Chapter 7 Systems of Particles and Rotational Motion Questions and Answers: NCERT Solutions for Class 11th Physics (Partnership Physics )

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Class 11 Physics (Partnership Physics ) Chapter 7: Systems of Particles and Rotational Motion - Questions and Answers of NCERT Book Solutions.

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1. Give the location of the centre of mass of a
sphere,
cylinder,
ring, and
cube, each of uniform mass density.
Does the centre of mass of a body necessarily lie inside the body?

Ans: The centre of mass (C.M.) can be defined as a point where the mass of a body is supposed to be concentrated.
For the above listed geometric shapes having a uniform mass density, the centre of mass lies at their respective geometric centres.
The centre of mass of a specific body need not necessarily lie inside of the body. For example, the centre of mass of bodies such as a ring, a hollow sphere, etc., lie outside the respective body.

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2. In the HCl HCl
molecule, the separation between the nuclei of the two atoms is about 1.27A0 1.27A0
20kg
20kg
. Find the approximate location of the CM of the molecule, given that a chlorine atom is about 35.5 35.5
times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in its nucleus.

Ans: The provided situation can be expressed as:
Distance between H and Cl atoms 1.27 0 A
Mass of H atom =m
Mass of Cl atom =35.5m
Let the centre of mass of the given system be at a distance x from the
Cl
atom.
Distance between the centre of mass and the H atom =(1.27−x)
Let us suppose that the centre of mass of the given molecule lies at the origin. Therefore, it can be written as: m(1.27−x)+35.5mx
m+35.5m
=0
⇒m(1.27−x)+35.5mx=0
⇒1.27−x=−35.5x
⇒x=M
−1.27
(35.5−1)
=\[−0.37
0
A
Here, the negative sign gives an indication that the centre of mass lies at the left side of the molecule.
Therefore, the centre of mass of the
HCl
molecule lies
0.37
0
A
from the Cl atom.

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3. A child sits stationary at one end of a long trolley moving uniformly with a speed v
on a smooth horizontal floor. If the child gets up and runs about on the trolley in any manner, what is the speed of the CM of the (trolley + child) system?

Ans: There will not be any change in the speed of the centre of mass of the given system.
The child is running arbitrarily on a trolley that is moving forward with velocity v v
. However, the running of the child will have no effect on the velocity of the centre of mass of the trolley. This happens because the force due to the motion of the child is purely internal. Internal forces in a body produce no effect on the motion of the bodies on which they are acting. Because there is no external force involved in the (child + trolley) system, the child’s motion will not produce any change in the speed of the centre of mass of the trolley.

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4. Show that the area of the triangle contained between the vectors a a and b b is one half of the magnitude of a×b a×b

Ans: Let us consider two vectors OK−→−=|a⃗ |
OK→=|a→|
and OM−→−=∣∣b⃗ ∣∣
OM→=|b→|
, which are inclined at an angle θ θ
, as shown in the following figure.
In ΔOMN
ΔOMN
, we can express the relation:
sinθ=MNOM=MN∣∣b⃗ ∣∣
sin⁡θ=MNOM=MN|b→|
⇒MN=∣∣b⃗ ∣∣sinθ
⇒MN=|b→|sin⁡θ
Now,
∣∣a⃗ ×b⃗ ∣∣=|a⃗ |∣∣b⃗ ∣∣sinθ
|a→×b→|=|a→||b→|sin⁡θ
⇒∣∣a⃗ ×b⃗ ∣∣=OK⋅MN×22
⇒|a→×b→|=OK⋅MN×22
⇒∣∣a⃗ ×b⃗ ∣∣=2×AreaofΔOMK
⇒|a→×b→|=2×AreaofΔOMK
⇒AreaofΔOMK=12∣∣a⃗ ×b⃗ ∣∣

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5. Show that a⋅(b×c) a⋅(b×c)
is equal in magnitude to the volume of the parallelepiped formed on the three vectors, a a , b b and c c

