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Prof. Muhammad Amin 1
Chapter-5 circular motion
Type equation here.
Q. 5.1 Define and explain centripetal force and derive relation for it?
Ans.
Centripetal Force
“The force required to bend the normally straight path into circular path is called
centripetal or radial force. Its direction is always towards the centre of a circle.”
It is denoted by cF
Explanation
Let us take an example of a body attached to a string whirled on circular path as shown
in figure. Observations shows that if string is snapped it will not continue its motion on
circular path rather it will follow straight path.
Hence centripetal force must be furnished to move the body on circular path
Relation for Centripetal force
Consider a particle that moves from point A to B with uniform speed v on a circular path of
radius “r”. If ∆v is the change in velocity then acceleration is given by
𝑎 =∆v
∆𝑡 − − − − − (1)
Where ∆t the time is required by the particle to move from A to B. let v1⃗⃗ ⃗ and v2⃗⃗ ⃗ are velocities
at point A and B respectively as shown in figure (a)
If S is the distance covered by the particle from point A to b then
∆𝑡 =S
V
Putting this value in equation (1)
𝑎 =∆v
𝑆 (v) − − − − − (2)
Now we draw a triangle ∆DOE as shown in figure (b)
Such that
OD⃗⃗ ⃗⃗ ⃗ is parallel and equal to v1⃗⃗ ⃗
OE⃗⃗⃗⃗ ⃗ is parallel and equal to v2⃗⃗ ⃗
Circular motion Chapter
5
Important long questions
Prof. Muhammad Amin 2
Chapter-5 circular motion
As radius of the circle is perpendicular to its tangent, so v1⃗⃗ ⃗ and v2 ⃗⃗⃗⃗ ⃗ are perpendicular to OA
and OB.
⇒ < 𝐴𝑂𝐵 = < 𝐷𝑂𝐸
Moreover v1 = v2 = v and OA =OB
Hence for Isosceles triangles AOB and DOE , we can write
DE
DO=
AB
OB
∆v
v=
AB
r− − − − − −(3)
When ∆t → 0 then AB ≈ S
Equation (3) becomes
∆v =S
r (v)
Putting this value in equation (2)
𝑎 =
Sr (v)
𝑆 (v)
2
__________(4)c
va
r
Equation (4) represents centripetal acceleration.
In term of angular velocity, V = r
2 2 2
c
r ra
r r
2 __________(5)cor a r
Now
According to Newton’s second law of motion, the centripetal force is given by
______________ 6___c cF m a
Putting the value of ac from equation (4)
2
________________ 7_c
VF m
r
In term of angular velocity, using equation (5) in equation (6)
2 r ______________ 8___cF m
Equations 7 and 8 represents centripetal acceleration.
Prof. Muhammad Amin 3
Chapter-5 circular motion
Q. 5.2 what is meant by moment of inertia. Explain its physical significance.
Ans. Moment of Inertia
“The analogous of mass in rotation motion is called moment of inertia”
OR
Moment of inertia of a particle is “the product of its mass and square of the
perpendicular distance from axis of rotation”
Mathematically
𝐼 = 𝑚𝑟2
Its S.I unit is kg m2
Explanation
Consider a mass attached to a light rod, it can rotate about the
point ‘o’ as shown in figure
The mass of the rod is negligible.A force F is acting on the
mass perpendicular to the road and hence, this will accelerate
the mass according to second law of motion
a _________________ 1c TF m
This force will rotate the mass about ‘o’. Tangential acceleration ‘aT’ is related to
angular acceleration ‘’ by the equation
Ta r Putting in (1)
__________(2)F mr
Multiplying by r
2rF mr
Thus 2rF mr I T is analogue of the Newton’s second law of motion F = ma.
Moment of inertia is defined as “the product of mass of the particle and square of its
perpendicular distance from the axis of rotation.”
i.e. I = m r2
Moment of inertia of a rigid body
A body is made of n – small particles of masses m1, m2, m3,……………… and if the
distances of the particles from the axis of rotation ‘o’ be r 1,r2,……. rn,
Prof. Muhammad Amin 4
Chapter-5 circular motion
Then moment of inertial of the body is.
