
Y
O
u
P (x,y)
x
y
r
u
nb
b
nh
h
160 Motion in Two Dimension
Sample problems based on velocity
Problem 69. A body is projected horizontally from the top of a tower with initial velocity 18 ms–1. It hits
the ground at angle 45o. What is the vertical component of velocity when it strikes the
ground
(a) 9 ms–1 (b) 9
2
ms–1 (c) 18 ms–1 (d) 18
2
ms–1
Solution : (c) When the body strikes the ground
18
45tan y
x
y
ov
v
v
= 1
./18 smvy
Problem 70. A man standing on the roof of a house of height h throws one particle vertically downwards
and another particle horizontally with the same velocity u. The ratio of their velocities
when they reach the earth’s surface will be
(a)
uugh :2 2
(b) 1 : 2 (c) 1 : 1 (d)
ghugh 2:2 2
Solution : (c) For first particle :
ghuv 2
22
ghuv 2
2
For second particle :
ghughuvvv yx 22 2
2
222
So the ratio of velocities will be 1 : 1.
Problem 71. A staircase contains three steps each 10 cm high and 20 cm wide. What should be the
minimum horizontal velocity of a ball rolling off the uppermost plane so as to hit directly
the lowest plane
(a) 0.5 m/s (b) 1 m/s (c) 2 m/s (d) 4 m/s
Solution : (c) Formula for this condition is given by
2
2
2
gb
hu
n
secmseccmu
u/2/200
2010
102
32
2
2
u = 18
45o
vy
vx
u
h
u
v
v
vx
vy
b
h
u
where h = height of each step, b = width
of step, u = horizontal velocity of
projection, n = number of step.
Motion in Two Dimension 161
(4) Time of flight : If a body is projected horizontally from a height h with velocity u and
time taken by the body to reach the ground is T, then
2
2
1
0gTh
(for vertical motion)
g
h
T2
Sample problems based on time of flight
Problem 72. Two bullets are fired simultaneously, horizontally and with different speeds from the same
place. Which bullet will hit the ground first [Orissa JEE 2003]
(a) The faster one (b) Depends on their mass
(c) The slower one (d) Both will reach simultaneously
Solution : (d)
Problem 73. An aeroplane is flying at a height of 1960 m in horizontal direction with a velocity of 360
km/hr. When it is vertically above the point. A on the ground, it drops a bomb. The bomb
strikes a point B on the ground, then the time taken by the bomb to reach the ground is
(a)
220
sec (b) 20 sec (c)
210
sec (d) 10 sec
Solution : (b)
8.9
196022
g
h
t
= 20 sec
(5) Horizontal range : Let R is the horizontal distance travelled by the body
2
0
2
1TuTR
(for horizontal motion)
g
h
uR 2
Sample problems based on horizontal range
Problem 74. A bomb is dropped on an enemy post by an aeroplane flying with a horizontal velocity of 60
km/hr and at a height of 490 m. How far the aeroplane must be from the enemy post at
time of dropping the bomb, so that it may directly hit the target. (g = 9.8 m/s2)
(a)
m
3
100
(b)
m
3
500
(c)
m
3
200
(d)
m
3
400
Solution : (b)
tuS
g
h
u2
8.9
4902
18
5
60
m
3
500
Problem 75. A body is thrown horizontally with velocity
gh2
from the top of a tower of height h. It
strikes the level ground through the foot of tower at a distance x from the tower. The value
of x is
(a) h (b)
2
h
(c) 2 h (d)
3
2h
u
h
Range
162 Motion in Two Dimension
Solution : (c)
g
h
ux 2
g
h
gh 2
2
hx 2
Problem 76. An aeroplane moving horizontally with a speed of 720 km/h drops a food packet,
while flying at a height of 396.9 m. The time taken by a food packet to reach the
ground and its horizontal range is (Take g = 9.8 m/sec2) [AFMC 2001]
(a) 3 sec and 2000 m (b) 5 sec and 500 m (c) 8 sec and 1500 m (d) 9 sec and 1800 m
Solution : (d) Time of descent
g
h
t2
8.9
9.3962
sect 9
and horizontal distance
tuS
9
18
5720
S
m1800
(6) If projectiles A and B are projected horizontally with different initial velocity from
same height and third particle C is dropped from same point then
(i) All three particles will take equal time to reach the ground.
(ii) Their net velocity would be different but all three particle possess same vertical
component of velocity.
(iii) The trajectory of projectiles A and B will be straight line w.r.t. particle C.
(7) If various particles thrown with same initial velocity but indifferent direction then
h
C
A
B
h
A
A
A
u
E
E
u
B
C
D
u
u
u
ghu2
h
x
u = 720 km/h
s
Motion in Two Dimension 163
(i) They strike the ground with same speed at different times irrespective of their initial
direction of velocities.
(ii) Time would be least for particle E which was thrown vertically downward.
(iii) Time would be maximum for particle A which was thrown vertically upward.
3.8 Projectile Motion on an Inclined Plane.
Let a particle be projected up with a speed u from an inclined plane which makes an angle
with the horizontal velocity of projection makes an angle
with the inclined plane.
We have taken reference x-axis in the direction of plane.
Hence the component of initial velocity parallel and perpendicular to the plane are equal to
cosu
and
sinu
respectively i.e.
cos
|| uu
and
sinuu
.
The component of g along the plane is
sing
and
perpendicular to the plane is
cosg
as shown in the figure i.e.
sin
|| ga
and
cosga
.
Therefore the particle decelerates at a rate of
sing
as it
moves from O to P.
(1) Time of flight : We know for oblique projectile motion
g
u
T
sin2
or we can say
a
u
T2
Time of flight on an inclined plane
cos
sin2
g
u
T
(2) Maximum height : We know for oblique projectile motion
g
u
H2
sin 22
or we can say
a
u
H2
2
Maximum height on an inclined plane
cos2
sin 22
g
u
H
(3) Horizontal range : For one dimensional motion
2
2
1atuts
Horizontal range on an inclined plane
2
|||| 2
1TaTuR
2
sin
2
1
cos TgTuR
2
cos
sin2
sin
2
1
cos
sin2
cos
g
u
g
g
u
uR
By solving
2
2
cos
)cos(sin
2
g
u
R
O
Y
X
u
t =0
t =T
g
ay= g cos
ax=–g sin
P
164 Motion in Two Dimension
(i) Maximum range occurs when
24
(ii) The maximum range along the inclined plane when the projectile is thrown upwards is
given by
)sin1(
2
max
g
u
R
(iii) The maximum range along the inclined plane when the projectile is thrown downwards
is given by
)sin1(
2
max
g
u
R
Sample problem based on inclined projectile
Problem 77. For a given velocity of projection from a point on the inclined plane, the maximum range
down the plane is three times the maximum range up the incline. Then, the angle of
inclination of the inclined plane is
(a) 30o (b) 45o (c) 60o (d) 90o
Solution : (a) Maximum range up the inclined plane
)sin1(
)( 2
max
g
u
Rup
Maximum range down the inclined plane
)sin1(
)( 2
max
g
u
Rdown
and according to problem :
)sin1(
3
)sin1(
22
g
u
g
u
By solving
= 30o
Problem 78. A shell is fired from a gun from the bottom of a hill along its slope. The slope of the hill is
= 30o, and the angle of the barrel to the horizontal
= 60o. The initial velocity v of the shell
is 21 m/sec. Then distance of point from the gun at which shell will fall
(a) 10 m (b) 20 m (c) 30 m (d) 40 m
Solution : (c) Here u = 21 m/sec,
= 30o,
=
–
= 60o – 30o = 30o
Maximum range
2
2
cos
)cos(sin
2
g
u
R
m
o
oo 30
30cos8.9
60cos30sin)21(2
2
2
Problem 79. The maximum range of rifle bullet on the horizontal ground is 6 km its maximum range on an
inclined of 30o will be
(a) 1 km (b) 2 km (c) 4 km (d) 6 km
Solution : (c) Maximum range on horizontal plane
km
g
u
R6
2
(given)
Maximum range on a inclined plane
)sin1(
2
max
g
u
R
Putting
= 30o
.46
3
2
3
2
)30sin1(
22
max km
g
u
g
u
Ro
164 Motion in Two Dimension
CIRCULAR MOTION
Circular motion is another example of motion in two dimensions. To create circular motion
in a body it must be given some initial velocity and a force must then
act on the body which is always directed at right angles to
instantaneous velocity.
Since this force is always at right angles to the displacement due to
the initial velocity therefore no work is done by the force on the
particle. Hence, its kinetic energy and thus speed is unaffected. But due
to simultaneous action of the force and the velocity the particle follows
resultant path, which in this case is a circle. Circular motion can be classified into two types –
Uniform circular motion and non-uniform circular motion.
3.9 Variables of Circular Motion.
(1) Displacement and distance : When particle moves in a circular path describing an
angle
during time t (as shown in the figure) from the position A to the position B, we see that
the magnitude of the position vector
r
(that is equal to the radius of the circle) remains
constant. i.e.,
rrr 21
and the direction of the position vector changes from time to time.
(i) Displacement : The change of position vector or the displacement
r
of the particle
from position A to the position B is given by referring the figure.
12 rrr
12 rrrr
cos2 21
2
2
2
1rrrrr
Putting
rrr 21
we obtain
cos.2
22 rrrrr
2
sin22cos12 222
rrr
2
sin2
rr
(ii) Distance : The distanced covered by the particle during the time t is given as
d = length of the arc AB = r
(iii) Ratio of distance and displacement :
2/sin2
r
r
r
d
)2/(cosec
2
O
1
r
2
r
A
B
1
v
2
v
1
r
2
r
O
r
A
B
F
F
F
F
1
v
2
v
3
v
4
v
Motion in Two Dimension 165
Sample problems based on distance and displacement
Problem 80. A particle is rotating in a circle of radius r. The distance traversed by it in completing half
circle would be
(a) r (b)
r
(c)
r
2
(d) Zero
Solution : (b) Distance travelled by particle = Semi-circumference =
r.
Problem 81. An athlete completes one round of a circular track of radius 10 m in 40 sec. The distance
covered by him in 2 min 20 sec is [Kerala PMT 2002]
(a) 70 m (b) 140 m (c) 110 m (d) 220 m
Solution : (d)
5.3
sec40
sec140
periodTime
tionmoof Total time
)(revolutionof No. n
Distance covered by an athlete in revolution
)2( rn
)2(5.3 r
10
7
22
25.3
= 220 m.