Ans: A parallelepiped with origin O
O and sides a a , b b , and c c is depicted in the following figure.
Volume of the given parallelepiped =abc
=abc
And
OA−→−=a⃗
OA→=a→
OB−→−=b⃗
OB→=b→
OC−→−=c⃗
OC→=c→
Let us suppose that n^ n^ be a unit vector perpendicular to both b⃗ b→ and c⃗ c→. Therefore,
n^ n^ and a⃗
a→ have the same direction.
b⃗ ×c⃗ =bcsinθn^
b→×c→=bcsin⁡θn^
⇒b⃗ ×c⃗ =bcsin90∘n^
⇒b→×c→=bcsin⁡90∘n^
⇒b⃗ ×c⃗ =bcn^
⇒b→×c→=bcn^
Now,
a⃗ (b⃗ ×c⃗ )=a⋅(bcn^)
a→(b→×c→)=a⋅(bcn^)
⇒a⃗ (b⃗ ×c⃗ )=abccosθn^
⇒a→(b→×c→)=abccos⁡θn^
⇒a⃗ (b⃗ ×c⃗ )=abccos0∘n^
⇒a→(b→×c→)=abccos⁡0∘n^
⇒a⃗ (b⃗ ×c⃗ )=abccos0∘
⇒a→(b→×c→)=abccos⁡0∘
⇒a⃗ (b⃗ ×c⃗ )=abc
⇒a→(b→×c→)=abc
⇒a⃗ (b⃗ ×c⃗ )=abc=Volumeoftheparallelepiped

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6. Find the components along the x x , y y , z z axes of the angular momentum l
l of a particle, whose position vector is r r with components x
x , y y , z z and momentum is p
p with components px
px , py py and pz pz . Show that if the particle moves only in the x-y plane the angular momentum has only a z-component.

Ans: Linear momentum of the particle, p⃗ =pxi^+pyj^+pzk^ p→=pxi^+pyj^+pzk^
Position vector of the particle, r⃗ =xi^+yj^+zk^ r→=xi^+yj^+zk^
Angular momentum,
l⃗ =r⃗ ×p⃗ l→=r→×p→
⇒l⃗ =(xi^+yj^+zk^)×(pxi^+pyj^+pzk^) ⇒l→=(xi^+yj^+zk^)×(pxi^+pyj^+pzk^)
⇒I⃗ =⎛⎝⎜i^xpxj^ypyk^zpz⎞⎠⎟⇒I→=(i^j^k^xyzpxpypz)
Now,
lxi^+lyj^+lzk^=i^(ypz−zpy)−j^(xpz−zpx)+k^(xpy−zpx) lxi^+lyj^+lzk^=i^(ypz−zpy)−j^(xpz−zpx)+k^(xpy−zpx)
On comparing the coefficients of i^
i^ , j^ j^ and k^ k^ , we can write:
lx=ypz−zpy lx=ypz−zpy
, ly=zpx−xpz ly=zpx−xpz
, lz=xpy−ypx lz=xpy−ypx
…… (1) The particle is moving in the x-y plane. Therefore, the z-component of the position vector and linear momentum vector is becoming zero, i.e., z=pz=0 z=pz=0
Thus, equation (1) reduces to:
lx=0 lx=0
ly=0 ly=0
lz=xpy−ypx lz=xpy−ypx
Hence, when the particle is subject to move in the x-y plane, the direction of angular momentum will be along the z-direction.

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7. Two particles, each of mass m m and speed v
v , travel in opposite directions along parallel lines separated by a distance d d
. Show that the vector angular momentum of the two-particle system is the same whatever be the point about which the angular momentum is taken.

Ans: Let us suppose that at a certain instant two particles be at points P
P and Q Q , as shown in the given figure.
(image will be uploaded soon) Angular momentum of the system about point P P
can be given as: LP−→=mv×0+mv×d LP→=mv×0+mv×d ⇒LP−→=mvd ⇒LP→=mvd …… (1)
Angular momentum of the system about point Q can be given as:
LQ−→=mv×d+mv×0 LQ→=mv×d+mv×0
⇒LQ−→=mvd ⇒LQ→=mvd …… (2)
Let us consider a point R R , which is at a distance y
y from point Q
Q , i.e., QR=yQR=y
⇒PR=d−y⇒PR=d−y
Angular momentum of the system about point R R
can be given as:
LR−→=mv×(d−y)+mv×y LR→=mv×(d−y)+mv×y
⇒LR−→=mvd−mvy+mvy ⇒LR→=mvd−mvy+mvy
⇒LR−→=mvd ⇒LR→=mvd
…… (3) On comparing equations (1), (2), and (3), we get:
L→P=L→Q=L→R L→P=L→Q=L→R
…… (4) We can hence infer from equation (4) that the angular momentum of a system is independent of the point about which it is taken.

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8. A non-uniform bar of weight W W is suspended at rest by two strings of negligible weight as shown in Fig.7.39. The angles made by the strings with the vertical are 36.9∘
36.9∘ and 53.1∘ 53.1∘
respectively. The bar is 2m 2m
long. Calculate the distance d d
of the centre of gravity of the bar from its left end.