2 2 21 1 2 2 ....... n nI m r m r m r
2
1
i n
i ii
or I m r
Physical Significance
For a body in linear motion, the acceleration is directly proportional to the force acting
upon the body.
. . ( tan )F
i e F ma or m Cons ta
Where m is the mass of a moving body which is direct measure of inertia similarly for a
rotating body, the angular acceleration is directly proportional to the torque acting on it
or the ration of the torque to the angular acceleration is constant.
tanT
Cons t
So moment of inert of a body is analogous to the mass in linear motion. Hence
moment of inertia may be described as rotational mass of a body.
Q. 5.3 (a) Define rotational kinetic energy and derive a relation for it?
(b) Discuss rotational kinetic energy of a disc and a hop. Also derive the relation
for the velocities of a disc and hoop start moving down an inclined plane.
Ans. Rotational Kinetic Energy
The energy due to the spinning of a body about an axis is called rotational kinetic
energy.
Explanation
Consider a body is spinning about an axis with constant angular velocity . Everybody
is made up of large number of tiny pieces. Each piece of the body is moving in a
circular path and therefore, has some kinetic energy. The sum of the kinetic energies of
these pieces will give the kinetic energy of the whole body.If a piece of mass m i is at a
distance r i from the axis of rotation as shown in fig.
The linear speed of the piece is given by
i iV r
The kinetic energy of this piece is
21. .
2i ii
K E m V
21
. .2
i iiK E m r
2 21
2i im r
Let us suppose that the tiny pieces of the body having masses m1, m2, m3………… at
distances r1, r2, r3…….
Prof. Muhammad Amin 5
Chapter-5 circular motion
The K.E of the first piece 2 22 1
1
2m r
The K.E of the second piece 2 22 2
1
2m r
Total K.E 2 2 22 1 2 2
1.....
2m r m r
Where 2 21 1 2 2 .....m r m r I
I = moment of inertia of the body about the axis of rotation.
2
.
1. . . .
2rotTotal rotational K E K E I
(b) Rotational Kinetic Energy of a Disc And Hoop
Let us use the idea of rotational K.E to compare the velocity with which a disc and a
hoop reach the bottom of an inclined plane.
Figure shows that disc and the hoop start moving down an inclined plane of height h,
their motion consists of rotational and translational motions.
If no energy is lost in friction, the total K.E of the disc or hoop on reaching the bottom
of the incline must be equal to its potential energy at the top P.E at the top = rotational
K.E.at bottom + translational K.E at the bottom.
2 21 1____________(1)
2 2mgh I mV
Where and V are the angular and linear speeds at the bottom m is the mass of the
rolling body.
For Disc
21,
2I mr Putting in equation (1)
Where r is its radius.
2 2 21 1 1
2 2 2mgh mr mV
2 2But V r
2 2 2or V r
2 21 1 1
2 2 2mgh mV V
or m gh m 2 1 1
4 2V
2 3
4gh V
Prof. Muhammad Amin 6
Chapter-5 circular motion
2 4 4
3 3
gh ghV or V
This is the speed of disc on reaching the bottom of inclined plane
For Hoop
2I m r
Putting in equation (1)
2 2 21 1
2 2mgh mr mV
2 2 2V r
2 21 1
2 2mgh mV mV
mgh m 2V
2or V gh
or V gh
This is the speed of hoop on reaching the bottom of inclined plane
Q. 5.4 An object is suspended from a spring balance by a string and spring balance in a
lift. Find the apparent weight of the object in the following cases.
(a) When the lift is at rest.
(b) When the lift moves up with uniform acceleration.
(c) When the lift moves down with uniform acceleration.
(d) When the lift falling freely under the action of gravity.
Ans. Real weight
The force with which an object is attracted towards the center of earth is called its real
weight i.e W = mg
Apparent Weight
The force which prevents a body from falling in a certain frame of reference is known
as its apparent weight
Example of an elevator
Consider an object of mass m is suspended from a spring balance
by means of a string and spring balance is attached to the ceiling
of a lift as shown in figure.