Problem 82. A wheel covers a distance of 9.5 km in 2000 revolutions. The diameter of the wheel is
[RPMT 1999; BHU 2000]
(a) 15 m (b) 7.5 m (c) 1.5 m (d) 7.5 m
Solution : (c) Distance
)2( rn
)(2000105.9 3D
.5.1
2000
105.9 3mD
(2) Angular displacement (
) : The angle turned by a body moving on a circle from some
reference line is called angular displacement.
(i) Dimension = [M0L0T0] (as
=
radius)/arc
.
(ii) Units = Radian or Degree. It is some times also specified in terms of fraction or
multiple of revolution.
(iii)
Revolution1360rad2 o
(iv) Angular displacement is a axial vector quantity.
Its direction depends upon the sense of rotation of the object and can be given by Right
Hand Rule; which states that if the curvature of the fingers of right hand
represents the sense of rotation of the object, then the thumb, held
perpendicular to the curvature of the fingers, represents the direction of
angular displacement vector.
(v) Relation between linear displacement and angular displacement
rs
or
rs
Sample problem based on angular displacement
r
O
S
166 Motion in Two Dimension
Problem 83. A flywheel rotates at a constant speed of 3000 rpm. The angle described by the shaft in
radian in one second is
(a) 2
(b) 30
(c) 100
(d) 3000
Solution : (c) Angular speed
rpm3000
rps50
sec/250 rad
sec/100 rad
i.e. angle described by the shaft in one second is
100
rad.
Problem 84. A particle completes 1.5 revolutions in a circular path of radius 2 cm. The angular
displacement of the particle will be – (in radian)
(a)
6
(b)
3
(c)
2
(d)
Solution : (b) 1 revolution mean the angular displacement of
rad
2
5.1
revolution means
25.1
=
rad
3
.
(3) Angular velocity (
) : Angular velocity of an object in circular motion is defined as the
time rate of change of its angular displacement.
(i) Angular velocity
=
dt
d
t
Lt
t
0
takentime
tracedangle
dt
d
(ii) Dimension : [M0L0T–1]
(iii) Units : Radians per second (rad.s–1) or Degree per second.
(iv) Angular velocity is an axial vector.
(v) Relation between angular velocity and linear velocity
rv
Its direction is the same as that of
. For anticlockwise rotation of the point object on the
circular path, the direction of
, according to Right hand rule is along the axis of circular path
directed upwards. For clockwise rotation of the point object on the circular path, the direction
of
is along the axis of circular path directed downwards.
Note : It is important to note that nothing actually moves in the direction of the angular
velocity vector
. The direction of
simply represents that the rotational motion
is taking place in a plane perpendicular to it.
(vi) For uniform circular motion
remains constant where as for non-uniform motion
varies
with respect to time.
Sample problems based on angular velocity
Problem 85. A scooter is going round a circular road of radius 100 m at a speed of 10 m/s. The angular
speed of the scooter will be [Pb. PMT 2002]
Motion in Two Dimension 167
(a) 0.01 rad/s (b) 0.1 rad/s (c) 1 rad/s (d) 10 rad/s
Solution : (b)
secrad
r
v/1.0
100
10
Problem 86. The ratio of angular velocity of rotation of minute hand of a clock with the angular velocity
of rotation of the earth about its own axis is
(a) 12 (b) 6 (c) 24 (d) None of these
Solution : (c)
min60
2rad
handMinute
and
min6024
2
24
2rad
hr
Rad
Earth
1:24
Earth
handMinute
Problem 87. A particle P is moving in a circle of radius ‘a’ with a uniform speed v. C is the centre of the
circle and AB is a diameter. When passing through B the angular velocity of P about A and C
are in the ratio [NCERT 1982]
(a) 1 : 1 (b) 1 : 2 (c) 2 : 1 (d) 4 : 1
Solution : (b) Angular velocity of P about A
a
v
A2
Angular velocity of P about C
a
v
C
2:1
C
A
Problem 88. The length of the seconds hand of a watch is 10 mm. What is the change in the angular
speed of the watch after 15 seconds
(a) Zero (b)
1
)2/10(
mms
(c)
1
)/20(
mms
(d)
1
210
mms
Solution : (a) Angular speed of seconds hand of watch is constant and equal to
sec
rad
60
2
secrad /
30
. So
change in angular speed will be zero.
(4) Change in velocity : We want to know the magnitude and direction of the change in
velocity of the particle which is performing uniform circular motion as it moves from A to B
during time t as shown in figure. The change in velocity vector is given as
12 vvv
or
12 vvv
cos2 21
2
2
2
1vvvvv
For uniform circular motion
vvv 21
So
cos12 2 vv
2
sin2
v
O
A
B
1
v
2
v
1
v
1
v
2
v
12 vvv
P
A
C
a
B
168 Motion in Two Dimension
The direction of
v
is shown in figure that can be given as
2/90
2
180
o
o
Note : Relation between linear velocity and angular velocity.
In vector form
rv
Sample problems based on velocity
Problem 89. If a particle moves in a circle describing equal angles in equal times, its velocity vector
[CPMT 1972, 74; JIPMER 1997]
(a) Remains constant (b) Changes in magnitude
(c) Changes in direction (d) Changes both in
magnitude and direction
Solution : (c) In uniform circular motion velocity vector changes in direction but its magnitude always
remains constant.
|||||||| 4321 vvvv
= constant
Problem 90. A body is whirled in a horizontal circle of radius 20 cm. It has angular velocity of 10 rad/s.
What is its linear velocity at any point on circular path [CBSE PMT 1996; JIPMER 2000]
(a) 10 m/s (b) 2 m/s (c) 20 m/s (d)
2
m/s
Solution : (b)
rv
sm/2102.0
Problem 91. The linear velocity of a point on the equator is nearly (radius of the earth is 6400 km)
(a) 800 km/hr (b) 1600 km/hr (c) 3200 km/hr (d) 6400 km/hr
Solution : (b)
hr
rad
kmwrv 24
2
6400
hrkm /1675
hrkm /1600
Problem 92. A particle moves along a circle with a uniform speed v. After it has made an angle of 60o its
speed will be
[RPMT 1998]
(a)
2v
(b)
2
v
(c)
3
v
(d) v
Solution : (d) Uniform speed means speed of the particle remains always constant.
1
v
2
v
3
v
4
v
Motion in Two Dimension 169
Problem 93. A particle is moving along a circular path of radius 2 m and with uniform speed of 5 ms–1.
What will be the change in velocity when the particle completes half of the revolution
(a) Zero (b) 10 ms–1 (c)
1
210
ms
(d)
1
2/10
ms
Solution : (b)
2
sin2
vv
2
180
sin52 o
o
90sin52
sm /10
Problem 94. What is the value of linear velocity, if
kji ˆ
ˆ
4
ˆ
3
and
kjir ˆ
6
ˆ
6
ˆ
5
[Pb. PMT 2000]
(a)
kji ˆ
3
ˆ
2
ˆ
6
(b)
kji ˆ
2
ˆ
13
ˆ
18
(c)
kji ˆ
6
ˆ
13
ˆ
4
(d)
kji ˆ
8
ˆ
2
ˆ
6
Solution : (b)
665
143
ˆ
ˆ
ˆ
kJi
rv
kJiv ˆ
)2018(
ˆ
)518()624(
kJi ˆ
2
ˆ
13
ˆ
18
Problem 95. A particle comes round circle of radius 1 m once. The time taken by it is 10 sec. The average
velocity of motion is [JIPMER 1999]
(a) 0.2
m/s (b) 2
m/s (c) 2 m/s (d) Zero
Solution : (d) In complete revolution total displacement becomes zero. So the average velocity will be
zero.
Problem 96. Two particles of mass M and m are moving in a circle of radii R and r. If their time-periods
are same, what will be the ratio of their linear velocities [CBSE PMT 2001]
(a) MR : mr (b) M : m (c) R : r (d) 1 : 1
Solution : (c)
22
11
2
1
r
r
v
v
. Time periods are equal i.e.
21
r
R
r
r
v
v
2
1
2
1
(5) Time period (T) : In circular motion, the time period is defined as the time taken by
the object to complete one revolution on its circular path.
(i) Units : second.
(ii) Dimension : [M0L0T]
(iii) Time period of second’s hand of watch = 60 second.
(iv) Time period of minute’s hand of watch = 60 minute
(v) Time period of hour’s hand of watch = 12 hour
(6) Frequency (n) : In circular motion, the frequency is defined as the number of
revolutions completed by the object on its circular path in a unit time.
180o
1
v
2
v
170 Motion in Two Dimension
(i) Units : s–1 or hertz (Hz).
(ii) Dimension : [M0L0T–1]
Note : Relation between time period and frequency : If n is the frequency of revolution of
an object in circular motion, then the object completes n revolutions in 1 second.
Therefore, the object will complete one revolution in 1/n second.
nT /1
Relation between angular velocity, frequency and time period : Consider a point
object describing a uniform circular motion with frequency n and time period T.
When the object completes one revolution, the angle traced at its axis of circular
motion is 2
radians. It means, when time t = T,
2
radians. Hence, angular
velocity
n
Tt
2
2
(
T = 1/n)
n
T
2
2
If two particles are moving on same circle or different coplanar concentric circles
in same direction with different uniform angular speeds
A and
B respectively, the
angular velocity of B relative to A will be
AB
rel
So the time taken by one to complete one revolution around O with respect to the
other (i.e., time in which B complete one revolution around O with respect to the
other (i.e., time in which B completes one more or less revolution around O than A)
21
21
12rel
22
TT
TT
T
2
as T
Special case : If
0, rel
AB
and so T = ., particles will maintain their position
relative to each other. This is what actually happens in case of geostationary
satellite (
1 =
2 = constant)
(7) Angular acceleration (
) : Angular acceleration of an object in circular motion is
defined as the time rate of change of its angular velocity.
(i) If
be the change in angular velocity of the object in time interval t and t + t, while
moving on a circular path, then angular acceleration of the object will be
dt
d
t
Lt
t
0
2
2
dt
d
(ii) Units : rad. s–2
(iii) Dimension : [M0L0T–2]
(iv) Relation between linear acceleration and angular acceleration
ra
Motion in Two Dimension 171
(v) For uniform circular motion since
is constant so
0 dt
d
(vi) For non-uniform circular motion
0
Note : Relation between linear (tangential) acceleration and angular acceleration
ra
For uniform circular motion angular acceleration is zero, so tangential acceleration
also is equal to zero.