Ans: The free body diagram of the bar can be drawn as shown in the given figure. Length of the bar is given, l=2m l=2m
T1 T1 and T2 T2 are the tensions generated in the left and right strings respectively.
At translational equilibrium, we can express:
T1sin36.9∘=T2sin53.1∘
T1sin⁡36.9∘=T2sin⁡53.1∘
⇒T1T2=sin53.1∘sin36.9∘
⇒T1T2=sin⁡53.1∘sin⁡36.9∘
⇒T1T2=0.8000.600=43
⇒T1T2=0.8000.600=43
⇒T1=43T2
⇒T1=43T2
On taking the torque about the centre of gravity, for rotational equilibrium, we can write:
T1cos36.9∘×d=T2cos53.1∘(2−d)
T1cos⁡36.9∘×d=T2cos⁡53.1∘(2−d)
⇒T1×0.800×d=T20.600(2−d)
⇒T1×0.800×d=T20.600(2−d)
⇒43×T2×0.800×d=T2(0.600×2−0.600d)
⇒43×T2×0.800×d=T2(0.600×2−0.600d)
⇒1.067d+0.6d=1.2
⇒1.067d+0.6d=1.2
⇒d=1.21.67
⇒d=1.21.67
⇒d=0.72m
⇒d=0.72m
Therefore, the centre of gravity of the given bar lies 0.72m 0.72m from the left end of the bar.

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9. A car weighs 1800kg 1800kg
. The distance between its front and back axles is 1.8m 1.8m
. Its centre of gravity is 1.05m 1.05m
behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel.

Ans: Given that,
Mass of the car is given as, m=1800kg m=1800kg
Distance between the front and back axles, d=1.8m d=1.8m
Distance between the centre of gravity and the back axle =1.05m =1.05m
The different forces acting on the car are shown in the given figure:
(image will be uploaded soon) The forces in the figure, Rf
Rf and Rb Rb
are the forces exerted by the level ground on the front wheels and back wheels respectively.
At translational equilibrium we can write:
Rf+Rb=mg
Rf+Rb=mg ⇒Rf+Rb=1800×9.8
⇒Rf+Rb=1800×9.8
⇒Rf+Rb=17640N ⇒Rf+Rb=17640N …... (1)
For rotational equilibrium, on taking the torque about the centre of gravity, we can write:
Rf(1.05)=Rb(1.8−1.05)
Rf(1.05)=Rb(1.8−1.05)
⇒Rf×(1.05)=Rb×(0.75)
⇒Rf×(1.05)=Rb×(0.75)
⇒RfRb=0.751.05=57
⇒RfRb=0.751.05=57
⇒RbRf=75
⇒RbRf=75
⇒Rb=1.4Rf
⇒Rb=1.4Rf
…… (2) Solving equations (1) and (2), we obtain:
1.4Rf+Rf=17640N
1.4Rf+Rf=17640N
⇒Rf=176402.4N=7350N
⇒Rf=176402.4N=7350N
⇒Rb=(17640−7350)N=10290N
⇒Rb=(17640−7350)N=10290N
Therefore, the force exerted on each front wheel can be given as 73502N=3675N 73502N=3675N
and
the force exerted on each back wheel can be given as 102902N=5145N

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10.Find the moment of inertia of a sphere about a tangent to the sphere, given the moment of inertia of the sphere about any of its diameters to be 2MR25
2MR25
where M M is the mass of the sphere and R R
is the radius of the sphere.

Ans: The moment of inertia (M.I.) of a sphere about its diameter can be given as: 2MR25 2MR25
According to the theorem of parallel axes, the moment of inertia of a body about any axis is same as the sum of the moment of inertia of a certain body about a parallel axis passing through its centre of mass and the product of its mass and the square of the distance between the two parallel axes.
The moment of inertia about a tangent of the sphere can be expressed as:

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11. Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry, and the sphere is free to rotate about an axis passing through its centre. Which of the two will acquire a greater angular speed after a given time?