The tension in the string is always equal in magnitude to the
weight of the object.
The reading of the spring balance i.e. tension T will
represent the apparent weight of the object.
Prof. Muhammad Amin 7
Chapter-5 circular motion
(a) When the lift is at rest
According to Newton’s second law of motion, the resultant force
on the object is zero.
So a = o
T ma
T m o
T o
________________(1)T
From equation (1), it is found that the apparent weight of the object is equal to its real
weight.
(b) When the lift moves up with uniform acceleration.
Let the lift moves upwards with an acceleration a.
T
T – = F (resultant force)
,mg F ma
T mg ma
____________(2)or T mg ma
This equation shows that the apparent weight of the object has increased by an amount
ma.
(c) When the lift moves down with uniform acceleration.
Let the lift moves downwards with an acceleration ‘a’ as shown in fig (b).
T
–T = F (resultant force)
,mg F ma
mg T ma
____________(3)or T mg ma
Equation (3) shows that the apparent weight of the object has
decreased by an amount ma.
(d) When the lift falling freely under the action of gravity.
If the cable supporting the lift breaks, the lift will fall down like
a free body with an acceleration equal to g.
, (3)a g Putting in equation
T mg mg
or T o
Hence the spring balance will read zero weight and the man in the lift will think that
the object has no weight i.e. it is in the state of weightlessness.
at rest
a=0
T=w
Acceleration downward
w-T=ma
T=w-ma
Acceleration upward
T-W=ma
T=w+ma
Prof. Muhammad Amin 8
Chapter-5 circular motion
Q. 5.5 Write short note on artificial gravity?
Ans. Artificial Gravity
Artificial gravity is produced by rotating a space craft about its own axis.
Explanation
In satellite all the objects become weightless. This produces a lot of difficulties for
astronauts in doing experiments. This problem is removed by producing artificial
gravity in the satellite. Rotating the satellite about its own axis produces the artificial
gravity.
Expression for Frequency
Consider a ring shaped space craft as shown in fig.
The outer radius of the space craft is R and it rotates around its own axis with angular
speed ω. At any point on the outer rim we have a centripetal acceleration is given by
the formula.
2
________________(1)c
Va V
R
Where V = linear speed of the space croft
As V = Rω Putting in equation (1)
2 22 ________(2)c
Ra R
R
If time period of space craft is T, then
2
TPutting in equation (2)
2 2
2
2 4ca R R
T T
2
2
14 _________(3)cor a R
T
Frequency can be expressed as
1f
TPutting in equation (3)
2 24ca R f
2
2
1
4 2c ca a
f or fR R
As force of gravity provides the centripetal acceleration, therefore, a c = g.
1
2
gf
R
When the space craft rotates with this frequency, the artificial gravity is produced in it.
Q. 5.6 What are geostationary orbits? Calculate the value of the radius measured from
the centre of the earth for a geostationary satellite?
Ans. Geostationary Orbit
A geo-stationary orbit is that in which the period of rotation of the satellite around the
earth is exactly equal to the period of rotation of the earth about its axis. A satellite
Prof. Muhammad Amin 9
Chapter-5 circular motion
with an orbital period of 24 hours will always be at the point above the earth’s surface.
Such satellites are called geostationary satellites or geosynchronous satellites.
They remain at a certain height 3600 Km above the equator of the earth and move in the
same direction as the earth is rotating.
Uses
Geostationary satellites are useful for worldwide communication. Weather observations
navigation and military purpose
Radius of Geostationary Orbit
The orbital speed of satellite is given by
___________(1)GM
Vr
Where
M = mass of the earth
G = gravitational constant.
r = radius of the orbit
This orbital speed is equal to the average speed of the satellite in one
2. . ______(2)
S ri e V
t T
Where T is the period of revolution of the satellite which is equal to one day
As the earth completes its one revolution in one day and the satellite will revolve
around the earth in one day, so the satellite at position A will always stay over the same
position A on the earth as shown in fig.