For non-uniform circular motion a 0 (because
0).
Sample problems based on angular acceleration
Problem 97. A body is revolving with a uniform speed v in a circle of radius r. The angular acceleration
of the body is
(a)
r
v
(b) Zero
(c)
r
v2
along the radius and towards the centre (d)
r
v2
along the radius and
away from the centre
Solution : (b) In uniform circular motion
constant so
0 dt
d
Problem 98. The linear acceleration of a car is 10m/s2. If the wheels of the car have a diameter of 1m,
the angular acceleration of the wheels will be
(a) 10 rad/sec2 (b) 20 rad/sec2 (c) 1 rad/sec2 (d) 2 rad/sec2
Solution : (b)
radius
onacceleratilinear
onacceleratiAngular
5.0
10
2
/20 secrad
Problem 99. The angular speed of a motor increases from 600 rpm to 1200 rpm in 10 s. What is the
angular acceleration of the motor
(a)
2
600
secrad
(b)
2
60
secrad
(c)
2
60
secrad
(d)
2
2
secrad
Solution : (d)
t
nn
t
)(2 1212
2
6010
)6001200(2
sec
rad
2
/2 secrad
3.10 Centripetal Acceleration.
(1) Acceleration acting on the object undergoing uniform circular motion is called
centripetal acceleration.
(2) It always acts on the object along the radius towards the centre of
the circular path.
(3) Magnitude of centripetal acceleration
r
T
rnr
r
v
a2
2
22
24
4
ac
v
172 Motion in Two Dimension
(4) Direction of centripetal acceleration : It is always the same as that of
. When t
decreases,
also decreases. Due to which
becomes more and more perpendicular to
.
When t 0,
becomes perpendicular to the velocity vector. As the velocity vector of the
particle at an instant acts along the tangent to the circular path, therefore
and hence the
centripetal acceleration vector acts along the radius of the circular path at that point and is
directed towards the centre of the circular path.
Sample problems based on centripetal acceleration
Problem 100. If a cycle wheel of radius 4 m completes one revolution in two seconds. Then acceleration
of the cycle will be
[Pb. PMT 2001]
(a)
22 /sm
(b)
22 /2 sm
(c)
22 /4 sm
(d)
2
/8 sm
Solution : (c) Given
mr 4
and
.sec2 ondsT
r
T
ac2
2
4
4
)2(
4
2
2
22 /4 sm
Problem 101. A stone is tied to one end of a spring 50 cm long is whirled in a horizontal circle with a
constant speed. If the stone makes 10 revolutions in 20 s, what is the magnitude of
acceleration of the stone [Pb. PMT 2000]
(a) 493 cm/sec2 (b) 720 cm/sec2 (c) 860 cm/sec2 (d) 990 cm/sec2
Solution : (a) Time period
revolutionof No.
Total time
10
20
sec2
r
T
ac.
4
2
2
2
2
2/)2/1(
)2(
4sm
2
/93.4 sm
2
/493 scm
Problem 102. A particle moves with a constant speed v along a circular path of radius r and completes the
circle in time T. What is the acceleration of the particle [Orissa JEE 2002]
(a)
mg
(b)
T
v
2
(c)
T
r2
(d)
T
v2
Solution : (b)
vr
r
v
ac 2
2
=
T
v
2
T
v
2
Problem 103. If the speed of revolution of a particle on the circumference of a circle and the speed gained
in falling through a distance equal to half the radius are equal, then the centripetal
acceleration will be
(a)
2
g
(b)
4
g
(c)
3
g
(d) g
Motion in Two Dimension 173
Solution : (d) Speed gain by body falling through a distance h is equal to
ghv2
2
2r
g
[As
2
r
h
given]
grv
g
r
v
2
Problem 104. Two cars going round curve with speeds one at 90 km/h and other at 15 km/h. Each car
experiences same acceleration. The radii of curves are in the ratio of [EAMCET (Med.) 1998]
(a) 4 : 1 (b) 2 : 1 (c) 16 : 1 (d) 36 : 1
Solution : (d) Centripetal acceleration =
2
2
2
1
2
1r
v
r
v
(given)
1
36
15
90 2
2
2
1
2
1
v
v
r
r
Problem 105. A wheel of radius
m20.0
is accelerated from rest with an angular acceleration of
2
/1 srad
.
After a rotation of
o
90
the radial acceleration of a particle on its rim will be
(a)
2
/sm
(b)
2
/5.0 sm
(c)
2
/0.2 sm
(d)
2
/2.0 sm
Solution : (d) From the equation of motion
Angular speed acquired by the wheel,

2
2
1
2
2
2
120
2
2
Now radial acceleration
2.0
2
r
2
/2.0 sm
3.11 Centripetal Force.
According to Newton's first law of motion, whenever a body moves in a straight line with
uniform velocity, no force is required to maintain this velocity. But when a body moves along a
circular path with uniform speed, its direction changes continuously i.e. velocity keeps on
changing on account of a change in direction. According to Newton's second law of motion, a
change in the direction of motion of the body can take place only if some external force acts on
the body.
Due to inertia, at every point of the circular path; the body tends to move along the tangent
to the circular path at that point (in figure). Since every body has
directional inertia, a velocity cannot change by itself and as such we
have to apply a force. But this force should be such that it changes the
direction of velocity and not its magnitude. This is possible only if the
force acts perpendicular to the direction of velocity. Because the
velocity is along the tangent, this force must be along the radius
(because the radius of a circle at any point is perpendicular to the
F
v
v
v
v
F
F
F
174 Motion in Two Dimension
tangent at that point). Further, as this force is to move the body in a circular path, it must acts
towards the centre. This centre-seeking force is called the centripetal force.
Hence, centripetal force is that force which is required to move a body in a circular path
with uniform speed. The force acts on the body along the radius and towards centre.
(1) Formulae for centripetal force :
2
2
222
24
4T
rm
rnmrm
r
mv
F
(2) Centripetal force in different situation
Situation
Centripetal Force
A particle tied to a string and whirled in a
horizontal circle
Tension in the string
Vehicle taking a turn on a level road
Frictional force exerted by the road on the tyres
A vehicle on a speed breaker
Weight of the body or a component of weight
Revolution of earth around the sun
Gravitational force exerted by the sun
Electron revolving around the nucleus in an atom
Coulomb attraction exerted by the protons in the
nucleus
A charged particle describing a circular path in a
magnetic field
Magnetic force exerted by the agent that sets up the
magnetic field
3.12 Centrifugal Force.
It is an imaginary force due to incorporated effects of inertia. When a body is rotating in a
circular path and the centripetal force vanishes, the body would leave the circular path. To an
observer A who is not sharing the motion along the circular path, the body appears to fly off
tangential at the point of release. To another observer B, who is sharing the motion along the
circular path (i.e., the observer B is also rotating with the body with the same velocity), the
body appears to be stationary before it is released. When the body is released, it appears to B,
as if it has been thrown off along the radius away from the centre by some force. In reality no
force is actually seen to act on the body. In absence of any real force the body tends to continue
its motion in a straight line due to its inertia. The observer A easily relates this events to be due
to inertia but since the inertia of both the observer B and the body is same, the observer B can
not relate the above happening to inertia. When the centripetal force ceases to act on the body,
the body leaves its circular path and continues to moves in its straight-line motion but to
observer B it appears that a real force has actually acted on the body and is responsible for
throwing the body radially out-words. This imaginary force is given a name to explain the
effects on inertia to the observer who is sharing the circular motion of the body. This inertial
force is called centrifugal force. Thus centrifugal force is a fictitious force which has
significance only in a rotating frame of reference.
Motion in Two Dimension 175
Sample problems based on centripetal and centrifugal force
Problem 106. A ball of mass 0.1 kg is whirled in a horizontal circle of radius 1 m by means of a string
at an initial speed of 10 r.p.m. Keeping the radius constant, the tension in the string is
reduced to one quarter of its initial value. The new speed is [MP PMT 2001]
(a) 5 r.p.m. (b) 10 r.p.m. (c) 20 r.p.m. (d) 14 r.p.m.
Solution : (a) Tension in the string
rnmrmT 222 4
2
nT
or
Tn
[As m and r are constant]
T
T
T
T
n
n4/
1
2
1
2
rpm
n
n5
2
10
2
1
2
Problem 107. A cylindrical vessel partially filled with water is rotated about its vertical central axis.
It’s surface will
[RPET 2000]
(a) Rise equally (b) Rise from the sides (c) Rise from the middle (d) Lowered equally
Solution : (b) Due to the centrifugal force.
Problem 108. A proton of mass 1.6 × 10–27 kg goes round in a circular orbit of radius 0.10 m under a
centripetal force of 4 × 10–13 N. then the frequency of revolution of the proton is about [Kerala PMT 2002]
(a) 0.08 × 108 cycles per sec (b) 4 × 108 cycles per sec
(c) 8 × 108 cycles per sec (d) 12 × 108 cycles per sec
Solution : (a)
NF 13
104
;
kgm27
106.1
;
mr 1.0
Centripetal force
rnmF 22
4
rm
F
n2
4
sec/1086cycles
sec/1008.0 8cycle
.