By comparing equation (1) and (2),
2 r GMwe have
T r
Taking square on both sides
2 2
2
4 r GM
T r
2 2 24 r r GM T
23
24
GMTor r
T = period of the earth = period of the satellite.
T = 234 hours
T = 24 60 60 S = 86 400 s
G = 6.67 10–11 N–m2 / Kg2
M = 6 1024 Kg
211 243
2
6.67 10 6 10 86400
4 3.14r
or r = (7.575 1022)1/3 = 4.23 107 m
or r = 4.23 104 Km = 42300 Km
Prof. Muhammad Amin 10
Chapter-5 circular motion
Which is the radius of a geostationary satellite from the centre of the earth.
If h is the height of the satellite above a particular point on the surface of the earth,
then
R = R + h
Or h = r – R
Radius of the earth = R = 6400 Km
h = 423 100 – 6400
h = 35900 Km 36000 Km
Prof. Muhammad Amin 11
Chapter-5 circular motion
5.1 Explain the difference between tangential velocity and the angular velocity. If one
of these is given for a wheel of known radius, how will you find the other?
Ans:
TANGENTIAL VELOCITY ANGUALR VELOCITY
Velocity of a body along the
tangent is known as tangential
velocity of linear velocity.
Its unit is m/s
Its direction is along tangent
,t
dv v r
t
Angular velocity of a body is the
rate of change of angular
displacement.
Its unit is radian/sec
Its direction is along the axis of
rotation.
v
t r
If one of them is given for a wheel of known radius, then other can be calculated using
the relation v = r
5.2 Explain what is meant by centripetal force and why it must be furnished to an
object if the object is to follow a circular path?
Ans: Definition
“The force needed to bend the normally straight path of the particle into a circular path
is called centripetal force”
Centripetal Force only changes the direction of motion
When a force acts perpendicular to the direction of motion of a
body then that force changes only the direction of motion of the
body. When a body moves in a circular path then at every instant
its direction of motion changes.
Hence it must be furnished so that direction of motion of body
changes continuously due to which straight path bends into
circular path.
5.3 What is meant by moment of inertia? Explain its significance.
Ans: Definition
“The rotational analogous of linear mass is called moment of inertia”
OR
Moment of inertia of a particle is “the product of its mass and square of the
perpendicular distance from axis of rotation”
Mathematically
𝐼 = 𝑚𝑟2
Its S.I unit is kg m2
Important short questions
Prof. Muhammad Amin 12
Chapter-5 circular motion
Significance
It plays same role in angular motion which inertia plays in linear motion. It resists
angular / circular motion as inertia resists linear motion.
It may be noted that moment of inertia depends not only on mass m but also on r 2.
5.4 What is meant by angular momentum? Explain the law of conservation of angular
momentum.
Ans: Angular Momentum
The moment of linear momentum of a body about a point is called angular momentum.
OR
The cross product of position vector r and linear momentum P of an object is called
angular momentum. It is denoted by L .
Mathematically, L r p
Its S.I unit kg m2 s-
Law of conservation of angular momentum:
It states that if no external torque acts on a system, the total angular momentum of the
system remains constant.
1 2
1 1 2 2
constant
I =I constant (For an isolated system)
TotalL L L
5.5 Show that orbital angular momentum L0 = mvr or L = I𝝎
Ans: Proof
Consider a boy of mass m is moving in a circular path of radius r with speed V, the
angular momentum of the body is given by
oL r P
The magnitude of angular momentum is given by
0
0
0
0
sin
( 90 )
sin 90
( sin 90 1)
,
,
o
o
o
L rp
L rp
L rp
As p mv
So
L rmv
Hence, Proved, L0 = mvr
Also 𝑣 = 𝑟 𝜔
⇒ L0 = m ( r ω) r = m r2 𝜔
L0= I 𝜔
5.6 Describe what should be the minimum velocity, for a satellite, to orbit close to the
Earth around it.