Problem 109. Three identical particles are joined together by a thread as shown in figure. All the three
particles are moving in a horizontal plane. If the velocity of the outermost particle is v0,
then the ratio of tensions in the three sections of the string is [UPSEAT 2003]
(a) 3 : 5 : 7 (b) 3 : 4 : 5 (c) 7 : 11 : 6 (d) 3 : 5 : 6
Solution : (d) Let the angular speed of the thread is
For particle ‘C’
lmT 3
2
3
For particle ‘B’
lmTT 2
2
32
lmT 5
2
2
For particle ‘C’
lmTT 2
21
lmT 6
2
1
l
l
l
O
A
B
C
T1
O
A
B
C
T2
T3
176 Motion in Two Dimension
6:5:3:: 123 TTT
Problem 110. A stone of mass 1 kg tied to the end of a string of length 1 m, is whirled in a horizontal
circle with a uniform angular velocity of 2 rad/s. The tension of the string is (in N) [KCET 1998]
(a) 2 (b)
3
1
(c) 4 (d)
4
1
Solution : (c)
NewtonrmT 4)1()2(1 22
Problem 111. A cord can bear a maximum force of 100 N without breaking. A body of mass 1 kg tied to
one end of a cord of length 1 m is revolved in a horizontal plane. What is the maximum
linear speed of the body so that the cord does not break
(a) 10 m/s (b) 20 m/s (c) 25 m/s (d) 30 m/s
Solution : (a) Tension in cord appears due to centrifugal force
r
vm
T2
and for critical condition this
tension will be equal to breaking force (100 N)
100
2
max r
vm
1
1100
2
max
v
smv /10
max
Problem 112. A mass is supported on a frictionless horizontal surface. It is attached to a string and
rotates about a fixed centre at an angular velocity
0. If the length of the string and angular
velocity are doubled, the tension in the string which was initially T0 is now [AIIMS 1985]
(a) T0 (b) T0/2 (c) 4T0 (d) 8T0
Solution : (d)
lmT 2
1
2
2
1
2
1
2l
l
T
T
l
l
T
T22 2
0
2
02 8TT
Problem 113. A stone is rotated steadily in a horizontal circle with a period T by a string of length l. If the
tension in the string is kept constant and l increases by 1%, what is the percentage change
in T
(a) 1% (b) 0.5% (c) 2% (d) 0.25%
Solution : (b) Tension
2
2
4
T
lm
2
Tl
or
lT
[Tension and mass are constant]
Percentage change in Time period
2
1
(percentage change in length) [If %
change is very small]
%)1(
2
1
%5.0
Problem 114. If mass speed and radius of rotation of a body moving in a circular path are all increased by
50%, the necessary force required to maintain the body moving in the circular path will
have to be increased by
(a) 225% (b) 125% (c) 150% (d) 100%
Solution : (b) Centripetal force
r
vm
F2
Motion in Two Dimension 177
If m, v and r are increased by 50% then let new force
2
22
'
2
r
r
v
v
m
m
F
r
vm 2
4
9
F
4
9
Percentage increase in force
%100
'
100
F
FF
F
F
%125%
4
500
Problem 115. Two masses
m
and M are connected by a light string that passes through a smooth hole O
at the centre of a table. Mass
m
lies on the table and M hangs vertically.
m
is moved round
in a horizontal circle with O as the centre. If
l
is the length of the string from O to
m
then
the frequency with which
m
should revolve so that M remains stationary is
(a)
lm
Mg
2
1
(b)
lm
Mg
1
(c)
Mg
lm
2
1
(d)
Mg
lm
1
Solution : (a) ‘m’ Mass performs uniform circular motion on the table. Let n is the frequency of
revolution then centrifugal force
lnm 22
4
For equilibrium this force will be equal to weight Mg
Mglnm
22
4
lm
Mg
n
2
1
Problem 116. A particle of mass M moves with constant speed along a circular path of radius r under the
action of a force F. Its speed is [MP PMT 2002]
(a)
m
Fr
(b)
r
F
(c)
rmF
(d)
rm
F
Solution : (a) Centripetal force
r
vm
F2
m
Fr
v
Problem 117. In an atom for the electron to revolve around the nucleus, the necessary centripetal force is
obtained from the following force exerted by the nucleus on the electron [MP PET 2002]
(a) Nuclear force (b) Gravitational force (c) Magnetic force (d) Electrostatic force
Solution : (d)
Problem 118. A motor cycle driver doubles its velocity when he is having a turn. The force exerted
outwardly will be
[AFMC 2002]
(a) Double (b) Half (c) 4 times (d)
4
1
times
M
l
m
M
m
O
l
178 Motion in Two Dimension
Solution : (c)
r
vm
F2
2
vF
or
2
1
2
1
2
v
v
F
F
4
22
v
v
12 4FF
Problem 119. A bottle of soda water is grasped by the neck and swing briskly in a vertical circle. Near
which portion of the bottle do the bubbles collect
(a) Near the bottom (b) In the middle of the bottle
(c) Near the neck (d) Uniformly distributed in the bottle
Solution : (c) Due to the lightness of the gas bubble they feel less centrifugal force so they get collect
near the neck of the bottle. They collect near the centre of circular motion i.e. near the neck
of the bottle.
Problem 120. A body is performing circular motion. An observer
1
O
is sitting at the centre of the circle
and another observer
2
O
is sitting on the body. The centrifugal force is experienced by the
observer
(a)
1
O
only (b)
2
O
only (c) Both by
1
O
and
2
O
(d) None of these
Solution : (b) Centrifugal force is a pseudo force, which is experienced only by that observer who is
attached with the body performing circular motion.
3.13 Work done by Centripetal Force.
The work done by centripetal force is always zero as it is perpendicular to velocity and
hence instantaneous displacement.
Work done = Increment in kinetic energy of revolving body
Work done = 0
Also W =
SF .
= F S cos
= FS cos 90o = 0
Example : (i) When an electron revolve around the nucleus in hydrogen atom in a particular
orbit, it neither absorb nor emit any energy means its energy remains constant.
(ii) When a satellite established once in a orbit around the earth and it starts revolving
with particular speed, then no fuel is required for its circular motion.
Sample problem based on work done
Problem 121. A particle does uniform circular motion in a horizontal plane. The radius of the circle is 20
cm. The centripetal force acting on the particle is 10 N. It’s kinetic energy is
(a) 0.1 Joule (b) 0.2 Joule (c) 2.0 Joule (d) 1.0 Joule
Solution : (d)
N
r
vm 10
2
(given)
rvm 10
2
22.010
S
F
90o
Motion in Two Dimension 179
Kinetic energy
.1)2(
2
1
2
12Joulevm
Problem 122. A body of mass 100 g is rotating in a circular path of radius r with constant velocity. The
work done in one complete revolution is [AFMC 1998]
(a) 100r Joule (b) (r/100) Joule (c) (100/r) Joule (d) Zero
Solution : (d) Because in uniform circular motion work done by the centripetal force is always zero.
Problem 123. A particle of mass m is describing a circular path of radius r with uniform speed. If L is the
angular momentum of the particle about the axis of the circle, the kinetic energy of the
particle is given by [CPMT 1995]
(a)
22 /mrL
(b)
22 2/ mrL
(c)
22 /2 mrL
(d)
Lmr 2
Solution : (b) Rotational kinetic energy
I
L
E2
2
2
2
2rm
L
(As for a particle
2
rmI
)
3.14 Skidding of Vehicle on a Level Road.
When a vehicle turns on a circular path it requires
centripetal force.
If friction provides this centripetal force then vehicle can
move in circular path safely if
Friction force Required centripetal force
r
mv
mg 2
rgvsafe
This is the maximum speed by which vehicle can turn in a circular path of radius r, where
coefficient of friction between the road and tyre is
.
Sample problem based on skidding of vehicle on a level road
Problem 124. Find the maximum velocity for overturn for a car moved on a circular track of radius
m100
.
The coefficient of friction between the road and tyre is
2.0
[CPMT 1996]
(a)
sm /14.0
(b)
sm /140
(c)
skm/4.1
(d)
sm /14
Solution : (d)
grv
max
101002.0
210
sm /14
Problem 125. When the road is dry and the coefficient of friction is
, the maximum speed of a car in a
circular path is
sm /10
. If the road becomes wet and
2
, what is the maximum speed
permitted
(a)
sm /5
(b)
sm /10
(c)
sm /210
(d)
sm /25
mg
m
2r
180 Motion in Two Dimension
Solution : (d)
v
2
1
2/
1
2
1
2
v
v
12 2
1vv
2
10
2v
sm /25
Problem 126. The coefficient of friction between the tyres and the road is 0.25. The maximum speed with
which a car can be driven round a curve of radius 40 m with skidding is (assume g = 10 ms–
2) [Kerala PMT 2002]
(a)
1
40
ms
(b)
1
20
ms
(c)
1
15
ms
(d)
1
10
ms
Solution : (d)
grv
max
104025.0
sm /10
3.15 Skidding of Object on a Rotating Platform.
On a rotating platform, to avoid the skidding of an object (mass m) placed at a distance r
from axis of rotation, the centripetal force should be provided by force of friction.
Centripetal force = Force of friction
m
2r =
mg
,)/(
max rg
Hence maximum angular velocity of rotation of the platform is
,)/( rg
so that object will
not skid on it.
3.16 Bending of a Cyclist.
A cyclist provides himself the necessary centripetal force by leaning inward on a horizontal
track, while going round a curve. Consider a cyclist of weight mg taking a turn of radius r with
velocity v. In order to provide the necessary centripetal force, the cyclist leans through angle
inwards as shown in figure.
The cyclist is under the action of the following forces :
The weight mg acting vertically downward at the centre of gravity of cycle and the cyclist.
The reaction R of the ground on cyclist. It will act along a line-making angle
with the
vertical.
The vertical component R cos
of the normal reaction R will balance the weight of the
cyclist, while the horizontal component R sin
will provide the necessary centripetal force to
the cyclist.
r
mv
R2
sin
…..(i)
and R cos
= mg …..(ii)
Dividing equation (i) by (ii), we have
mg
rvm
R
R2
cos
sin
mg
R cos
R sin
mv2/r
R
Motion in Two Dimension 181
or
rg
v2
tan
…..…(iii)
Therefore, the cyclist should bend through an angle
rg
v2
1
tan
It follows that the angle through which cyclist should bend will be greater, if
(i) The radius of the curve is small i.e. the curve is sharper
(ii) The velocity of the cyclist is large.
Note : For the same reasons, an ice skater or an aeroplane has to bend inwards, while
taking a turn.
Sample problem based on bending of cyclist
Problem 127. A boy on a cycle pedals around a circle of 20 metres radius at a speed of 20 metres/sec. The
combined mass of the boy and the cycle is
kg90
. The angle that the cycle makes with the
vertical so that it may not fall is
)sec/8.9( 2
mg
[MP PMT 1995]
(a)
o
25.60
(b)
o
90.63
(c)
o
12.26
(d)
o
00.30
Solution : (b)
,20 mr
,/20 smv
2
/8.9,90 smgkgm
(given)
)2(tan
1020
2020
tantan 11
2
1
gr
v
90.63
Problem 128. If a cyclist moving with a speed of
sm /9.4
on a level road can take a sharp circular turn of
radius
m4
, then coefficient of friction between the cycle tyres and road is [AIIMS 1999]
(a) 0.41 (b) 0.51 (c) 0.71 (d) 0.61
Solution : (d)
,/9.4 smv
mr 4
and
2
/8.9 smg
(given)
8.94
9.49.4
2
gr
v
61.0
Problem 129. A cyclist taking turn bends inwards while a car passenger taking same turn is thrown
outwards. The reason is
[NCERT 1972]
(a) Car is heavier than cycle
(b) Car has four wheels while cycle has only two
(c) Difference in the speed of the two
(d) Cyclist has to counteract the centrifugal force while in the case of car only the passenger is
thrown by this force
Solution : (d)
3.17 Banking of a Road.
182 Motion in Two Dimension
For getting a centripetal force cyclist bend towards the centre of circular path but it is not
possible in case of four wheelers.
Therefore, outer bed of the road is raised so that a vehicle moving on it gets automatically
inclined towards the centre.
In the figure (A) shown reaction R is resolved into two components, the component R cos
balances weight of vehicle
mgR
cos
……(i)
and the horizontal component R sin
provides necessary centripetal force as it is directed
towards centre of desired circle
Thus
r
mv
R2
sin
...…(ii)
Dividing (ii) by (i), we have
gr
v2
tan
...... (iii)
or
rg
v
g
r
2
tan
...... (iv) [As v = r
]
If l = width of the road, h = height of the outer edge from the ground level then from the
figure (B)
l
h
x
h
tan
.......(v) [since
is very small]
From equation (iii), (iv) and (v)
rg
v2
tan
l
h
rg
v
g
r
2
Note : If friction is also present between the tyres and road then
tan1
tan
2
rg
v
Maximum safe speed on a banked frictional road
tan1
)tan(
rg
v
Sample problems based on banking of a road
Problem 130. For traffic moving at
hrkm /60
along a circular track of radius
km1.0
, the correct angle of
banking is
[MNR 1993]
(a)
1.0
)60(2
(b)
8.9100
)3/50(
tan 2
1
(c)
2
1
)3/50(
8.9100
tan
(d)
8.91.060tan 1
Solution : (b)
smhrkmv/
3
50
/60
,
,1001.0 mkmr
2
/8.9 smg
(given)
R cos
mg
R
R sin
Fig.
(A)
x
l
h
Fig.
(B)
Motion in Two Dimension 183
Angle of banking
gr
v2
tan
or
gr
v2
1
tan
8.9100
3/50
tan 2
1
Problem 131. A vehicle is moving with a velocity
v
on a curved road of width
b
and radius of curvature
R
. For counteracting the centrifugal force on the vehicle, the difference in elevation
required in between the outer and inner edges of the road is [EAMCET 1983; MP PMT 1996]
(a)
Rg
bv2
(b)
Rg
rb
(c)
Rg
vb 2
(d)
gR
vb
2
Solution : (a) For Banking of road
gr
v2
tan
and
l
h
tan
l
h
gr
v 2
gr
lv
h2
gR
bv2
[As
bl
and
Rr
given]
Problem 132. The radius of curvature of a road at a certain turn is
m50
. The width of the road is
m10
and
its outer edge is
m5.1
higher than the inner edge. The safe speed for such an inclination
will be
(a)
sm /5.6
(b)
sm /6.8
(c)
sm /8
(d)
sm /10
Solution : (b)
2
/10,10,50,5.1 smgmlmrmh
(given)
l
h
gr
v
2
l
hrg
v
10
10505.1
sm /6.8
Problem 133. Keeping the banking angle same to increase the maximum speed with which a car can
travel on a curved road by 10%, the radius of curvature of road has to be changed from
m20
to [EAMCET 1991]
(a)
m16
(b)
m18
(c)
m25.24
(d)
m5.30
Solution : (c)
gr
v2
tan
2
vr
(if
is constant)
2
2
1
2
1
21.1
v
v
v
v
r
r
= 1.21 r2 = 1.21 r1 = 1.21 20 = 24.2 m
Problem 134. The slope of the smooth banked horizontal road is
p
. If the radius of the curve be
r
, the
maximum velocity with which a car can negotiate the curve is given by
(a)
prg
(b)
prg
(c)
rgp/
(d)
rgp/
Solution : (b)
gr
v2
tan
gr
v
p2
grpv
3.18 Overturning of Vehicle.
When a car moves in a circular path with speed more than maximum speed then it
overturns and it’s inner wheel leaves the ground first
Weight of the car = mg
Speed of the car = v
Radius of the circular path = r
2a
G
R1
h
R2
mg
F
184 Motion in Two Dimension
Distance between the centre of wheels of the car = 2a
Height of the centre of gravity (G) of the car from the road level = h
Reaction on the inner wheel of the car by the ground = R1
Reaction on the outer wheel of the car by the ground = R2
When a car move in a circular path, horizontal force F provides the required centripetal
force
i.e.,
R
mv
F2
.......(i)
For rotational equilibrium, by taking the moment of forces R1, R2 and F about G
aRaRFh 21
.......(ii)
As there is no vertical motion so R1 + R2 = mg .......(iii)
By solving (i), (ii) and (iii)
ra
hv
gMR 2
12
1
.......(iv)
and
ra
hv
gMR 2
22
1
.......(v)
It is clear from equation (iv) that if v increases value of R1 decreases and for R1 = 0
g
ra
hv
2
or
h
gra
v
i.e. the maximum speed of a car without overturning on a flat road is given by
h
gra
v
Sample problems based on overturning of vehicle
Problem 135. The distance between two rails is
m5.1
. The centre of gravity of the train at a height of
m2
from the ground. The maximum speed of the train on a circular path of radius
m120
can be
(a)
sm /5.10
(b)
sm /42
(c)
sm /21
(d)
sm /84
Solution : (c) Height of centre of gravity from the ground h = 2m, Acceleration due to gravity g = 10
2
/sm
,
Distance between two rails 2a = 1.5m, Radius of circular path r = 120 m (given)
h
arg
v
max
2
75.012010
max
v
sm /2.21
Problem 136. A car sometimes overturns while taking a turn. When it overturns, it is [AFMC 1988]
(a) The inner wheel which leaves the ground first
(b) The outer wheel which leaves the ground first
(c) Both the wheels leave the ground simultaneously
Motion in Two Dimension 185
(d) Either wheel leaves the ground first
Solution : (a)
Problem 137. A car is moving on a circular path and takes a turn. If
1
R
and
2
R
be the reactions on the
inner and outer wheels respectively, then
(a)
21 RR
(b)
21 RR
(c)
21 RR
(d)
21 RR
Solution : (b) Reaction on inner wheel
ar
hv
g
M
R2
12
and Reaction on outer wheel
ar
hv
g
M
R2
22
21 RR
.
Problem 138. A train
A
runs from east to west and another train
B
of the same mass runs from west to
east at the same speed along the equator. A presses the track with a force
1
F
and
B
presses
the track with a force
2
F
(a)
21 FF
(b)
21 FF
(c)
21 FF
(d) The information is insufficient to find the relation between
1
F
and
2
F
Solution : (a) We know that earth revolves about its own axis from west to east. Let its angular speed is
e
and the angular speed of the train is
t
For train A : Net angular speed = (
te
) because the sense of rotation of train is opposite
to that of earth
So reaction of track
RmgmFR te 2
11 )(
For train B : Net angular speed =
)( te
because the sense of rotation of train is same as
that of earth
So reaction of track
RmgmFR te 2
22 )(
So it is clear that
21 FF
3.19 Motion of Charged Particle in Magnetic Field.
When a charged particle having mass m , charge q enters perpendicularly in a magnetic
field B, with velocity v then it describes a circular path of radius r.
Because magnetic force (qvB) works in the perpendicular direction of v and it provides
required centripetal force
Magnetic force = Centripetal force
qvB =
r
mv 2
radius of the circular path
qB
mv
r
3.20 Reaction of Road on Car.
v
F
q
B
186 Motion in Two Dimension
(1) When car moves on a concave bridge then
Centripetal force =
cosmgR
r
mv 2
and reaction
r
mv
mgR2
cos
(2) When car moves on a convex bridge
Centripetal force =
Rmg
cos
r
mv 2
and reaction
r
mv
mgR2
cos
Sample problem based on reaction of road
Problem 139. The road way bridge over a canal is in the form of an arc of a circle of radius
m20
. What is
the minimum speed with which a car can cross the bridge without leaving contact with the
ground at the highest point
)/8.9( 2
smg
(a)
sm /7
(b)
sm /14
(c)
sm /289
(d)
sm /5
Solution : (b) At the highest point of the bridge for critical condition
0
2 r
vm
mg
gm
r
vm
2
grv
max
208.9
196
sm /14
Problem 140. A car moves at a constant speed on a road as shown in the figure. The normal force exerted
by the road on the car is
A
N
and
B
N
when it is at the points
A
and
B
respectively
(a)
BA NN
(b)
BA NN
(c)
BA NN
(d) All possibilities are there
Solution : (c) From the formula
r
mv
mgN2
N r
As rA < rB
BA NN
Problem 141. A car while travelling at a speed of
hrkm /72
. Passes through a curved portion of road in the
form of an arc of a radius
m10
. If the mass of the car is
kg500
the reaction of the car at the
lowest point P is
(a)
kN25
(b)
kN50
(c)
kN75
(d) None of these
R
mg cos
mg
v
Concave bridge
R
mg cos
mg
v
Convex bridge
72
km/hr
B
A
Motion in Two Dimension 185
Solution : (a)
,/2072 sm
h
km
v
,10 mr
kgm500
(given)
Reaction at lowest point
r
vm
mgR2
10
)20(500
10500 2
N25000
KN25
3.21 Non-Uniform Circular Motion.
If the speed of the particle in a horizontal circular motion changes with respect to time,
then its motion is said to be non-uniform circular motion.
Consider a particle describing a circular path of radius r with centre at O. Let at an instant
the particle be at P and
be its linear velocity and
be its angular velocity.
Then,
r
…..(i)
Differentiating both sides of w.r.t. time t we have
dt
dr
r
dt
d
dt
d
…..(ii) Here,
,a
dt
dv
(Resultant acceleration)
ra
dt
d
(Angular acceleration)
ct aaa
.….(iii)
dt
dr
(Linear velocity)
Thus the resultant acceleration of the particle at P has two component accelerations
(1) Tangential acceleration :
rat
It acts along the tangent to the circular path at P in the plane of circular path.