Ans: Definition
“The minimum velocity necessary to put the satellite into an orbit close to the surface
of Earth is called critical velocity.”
Prof. Muhammad Amin 13
Chapter-5 circular motion
It is given by the relation.
69.8 6.4 10v gR
3 17.9 10 secv m
So, v = 7.9kms-1
5.7 State the direction of the following vectors in simple situation; angular momentum
and angular velocity.
Ans: The direction of angular momentum and angular velocity is determined by right hand
rule.
For Angular Momentum
We know that:
L r p
This shows that the direction of angular momentum is perpendicular to the plane
containing r and p .
In case of circular motion, angular momentum is perpendicular to the plane of circle
and is along axis of rotation.
For Angular velocity
The direction of angular velocity is perpendicular to the plane of the circle and is along
the axis of rotation.
5.8 Explain why an object, orbiting the Earth, is said to be freely falling. Use your
explanation to point out why objects appear weightless under certain
circumstances.
Ans: When an object is orbiting around the Earth, force of gravity provides the necessary
centripetal force, as no force is holding that object. Therefore, it is freely falling frame
of reference. The curvature of path of the object and the Earth prevent it from falling
on the surface of Earth.
In a freely falling frame of reference, the frame of reference and the objects inside the
frame of reference are falling at the same rate i.e. same acceleration a = g, therefore
they appear to be weightless.
5.9 When mud flies off the tyre of a moving bicycle, in what direction does it fly?
Explanation
Ans: When tyre of bicycle moves, then init ially mud rotates with the tyre. The adhesive
force between mud and the tyre provides necessary centripetal force to rotate the mud
along with tyre.
When speed of tyre increases, more centripetal force is required by mud to rotate with
tyre. Since, adhesive force is not enough to provide necessary centripetal force.
Therefore, the mud flies off the tyre in a direction of tangent to the circular path.
Prof. Muhammad Amin 14
Chapter-5 circular motion
5.10 Why does a diver change his body positions before and after diving in the pool?
Ans:
Diver changes his bosy positions before and after the diving of the pool for the
effective use of law of conservation of momentum.
i.e
𝐿 = 𝐼𝜔 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
When a diver jumps off the diving board, he has small angular velocity about the
horizontal axis through his centre of gravity. When he draws his legs and arms close to
his body, his moment of inertia decreases and to conserve angular momentum, his
angular velocity increases and he spins faster. This enables the diver to take extra
somersaults.
5.11 Explain how many minimum number of geo-stationary satellites are required for
global coverage of T.V transmission .
Ans: To cover the whole earth, three properly positioned geo stationary satellites are
required. Each satellite covers an angle of 120 0 longitude and three satellites will be
covering complete angle of 3600.
5.12 What is angular displacement? How can we determine its direction?
Ans. Angular Displacement
It is the angle subtended at the centre of a circle by a particle moving
along the circumference in a given time.
Direction
Its direction can be determined by right hand rule.
Grasp the axis of rotation in right hand with fingers curling in the
direction of rotation the thumb points in the direction of angular
displacement.
5.13 Prove that S r
Ans. Consider an arc of length s of a circle of radius r which subtends an angle at the
centre of circle. By the definition of radian, its value in radian is given as
Arc length
radradius
S
r
s r
Prof. Muhammad Amin 15
Chapter-5 circular motion
5.14 Show that I
Ans. Let the rigid body is made up of n small pieces of masses m1, m2,…i….mn at distances
r1, r2, ………….rn from the axis of rotation. If rigid body is rotating, then the magnitude of
torque acting on m1, is
2
1 1 1m r
Similarly torque on m2 is
2
2 2 2m r
2 2 2
1 1 2 2 ..........Total n nm r m r m r
2
1
n
i i
i
m r I
I
5.15 Convert kgm2s-1 into JS.
Ans. 2 1 skgm s
s
= 2
2
.kgm s
s
= 2
. .kgm m s
s
As 2/kg m s N
= N. m .s
As N.m = J
Hence
Kgm2s-1 = Js
5.16 Why does the coasting rotating system slow down as water drips into the beaker?
Ans. When water drips into the beaker, the mass of the contents in the beaker increases
which increases the moment of inertia. So the angular velocity decreases according to law
of conservation of momentum.