According to right hand rule since
and
r
are perpendicular to each other, therefore, the
magnitude of tangential acceleration is given by
.90sin|||| rrra o
t
(2) Centripetal (Radial) acceleration :
vac
It is also called centripetal acceleration of the particle at P.
It acts along the radius of the particle at P.
According to right hand rule since
and
are perpendicular to each other, therefore, the
magnitude of centripetal acceleration is given by
o
c
a90sin||||
=
rrr /)( 22
O
ac
a
at
P
186 Motion in Two Dimension
(3) Tangential and centripetal acceleration in different motions
Centripetal
acceleration
Tangential
acceleration
Net acceleration
Type of motion
ac = 0
at = 0
a = 0
Uniform translatory motion
ac = 0
at 0
a = at
Accelerated translatory
motion
ac 0
at = 0
a = ac
Uniform circular motion
ac 0
at 0
22 tc aaa
Non-uniform circular
motion
Note : Here at governs the magnitude of
v
while
c
a
its direction of motion.
(4) Force : In non-uniform circular motion the particle simultaneously possesses two
forces
Centripetal force :
2
2
mr
r
mv
maFcc
Tangential force :
tt maF
Net force :
maF
net
=
22 tc aam
Note : In non-uniform circular motion work done by centripetal force will be zero since
vFc
In non uniform circular motion work done by tangential of force will not be zero
since Ft 0
Rate of work done by net force in non-uniform circular = rate of work done by
tangential force
i.e.
vF
dt
dW
Pt
.
Sample problems based on non-uniform circular motion
Problem 142. The kinetic energy
k
of a particle moving along a circle of radius
R
depends on the
distance covered. It is given as K.E. = as2 where
a
is a constant. The force acting on the
particle is [MNR 1992; JIPMER 2001, 2002]
(a)
R
s
a2
2
(b)
2/1
2
2
12
R
s
as
(c)
as2
(d)
s
R
a2
2
Solution : (b) In non-uniform circular motion two forces will work on a particle
tc FF and
So the net force
22 tcNet FFF
….(i)
Motion in Two Dimension 187
Centripetal force
R
mv
Fc
2
R
as2
2
….(ii) [As kinetic energy
22
2
1asmv
given]
Again from :
22
2
1asmv
m
as
v2
22
m
a
sv 2
Tangential acceleration
dt
ds
ds
dv
dt
dv
at.
v
m
a
s
ds
d
at.
2
m
a
vat2
m
a
m
a
s22
m
as2
and
asmaFtt 2
….(iii)
Now substituting value of Fc and Ft in equation (i)
2
2
22
2as
R
as
FNet
2/1
2
2
12
R
s
as
Problem 143. A particle of mass
m
is moving in a circular path of constant radius
r
such that its
centripetal acceleration
c
a
is varying with time
t
as
22rtkac
, where
k
is a constant. The
power delivered to the particle by the forces acting on it is [IIT-JEE 1994]
(a)
trmk 22
2
(b)
trmk 22
(c)
3
524 trmk
(d) Zero
Solution : (b)
22 trkac
22
2trk
r
v
2222 trkv
trkv
Tangential acceleration
rk
dt
dv
at
As centripetal force does not work in circular motion.
So power delivered by tangential force
vFP t
vam t
=m(kr) krt
trmk 22
Problem 144. A simple pendulum is oscillating without damping. When the displacement of the bob is
less than maximum, its acceleration vector
a
is correctly shown in [IIT-JEE Screening 2002]
(a) (b) (c) (d)
Solution : (c)
onacceleratil centripeta
c
a
,
onaccelerati tangential
t
a
,
tcN aaa andof Resultantonaccelerati net
aN =
22 tc aa
Fc
Ft
a
a
a
a
t
a
N
a
c
a
188 Motion in Two Dimension
Problem 145. The speed of a particle moving in a circle of radius
m1.0
is
tv 0.1
where
t
is time in
second. The resultant acceleration of the particle at
st 5
will be
(a)
2
/10 sm
(b)
2
/100 sm
(c)
2
/250 sm
(d)
2
/500 sm
Solution : (c)
tv 0.1
2
/1 sm
dt
dv
at
and
1.0
)5( 22 r
v
ac
2
/250 sm
[At
smvt /5sec,5
]
22 tcN aaa
22 1)250(
2
/250 smaN
(approx.)
Problem 146. A particle moving along the circular path with a speed v and its speed increases by ‘g’ in
one second. If the radius of the circular path be r, then the net acceleration of the particle
is
(a)
g
r
v
2
(b)
2
2
2g
r
v
(c)
2
1
2
2
4
g
r
v
(d)
2
1
2
g
r
v
Solution : (c)
gat
(given) and
r
v
ac
2
and
22 ctN aaa
2
2
2g
r
v
2
2
4g
r
v
Problem 147. A car is moving with speed 30 m/sec on a circular path of radius 500 m. Its speed is
increasing at the rate of 2m/sec2. What is the acceleration of the car [Roorkee 1982; RPET 1996; MH CET 2002; MP PMT 2003]
(a) 2 m/s2 (b) 2.7 m/s2 (c) 1.8 m/s2 (d) 9.8 m/s2
Solution : (b) at = 2m/s2 and
2
2/8.1
500
3030 sm
r
v
ac
22222 /7.2)8.1(2 smaaa ct
.
Problem 148. For a particle in circular motion the centripetal acceleration is [CPMT 1998]
(a) Less than its tangential acceleration (b) Equal to its tangential acceleration
(c) More than its tangential acceleration (d) May be more or less than its tangential
acceleration
Solution : (d)
Problem 149. A particle is moving along a circular path of radius 3 meter in such a way that the distance
travelled measured along the circumference is given by
32
32 tt
S
. The acceleration of
particle when
sec2t
is
(a) 1.3 m/s2 (b) 13 m/s2 (c) 3 m/s2 (d) 10 m/s2
Solution : (b)
32
32 tt
s
2
tt
dt
ds
v
and
)( 2
tt
dt
d
dt
dv
at
t21
At t = 2 sec, v = 6 m/s and
2
/5 smat
,
2
2/12
3
36 sm
r
v
ac
Motion in Two Dimension 189
22 tcN aaa
22 )5()12(
2
/13 sm
.
3.22 Equations of Circular Motion.
For accelerated motion
For retarded motion
t
12
t
12
2
12
1tt
2
12
1tt
2
22
12
2
22
12
)12(
2
1 n
n
)12(
2
1 n
n
Sample problems based on equation of circular motion
Problem 150. The angular velocity of a particle is given by
235.1 2 tt
, the time when its angular
acceleration ceases to be zero will be
(a)
sec25
(b)
sec25.0
(c)
sec12
(d)
sec2.1
Solution : (b)
235.1 2 tt
and
t
dt
d65.1
t65.10
sec25.0
6
5.1 t
Problem 151. A wheel is subjected to uniform angular acceleration about its axis. Initially its angular
velocity is zero. In the first
sec2
, it rotates through an angle
1
. In the next
sec2
, it rotates
through an additional angle
2
. The ratio of
21 /
is [AIIMS 1982]
(a) 1 (b) 2 (c) 3 (d) 5
Solution : (c) From equation of motion
2
12
1tt
2
1)2(
2
1
0
= 2
…..(i) [As
,0
1
sec,2t
1
]
For second condition
2
21 )4(
2
1
0
[As
,0
1
sec,422 t
21
]
8
21
….(ii)
From (i) and (ii)
,2
1
6
2
3
1
2
Problem 152. If the equation for the displacement of a particle moving on a circular path is given by
5.02)( 3 t
, where
is in radians and
t
in seconds, then the angular velocity of the
particle after
sec2
from its start is
[AIIMS 1998]
Where
1 = Initial angular velocity of particle
2 = Final angular velocity of particle
= Angular acceleration of particle
= Angle covered by the particle in
time t
n = Angle covered by the particle in nth
second
190 Motion in Two Dimension
(a)
sec/8 rad
(b)
sec/12 rad
(c)
sec/24 rad
(d)
sec/36 rad
Solution : (c)
5.02 3 t
and
2
6t
dt
d
at t = 2 sec,
secrad /24)2(6 2
Problem 153. A grinding wheel attained a velocity of 20 rad/sec in 5 sec starting from rest. Find the
number of revolutions made by the wheel
(a)
25
rev/ sec (b)
1
rev/sec (c)
25
rev/sec (d) None of these
Solution : (c)
,0
1
,/20
2secrad
sect 5
2
12 /4
5
020 secrad
t
From the equation
22
1)5(.)4(
2
1
0
2
1 tt
rad50
rad
2
means 1 revolution.
50 Radian means
2
50
or
25
rev.
Problem 154. A grind stone starts from rest and has a constant angular acceleration of 4.0 rad/sec2. The
angular displacement and angular velocity, after 4 sec. will respectively be
(a) 32 rad, 16 rad/sec (b) 16 rad, 32 rad/sec (c) 64 rad, 32 rad/sec (d) 32 rad, 64 rad/sec
Solution : (a)
,0
1
,sec/4 2
rad
sect 4
Angular displacement
2
12
1tt
2
)4(4
2
1
0
.32 rad
Final angular
secradt /16440
12
Problem 155. An electric fan is rotating at a speed of 600 rev/minute. When the power supply is stopped,
it stops after 60 revolutions. The time taken to stop is
(a) 12 s (b) 30 s (c) 45 s (d) 60 s
Solution : (a)
,/10min/600
1secrevrev
rev60and0
2
From the equation

2
2
1
2
2
602)10(0 2
6
5
120
100
Again
t
12
t
1
0
sect 12
5
610
1
.
3.23 Motion in Vertical Circle.
This is an example of non-uniform circular motion. In this motion body is under the
influence of gravity of earth. When body moves from lowest point to highest point. Its speed
decrease and becomes minimum at highest point. Total mechanical energy of the body remains
conserved and KE converts into PE and vice versa.
Motion in Two Dimension 191
(1) Velocity at any point on vertical loop : If u is the initial velocity imparted to body at
lowest point then. Velocity of body at height h is given by
)cos1(22 22
glughuv
[As h = l – l cos
= l (1 – cos
)]
where l in the length of the string
(2) Tension at any point on vertical loop : Tension at general point P, According to
Newton’s second law of motion.