5.17 What will be effect on the speed of satellite by increasing its mass?
Ans. We know that GM
vr
This relation shows that speed of satellite is independent of the mass of the satellite.
Therefore change of mass will not affect its speed.
5.18 What is meant by INTELSAT?
Ans. It means international telecommunication satellite organization. It operates at
microwaves frequencies of 4, 6, 11 and 14 GHz and has a capacity of 30, 000 two way
telephone circuits plus three TV channels.
5.19 Differentiate between Real and apparent weight of a body.
Ans.
Real Weight Apparent Weight
(i) It means. gravitational pull of
earth on that object.
(ii) It is measured as W = mg
The force which prevent the body from freely
falling is called apparent weight.
It is measured as tension in the spring balance.
Prof. Muhammad Amin 16
Chapter-5 circular motion
5.20 If a person is falling in an elevator freely. What will be his weight? Measured by
himself.
Ans. The force required to prevent a body from free fall is called its apparent weight. If a
person is falling freely in an elevator, its apparent weight is zero. Therefore he will be
in weightless. It means. that a freely falling body in an elevator feels no w eight by
himself.
Since apparent weight in descending lift is given by
T = mg – ma
But g = a = 9.8m/s2
T = mg – mg
= 0 N
5.21 Find the velocity of a disc rolls down along an inclined plane of height 10m?
Ans. h = 10m
v = ?
v = 4
3
gh
= 4 9.8 10
3 10
v = 392
3
v = 11.43 m/s
5.22 Describe the uses of geostationary satellite?
Ans. Uses of geostationary satellites are
(i) Worldwide communication
(ii) Weather observations
(iii) Navigation
(iv) Other military uses
5.23 Is any work done by centripetal force?
Ans. No, in circular motion, centripetal force is always acting perpendicular to the
displacement of the moving body
.w F d
= F d cosθ θ = 90o
= F d cos 90o
w = 0
Hence, work done by the centripetal force is zero.
5.24 Prove that 1 Radian = 57.3o
Ans. If OP is rotating, the point P covers a distance S = 2πr in one revolution. P.
In radian it would be 2s r
r r
= 2π
1 Revolution = 2π rad = 360o
1rad = 360
2
o
d
Fc
0
P
1P
S
Prof. Muhammad Amin 17
Chapter-5 circular motion
1rad = 57.3o
5.25 Show that v = r ω
Ans. Suppose during the course of its motion, the point P moves
through a distance P1P2 = ∆s in a time interval ∆t during
which reference line OP has an angular displacement ∆θ
radian during this interval. ∆s and ∆θ are related by Eq.
S r
Dividing both sides by ∆t
S
rt t
………..(i)
In the limit when 0t the ratio /S t represents v, the magnitude of the velocity
with which point P is moving on the circumference of the circle. Similarly / t
represents the angular velocity of the reference line OP. So equation (i) becomes.
v r
NUMERICAL PROBLEMS
4.1 A tiny laser beam is directed from the Earth to the Moon. If the beam is to have a diameter
of 2.50 m at the Moon, how small must divergence angle be for the beam? The distance of
Moon from the Earth is 3.8 x 108m.
Solution:
Data: Diameter of beam = length of arc = S = 2.50 m
Distance of Moon from the Earth = radius of circular arc = r = 3.8 x 108m
To find: Divergence angle = = ?
Calculation:
Using the formula
S = r
or = S/r
Putting the values, we get
9
8
2.50θ 6.6 10
3.8 10
rad
= 6.6 x 10-9 rad Ans
4.1 A gramophone record turntable accelerates from rest to an angular velocity of 45.0 rev min-1
in 1.60s. What is its average angular acceleration?
Solution:
Data: Initial angular velocity = iω =0
final angular velocity = fω =45.0 rev min-1
45.0×2π
= =1.5π60
rads-1
Time = t = 1.60s
To Find: Average angular acceleration = = ?