Net force towards centre = centripetal force
l
mv
mgT2
cos
or
l
mv
mgT2
cos
)]cos32([ 2
glu
l
m
T
[As
)cos1(2
2
gluv
]
(3) Velocity and tension in a vertical loop at different positions
Position
Angle
Velocity
Tension
A
0o
u
mg
l
mu
2
B
90o
glu 2
2
mg
l
mu 2
2
C
180o
glu4
2
mg
l
mu 5
2
D
270o
glu 2
2
mg
l
mu 2
2
It is clear from the table that :
CBA TTT
and TB = TD
,3mgTT BA
mgTT CA 6
and
mgTT CB 3
(4) Various conditions for vertical motion :
Velocity at lowest point
Condition
gluA5
Tension in the string will not be zero at any of the point and body will
O
l
P
v
h
A
u
D
C
B
O
l
P
A
D
C
B
mg
mg cos
+ mv2/r
T
192 Motion in Two Dimension
continue the circular motion.
,5gluA
Tension at highest point C will be zero and body will just complete the
circle.
,52 glugl A
Particle will not follow circular motion. Tension in string become zero
somewhere between points B and C whereas velocity remain positive.
Particle leaves circular path and follow parabolic trajectory.
gluA2
Both velocity and tension in the string becomes zero between A and B
and particle will oscillate along semi-circular path.
gluA2
velocity of particle becomes zero between A and B but tension will not
be zero and the particle will oscillate about the point A.
Note : K.E. of a body moving in horizontal circle is same throughout the path but the K.E.
of the body moving in vertical circle is different at different places.
If body of mass m is tied to a string of length l and is projected with a horizontal
velocity u then :
Height at which the velocity vanishes is
g
u
h2
2
Height at which the tension vanishes is
g
glu
h3
2
(5) Critical condition for vertical looping : If the tension at C is zero, then body will just
complete revolution in the vertical circle. This state of body is known as critical state. The
speed of body in critical state is called as critical speed.
From the above table TC =
05
2 mg
l
mu
glu5
It means to complete the vertical circle the body must be projected with minimum velocity
of
gl5
at the lowest point.
(6) Various quantities for a critical condition in a vertical loop at different positions :
Quantity
Point A
Point B
Point C
Point D
Point P
Linear velocity (v)
gl5
gl3
gl
gl3
)cos23(
gl
Angular velocity (
)
l
g5
l
g3
l
g
l
g3
)cos23(
l
g
Tension in String
(T)
6 mg
3 mg
0
3 mg
)cos1(3
mg
Kinetic Energy (KE)
mgl
2
5
mgl
2
3
mgl
2
1
mgl
2
3
)cos23(
2
mgl
Motion in Two Dimension 193
Potential Energy
(PE)
0
mgl
2 mgl
mgl
)cos1(
mgl
Total Energy (TE)
mgl
2
5
mgl
2
5
mgl
2
5
mgl
2
5
mgl
2
5
(7) Motion of a block on frictionless hemisphere : A small block of mass m slides down
from the top of a frictionless hemisphere of radius r. The component of the force of gravity (mg
cos
) provides required centripetal force but at point B it's circular motion ceases and the block
lose contact with the surface of the sphere.
For point B, by equating the forces,
r
mv
mg 2
cos
.....(i)
For point A and B, by law of conservation of energy
Total energy at point A = Total energy at point B
K.E.(A) + P.E.(A) = K.E.(B) + P.E.(B)
0 + mgr =
mghmv
2
2
1
)(2 hrgv
.....(ii)
and from the given figure
cosrh
.....(iii)
By substituting the value of v and h from eqn (ii) and (iii) in eqn (i)
2
)(2 hrg
r
m
r
h
mg
)(2 hrh
rh 3
2
i.e. the block lose contact at the height of
r
3
2
from the ground.
and angle from the vertical can be given by
3
2
cos r
h
3
2
cos 1
.
Sample problems based on vertical looping
Problem 156. A small block is shot into each of the four tracks as shown below. Each of the tracks
rises to the same height. The speed with which the block enters the track is the same in all
cases. At the highest point of the track, the normal reaction is maximum in [IIT-JEE (Screening) 2001]
(a) (b) (c) (d)
Solution : (a) Normal reaction at the highest point of the path
mg
r
mv
R 2
v
v
v
v
h
mg
(r –
h)
B
r
A
194 Motion in Two Dimension
For maximum R, value of the radius of curvature (r ) should be minimum and it is minimum
in first condition.
Problem 157. A stone tied to string is rotated in a vertical circle. The minimum speed with which the
string has to be rotated
[EAMCET (Engg.) 1998; CBSE PMT 1999]
(a) Decreases with increasing mass of the stone (b) Is independent of the
mass of the stone
(c) Decreases with increasing in length of the string (d) Is independent of the
length of the string
Solution : (b)
rgv 5
for lowest point of vertical loop.
0
mv
i.e. it does not depends on the mass of the body.
Problem 158. A mass m is revolving in a vertical circle at the end of a string of length 20 cms. By how
much does the tension of the string at the lowest point exceed the tension at the topmost
point
(a) 2 mg (b) 4 mg (c) 6 mg (d) 8 mg
Solution : (c)
mgTT 6
pointHighestpointLowest
(Always)
Problem 159. In a simple pendulum, the breaking strength of the string is double the weight of the bob.
The bob is released from rest when the string is horizontal. The string breaks when it
makes an angle
with the vertical
(a)
)3/1(cos 1
(b)
o
60
(c)
)3/2(cos 1
(d)
o
0
Solution : (c) Let the string breaks at point B.
Tension
r
vm
mg B
2
cos
Breaking strength
mg
r
vm
mg B2cos 2
….(i)
If the bob is released from rest (from point A) then velocity acquired by it at point B
ghvB2
cos2 rgvB
....(ii) [As h= r cos
]
By substituting this value in equation (i)
mgrg
r
m
mg 2)cos2(cos
or
mgmg 2cos3
3
2
cos
3
2
cos 1
r
T
A
C
B
mg
mg cos
mv2/r
h = r cos
Motion in Two Dimension 195
Problem 160. A toy car rolls down the inclined plane as shown in the fig. It goes around the loop at the
bottom. What is the relation between
H
and
h
(a)
2
h
H
(b)
3
h
H
(c)
4
h
H
(d)
5
h
H
Solution : (d) When car rolls down the inclined plane from height H, then velocity acquired by it at the
lowest point
Hgv 2
….(i)
and for looping of loop, velocity at the lowest point should be
rgv 5
….(ii)
From eqn (i) and (ii)
grgHv52
2
5r
H
….(iii)
From the figure
rhH 2
2
hH
r
Substituting the value of r in equation (iii) we get
22
5hH
H
5
h
H
Problem 161. The mass of the bob of a simple pendulum of length
L
is
m
. If the bob is left from its
horizontal position then the speed of the bob and the tension in the thread in the lowest
position of the bob will be respectively.
(a)
gL2
and
mg3
(b)
mg3
and
gL2
(c)
mg2
and
gl2
(d)
gl2
and
mg3
Solution : (a) By the conservation of energy
Potential energy at point A = Kinetic energy at point B
2
2
1vmlmg
glv2
and tension
l
mv
mg 2
)2( gl
l
m
mgT
mgT3
Problem 162. A stone of mass m is tied to a string and is moved in a vertical circle of radius r making n
revolutions per minute. The total tension in the string when the stone is at its lowest point
is [Kerala (Engg.) 2001]
(a)
}900/)({ 22 rngm
(b)
)( 2
nrgm
(c)
)( nrgm
(d)
)( 22rngm
Solution : (a) Tension at lowest point
rwmmgT2
rnmmg 22
4
h
r
H
l
T
mg
mv2/
L
B
A
196 Motion in Two Dimension
If n is revolution per minute then
r
n
mmgT3600
42
2
900
22 rnm
mg
900
22 rn
gm
Problem 163. A particle is kept at rest at the top of a sphere of diameter 42m. When disturbed slightly, it
slides down. At what height
h
from the bottom, the particle will leave the sphere [IMS-BHU 2003]
(a) 14 m (b) 28 m (c) 35 m (d) 7 m
Solution : (c) Let the particle leave the sphere at height ‘h’ from the bottom
We know for given condition
rx 3
2
and
rrxrh 3
2
r
3
5
m3521
3
5
[As r= 21 m]
Problem 164. A bucket tied at the end of a 1.6 m long string is whirled in a vertical circle with constant
speed. What should be the minimum speed so that the water from the bucket does not spill,
when the bucket is at the highest position (Take g = 10 m/sec2) [AIIMS 1987]
(a)
sec/4 m
(b)
secm /25.6
(c)
secm /16
(d) None of these
Solution : (a)
smrgv /4166.110
Problem 165. The ratio of velocities at points A, B and C in vertical circular motion is
(a)
25:9:1
(b)
3:2:1
(c)
5:3:1
(d)
5:3:1
Solution : (d)
rgrgrgvvv cBA 5:3:::
5:3:1
Problem 166. The minimum speed for a particle at the lowest point of a vertical circle of radius R, to
describe the circle is ‘v’. If the radius of the circle is reduced to one-fourth its value, the
corresponding minimum speed will be
[EAMCET (Engg.) 1999]
(a)
4
v
(b)
2
v
(c) 2v (d) 4v
Solution : (b)
rgv 5
rv
So
2
14/
1
2
1
2 r
r
r
r
v
v
2/
2vv
A
O
C
r
x
h
v
Motion in Two Dimension 197
Problem 167. A body slides down a frictionless track which ends in a circular loop of diameter D, then the
minimum height h of the body in term of D so that it may just complete the loop, is [AIIMS 2000]
(a)
2
5D
h
(b)
4
5D
h
(c)
4
3D
h
(d)
4
D
h
Solution : (b) We know
rh 2
5
22
5D
4
5D
[For critical condition of vertical looping]
Problem 168. A can filled with water is revolved in a vertical circle of radius 4m and the water just does
not fall down. The time period of revolution will be [CPMT 1985; RPET 1999]
(a) 1 sec (b) 10 sec (c) 8 sec (d) 4 sec
Solution : (d) At highest point
rmmg 2
r
T
g2
2
4
2
244
10 T
16
2T
secT 4
Problem 169. A particle is moving in a vertical circle. The tensions in the string when passing through
two positions at angles 30o and 60o from vertical (lowest position) are T1 and T2
respectively, Then [Orissa JEE 2002]
(a)
21 TT
(b)
21 TT
(c)
21 TT
(d)
21 TT
Solution : (b)
r
mv
mgT2
cos
As
increases T decreases
So
21 TT
Problem 170. A mass of
kg2
is tied to the end of a string of length 1m. It is, then, whirled in a vertical
circle with a constant speed of
1
5
ms
. Given that
2
10
msg
. At which of the following
locations of tension in the string will be
N70
(a) At the top (b) At the bottom
(c) When the string is horizontal (d) At none of the above locations
Solution : (b) Centrifugal force
Newton
r
mv
F50
1
)5(2 22
NewtonmgWeight 20102
Tension = 70 N (sum of above two forces)
i.e. the mass is at the bottom of the vertical circular path
Problem 171. With what angular velocity should a 20 m long cord be rotated such that tension in it, while
reaching the highest point, is zero [RPMT 1999]
(a) 0.5 rad/sec (b) 0.2 rad/sec (c) 7.5 rad/sec (d) 0.7 rad/sec
r
T
mg cos
+
mv2/r
mg
r
mv
mg
2
T
198 Motion in Two Dimension
Solution : (d)
r
g
5.0
20
10
secrad /7.0
Problem 172. A body of mass of
g100
is attached to a
m1
long string and it is revolving in a vertical
circle. When the string makes an angle of
o
60
with the vertical then its speed is
sm /2
. The
tension in the string at
o
60
will be
(a)
N89
(b)
N89.0
(c)
N9.8
(d)
N089.0
Solution : (b)
r
mv
mgT2
cos
1
)2(1.0
60cos8.91.0 2
4.049.0
Newton89.0
Problem 173. A body of mass
kg2
is moving in a vertical circle of radius
m2
. The work done when it
moves from the lowest point to the highest point is
(a)
J80
(b)
J40
(c)
J20
(d) 0
Solution : (a) work done = change in potential energy
mgr2
21022
J80
Problem 174. A body of mass
m
is tied to one end of a string of length
l
and revolves vertically in a
circular path. At the lowest point of circle, what must be the
..EK
of the body so as to
complete the circle [RPMT 1996]
(a)
mgl5
(b)
mgl4
(c)
mgl5.2
(d)