Calculation:
f iω -ωα=
t
Putting the values, we get
Numerical Problems
Prof. Muhammad Amin 18
Chapter-5 circular motion
1.5π-0
α= =2.951.60
rads-2
α = 2.95rad s-2 Ans
4.1 A body of moment of inertia I = 0.80 kg m2 about a fixed axis, rotates with a constant
angular velocity of 100 rad s-1. Calculate its angular momentum L and the torque to sustain
this motion.
Solution:
Data: Moment of inertia = I = 0.80 Kgm2
Angular velocity = = 100 rads-1
To Find: Angular momentum = L = ?
Torque = = ? Calculation: L = I Putting the values, we have L = 0.80 x 100 = 80 kgm2s-1
or L = 80 kgm2s-1 x s
s
= 80 kgm2s-2 s L = 80 Js Ans. Torque can be found by using the relation τ=Iα
Since the body has constant angular velocity, therefore angular acceleration is zero. Thus,
τ= I×0=0
τ = 0 Ans
4.1 Consider the rotating cylinder as shown in Fig. Suppose that m = 5.0 kg, F = 0.60 N and r = 0.20m. Calculate (a) the torque acting on the cylinder, (b) the angular acceleration of the
cylinder. (Moment of inertia of cylinder = 1
2mr2)
Solution: Data: Mass of cylinder = m = 5.0 kg Force acting on cylinder = F = 0.60 N Distance = r = 0.20m To Find: Torque acting on cylinder = = ?
Angular acceleration of cylinder = a = ? Calculation:
Using the formula of torque τ= r Fsinθ
r and F are perpendicular to each other. τ=r Fsin90
Putting the values, we get
τ = 0.20×0.60×1
.τ = 0.12Nm Ans
Now using the formula τ = Iα
Where I is the moment of inertia of cylinder which is equal to 21mr
2
Therefore,
21τ = mr α
2
Putting the values, we have
21τ= ×5.0×(0.20) α
2
0.12 = 0.1
or -2α=0.12/0.1=1.2rads
-2α = 1.2rads Ans
4.1 Calculate the angular momentum of a star of mass 2.0 x 1030 kg and radius 7.0 x 105 km.
Prof. Muhammad Amin 19
Chapter-5 circular motion
If it makes one complete rotation about its axis once in 20 days, what is its kinetic energy?
Solution:
Data: Mass of a star = M = 2.0 x 1030
Radius of star = R = 7.0 x 105 km
= 7.0 x 108 m
Time period of star = T = 20 days
= 20 x 24 x 60 x 60s
= 1728000s
To find: Angular momentum of star = L = ?
Kinetic energy = K. E. = ?
Calculation:
Using the formula
L = I
Where 2
I=5
MR2 for a sphere
And 2π
ω=T
Thus, 22 2πL= MR ×
5 T
Putting the values, we have
30 8 22×2×10 ×(7×10 ) ×2×3.14
L =5×1728000
46
3
2×2×7×7×2×3.14×10=
5×1728×10
43=0.142462963×10 42 -1 -1 -1L =1.4×10 kgms ×s /s
42L =1.4×10 Js Ans.
(b) Using the formula of kinetic energy K.E. = 2Iω
2
Putting the values of I and , we have
K.E. = 2 21 2× ×MR ×(2π/T)
2 5
or K.E. = 30 8 2 21 2× ×2×10 (7×10 ) (2π/20×24×3600)
2 5
30 16 -111= ×2×10 ×49×10 ×(1.3×10 )
5
35= 25.48×10 J
35= 25×10 ×10/10
K.E. = 2.5x 1036J Ans.
4.1 A 1000 kg car traveling with a speed of 144 km h-1 round a curve of radius 100 m. Find the
necessary centripetal force.
Solution:
Data: Mass of the car = M = 1000 kg
Speed of car = v = 144 kmh-1
-1144×1000= =40ms
60×60
Radius of the curve = r = 100m
Prof. Muhammad Amin 20
Chapter-5 circular motion
To Find: Centripetal force = F = ?
Calculation:
Using the formula 2
c
mvF =
r
Putting the values, we get 2
4
c
1000×(40)F = =1.60×10 N
100
Fc = 1.60 x 104 N Ans
4.1 What is the least speed at which an aeroplane can execute a vertical loop of 1.0 km radius
so that there will be no tendency for the pilot to fall down at the highest point?
Solution:
Data: Radius of loop = r = 1.0 km = 1000 m
Acceleration due to gravity = g = 9.8 ms-2
To find: Speed of aeroplane = v = ?
Calculation:
When an aeroplane executes a circular loop, the centripetal acceleration is supplied by gravity and
we have
or
2
c
2
a =g=v /r
v =rg
v= rg
Putting the values, we get
v = 1000 x 9.8
v = 99 ms-1 Ans.
4.1 The Moon orbits the Earth so that the same side always faces the Earth. Determine the ratio
of its spin angular momentum (about its own axis) and its orbital angular momentum. (In
this case, treat the Moon as a particle orbiting the Earth). Distance between the Earth and
the Moon is 3.85 x 108 m. Radius of the Moon is 1.74 x 106m.
Solution:
Data: Distance between Earth and Moon = r = 3.85 x 108 m
Radius of Moon = R = 1.74 x 106 m
To find: Ration of spin and orbital angular momentum = Ls/Lo = ?
The spin angular momentum of the Moon about its axis is
sL =Iω
When
22I= MR
5
Thus,
2
s
2L = MR ω.................(1)
5
The orbital angular momentum is given by
2
oL =Mr ω.................(2)
Where angular speed in both cases is the same and also the time to complete one rotation around
Earth and one rotation around its axis is the same.
Dividing equation (1) by equation (2) we get
2 2
s
2 2
o
L 2MR ω 2R= = .................(3)
L 5Mr ω 5r
Putting the values, we get
Prof. Muhammad Amin 21
Chapter-5 circular motion
6 2
s
8 2
o
12 12
17 17
-6
L 2×(1.74×10 )=
L 5×(3.85×10 )
2×3.03×10 6.06×10= =
5×1.48×10 7.4×10
=8.2×10
-6s
o
L= 8.2×10 Ans.
L
4.1 The Earth rotates on its axis once a day. Suppose, by some process the earth contracts so
that its radius is only half as large as at present. How fast will it be rotating then? (For
sphere I = 2/5 MR2).
Solution:
Data: Time period = T1 = 24 hours
Moment of inertia of sphere = I1 = 2
12MR
5
Radius of earth 12
RR =
2
To find: How fast will the earth be rotating i.e. T2 = ?
Calculation:
According to the law of conservation of momentum
1 1 2 2I ω =I ω
Or 1 2
2 1
ω I= ...............(1)
ω I
For sphere, I1 = 2/5 Mr12
And
2 2
2 12 2
2MR (R )2MI = =
5 5 (2)
As we have,
1 2
1 2
2π 2πω = andω =
T T
Putting these values in equation, (1) we get
2
1 1
2
2 1
2π/T 2/5M(R /2)=
2π/T 2/5MR
Or T2/T1 = 1
4
Or T2 = 1T
4
As we know that
T1 = 1 day = 24 hours
T2 = 24/4 = 6 hours
T2 = 6 hours Ans.
Q.5.10 What should be the orbital speed to launch a satellite in a circular orbit 900 km above the
surface of the Earth?
(Take mass of the earth as 6.0 1024 kg and its radius as 6400 km)
Solution:
Data: height above Earth surface = 900 km = 9 105 m
Radius of Earth = 6400 km = 6.4 106m
Mass of Earth = 6.0 1024 kg
Gravitational constant = G = 6.67 10-11 Nm2 /kg2
Prof. Muhammad Amin 22
Chapter-5 circular motion
To find: Orbital velocity = v = ?
Calculation: Using the formula
Gm
v=R +h
-11 24
6 6
6.67×10 6×10v =
6.4×10 +0.9×10
v = 7.4 km/s Ans