mgl2
Solution : (c) Minimum velocity at lowest point to complete vertical loop
gl5
So minimum kinetic energy
)(
2
12
vm
2
)5(
2
1glm
mgl
2
5
mgl5.2
3.24 Conical Pendulum.
This is the example of uniform circular motion in horizontal plane.
A bob of mass m attached to a light and in-extensible string rotates in a horizontal circle of
radius r with constant angular speed
about the vertical. The string makes angle
with vertical
and appears tracing the surface of a cone. So this arrangement is called conical pendulum.
The force acting on the bob are tension and weight of the bob.
From the figure
r
mv
T2
sin
….(i)
and
mgT
cos
….(ii)
(1) Tension in the string :
2
2
1
rg
v
mgT
22
cos rl
mglmg
T
[As
l
rl
l
h22
cos
]
O
S
T cos
T
P
mg
T sin
mv2/r
O
P
r
l
h
S
Motion in Two Dimension 199
(2) Angle of string from the vertical :
rg
v2
tan
(3) Linear velocity of the bob :
tangrv
(4) Angular velocity of the bob :
cos
tan l
g
h
g
r
g
(5) Time period of revolution :
tan
222
cos
222
g
r
g
rl
g
h
g
l
TP
Sample problems based on conical pendulum
Problem 175. A point mass m is suspended from a light thread of length l, fixed at O, is whirled in a
horizontal circle at constant speed as shown. From your point of view, stationary with
respect to the mass, the forces on the mass are [AMU (Med.) 2001]
(a) (b) (c) (d)
Solution : (c) Centrifugal force (F) works radially outward,
Weight (w) works downward
Tension (T) work along the string and towards the point of suspension
Problem 176. A string of length L is fixed at one end and carries a mass M at the other end. The
string makes 2/
revolutions per second around the vertical axis through the fixed end as
shown in the figure, then tension in the string is [BHU 2002]
(a) ML (b) 2 ML (c) 4 ML (d) 16 ML
Solution : (d)
RMT 2
sin
..... (i)
S
T
M
L
R
T
F
W
T
F
W
F
T
F
W
T
W
T
L
R
m
2R
m
l
O
200 Motion in Two Dimension
sinsin 2LMT
..... (ii)
From (i) and (ii)
LMT 2
LnM 22
4
LM 2
22
4
ML16
Problem 177. A string of length
m1
is fixed at one end and a mass of
gm100
is attached at the other end.
The string makes
/2
rev/sec around a vertical axis through the fixed point. The angle of
inclination of the string with the vertical is (
2
sec/10 mg
)
(a)
8
5
tan 1
(b)
5
8
tan 1
(c)
5
8
cos 1
(d)
8
5
cos 1
Solution : (d) For the critical condition, in equilibrium
rmT 2
sin
and
mgT
cos
g
r
2
tan
g
rn22
4
10
1.)/2(4 22
5
8
Sample problems (Miscellaneous)
Problem 178. If the frequency of the rotating platform is f and the distance of a boy from the centre is r,
which is the area swept out per second by line connecting the boy to the centre
(a)
rf
(b)
rf
2
(c)
fr2
(d)
fr2
2
Solution : (c) Area swept by line in complete revolution
2
r
If frequency of rotating platform is f per second, then Area swept will be
fr2
per second.
Problem 179. Figure below shows a body of mass M moving with uniform speed v along a circle of radius
R. What is the change in speed in going from P1 to P2
(a) Zero (b)
v2
(c)
2/v
(d)
v2
Solution : (a) In uniform circular motion speed remain constant.
change in speed is zero.
Problem 180. In the above problem, what is change in velocity in going from P1 to P2
(a) Zero (b)
v2
(c)
2/v
(d) 2 v
T
m
2R
T sin
T cos
mg
R
v
P2
P3
P1
P4
Motion in Two Dimension 201
Solution : (b) Change in velocity
)2/sin(2
v
2
90
sin2v
45sin2v
2
2v
v2
Problem 181. In the above problem, what is the change in angular velocity in going from P1 to P2
(a) Zero (b)
Rv /2
(c)
Rv 2/
(d)
Rv /2
Solution : (a) Angular velocity remains constant, so change in angular velocity = Zero.
Problem 182. A particle of mass
m
is fixed to one end of a light spring of force constant
k
and
unstretched length
l
. The system is rotated about the other end of the spring with an
angular velocity
, in gravity free space. The increase in length of the spring will be
(a)
k
lm 2
(b)
2
2
mk
lm
(c)
2
2
mk
lm
(d) None of these
Solution : (b) In the given condition elastic force will provides the required centripetal force
rmxk 2
)(
2xlmxk
xmlmxk 22
lmmkx 22 )(
2
2
mk
lm
x
Problem 183. A uniform rod of mass
m
and length
l
rotates in a horizontal plane with an angular
velocity
about a vertical axis passing through one end. The tension in the rod at a
distance
x
from the axis is
(a)
xm 2
2
1
(b)
l
x
m2
2
2
1
(c)
l
x
lm 1
2
12
(d)
][
2
122
2xl
l
m
Solution : (d) Let rod AB performs uniform circular motion about point A. We have to calculate the
tension in the rod at a distance x from the axis of rotation. Let mass of the small segment at
a distance x is dm
So
xdmdT 2
xdx
l
m2
.
l
m2
[x d x]
Integrating both sides
l
x
l
x
dxx
l
m
dT 2
l
x
x
l
m
T
2
22
22
2
2xl
l
m
T
Problem 184. A long horizontal rod has a bead which can slide along its length, and initially placed at a
distance
L
from one end
A
of the rod. The rod is set in angular motion about A with
constant angular acceleration
. If the coefficient of friction between the rod and the bead
is
, and gravity is neglected, then the time after which the bead starts slipping is [IIT-JEE (Screening) 2000]
m
l
r = (l+x)
m
A
dx
x
202 Motion in Two Dimension
(a)
(b)
(c)

1
(d) Infinitesimal
Solution : (a) Let the bead starts slipping after time t
For critical condition
Frictional force provides the centripetal force
RLm
2
mLam t
m (
t)2L =
mL
t
(As
=
t)
Problem 185. A smooth table is placed horizontally and an ideal spring of spring constant
mNk /1000
and
unextended length of
m5.0
has one end fixed to its centre. The other end is attached to a mass of
kg5
which is moving in a circle with constant speed
sm /20
. Then the tension in the spring and
the extension of this spring beyond its normal length are
(a)
mN 5.0,500
(b)
mN 6.0,600
(c)
mN 7.0,700
(d)
mN 8.0,800
Solution : (a)
,1000k
,5kgm
,5.0 ml
smv /20
(given)
Restoring force = kx
r
vm 2
xl
vm
2
x
x
5.0
)20(5
1000 2
mx 5.0
and Tension in the spring = kx
2
1
1000
N500
Problem 186. A particle describes a horizontal circle at the mouth of a funnel type vessel as shown in
figure. The surface of the funnel is frictionless. The velocity
v
of the particle in terms of
r
and
will be
(a)
tan/rgv
(b)
tanrgv
(c)
cotrgv
(d)
cot/rgv
Solution : (c) For uniform circular motion of a particle
cos
2R
r
vm
….(i)
and
sinRmg
….(ii)
Dividing (i) by (ii)
cot
2
rg
v
cotrgv
Problem 187. Figure shows a smooth track, a part of which is a circle of radius R. A block of mass m is
pushed against a spring constant
k
fixed at the left end and is then released. Find the
initial compression of the spring so that the block presses the track with a force mg when it
reaches the point P [see. Fig], where the radius of the track is horizontal
r
h
r
R sin
R cos
R
r
mv2
m
k
P
R
L
L
A
Motion in Two Dimension 203
(a)
k
mgR
3
(b)
mk
gR3
(c)
k
mgR3
(d)
kR
mg3
Solution : (c) For the given condition, centrifugal force at P should be equal to mg
i.e.
mg
R
mv P
2
RgvP
From this we can easily calculate the required velocity at the lowest point of circular track.
gRvv Lp 2
22
(by using formula :
ghuv 2
22
)
gRgRRggRvv PL 322
2
It means the block should possess kinetic energy
2
2
1L
mv
gRm3
2
1
And by the law of conservation of energy
Rgmkx 3
2
1
2
12
k
Rgm
x3
.