Ans: Let us assume that m m and r r be the respective mass and radius of the hollow cylinder and the solid sphere. The moment of inertia of the hollow cylinder about its standard axis can be given as,
I1=mr2 I1=mr2 The moment of inertia of the solid sphere about an axis that passes through its centre can be given as,
I2=25mr2 I2=25mr2
The formula for torque in terms of angular acceleration and moment of inertia can be expressed as:
τ=Iα τ=Iα Where,
τ= τ= Torque
α= α= Angular acceleration
I= I= Moment of inertia
For the hollow cylinder the expression can be given as,
τ1=I1α1 τ1=I1α1
For the solid sphere the expression can be given as,
τ2=I2α2 τ2=I2α2
As an equal amount of torque is applied to both the bodies it can be stated as, α2α1=I1I2=mr225mr2=25 α2α1=I1I2=mr225mr2=25
α2>α1 α2>α1
…… (1) Using the relation ω=ω0+αt ω=ω0+αt
Where,
α= α= Angular acceleration
t= t= Time of rotation
ω0= ω0= Initial angular velocity
ω= ω= Final angular velocity
For equal ω0
ω0
and t
t
, we have: ω=α ω=α …... (2) From equations (1) and (2), we can conclude:
ω2>ω1 ω2>ω1 Therefore, the angular velocity
) of the solid sphere will be greater than that of the hollow cylinder.

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12. A solid cylinder of mass 20kg 20kg rotates about its axis with angular speed 100rads−1
100rads−1 . The radius of the cylinder is 0.25m
0.25m . What is the kinetic energy associated with the rotation of the cylinder? What is the magnitude of angular momentum of the cylinder about its axis?

Ans: Mass of the cylinder is given, m=20kg m=20kg
Angular speed of the cylinder, ω=100rads−1 ω=100rads−1
Radius of the solid cylinder, r=0.25m r=0.25m
The moment of inertia of the solid cylinder can be expressed as:
I=12mr2 I=12mr2
⇒I=12×20kg×(0.25)2
⇒I=12×20kg×(0.25)2
⇒I=6.25kgm2 ⇒I=6.25kgm2
Kinetic energy of the cylinder=12Iω2 =12Iω2
⇒K.E.=12×6.25×(100)2
⇒K.E.=12×6.25×(100)2 Angular Momentum of the cylinder,
L=Iω L=Iω
⇒L=6.25×100
⇒L=6.25×100
⇒L=62.5Js

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13. A child stands at the centre of a turntable with his two arms outstretched. The turntable is set rotating with an angular speed of 40rev/min
40rev/min . How much is the angular speed of the child if he folds his hands back and thereby reduces his moment of inertia to 25
25 times the initial value? Assume that the turntable rotates without friction.

Ans: Given that,
Initial angular velocity of turntable, ω1=40rev/min ω1=40rev/min
Final angular velocity of the given turntable is ω2 ω2
The moment of inertia of the child with stretched hands can be given as I1 I1
The moment of inertia of the child with folded hands can be given as I2 I2
The two moments of inertia are related to each other as follows:
I2=25I1
I2=25I1
Since no external force acts on the child, the angular momentum L L
is not varying. Therefore, for the two circumstances, we can write:
I1ω1=I2ω2 I1ω1=I2ω2
⇒ω2=I1I2ω1 ⇒ω2=I1I2ω1
⇒ω2=I125I1×40=52×40 ⇒ω2=I125I1×40=52×40
⇒ω2=100rev/min

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14. A rope of negligible mass is wound round a hollow cylinder of mass 3kg 3kg and radius 40cm 40cm
. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30N 30N
? What is the linear acceleration of the rope? Assume that there is no slipping.

Ans: Given that,
Mass of the hollow cylinder is given as, m=3kg m=3kg
Radius of the hollow cylinder is given as, r=40cm=0.4m r=40cm=0.4m
Applied force on the given rope is given as, F=30N F=30N
The moment of inertia of the hollow cylinder about its geometric axis can be given as:
I=mr2I=mr2
⇒I=3×(0.4)2⇒I=3×(0.4)2
⇒I=0.48kgm2⇒I=0.48kgm2
Torque acting on the rope,
τ=F×r τ=F×r
⇒τ=30×0.4⇒τ=30×0.4
⇒τ=12Nm⇒τ=12Nm
For angular acceleration α α , torque can also be given by the expression:
τ=Iα τ=Iα
⇒α=τI=120.48 ⇒α=τI=120.48
⇒α=25rads−2
⇒α=25rads−2
Linear acceleration of the rope can be stated as =ra=0.4×25=10ms−2

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15. To maintain a rotor at a uniform angular speed of 200rads−1 200rads−1
an engine needs to transmit a torque of 180Nm 180Nm
. What is the power required by the engine? (Note: uniform angular velocity in the absence of friction implies zero torque. In practice, applied torque is needed to counter frictional torque). Assume that the engine is 100 100 efficient.

Ans: Given that,
Angular speed of the rotor is given as, 200rad/s 200rad/s
Torque required by the rotor of the engine is given as 180Nm 180Nm
. The power of the rotor (P) (P)
can be expressed in the relation of torque and angular speed by the formula P=τω P=τω
⇒P=180×200=30×103
⇒P=180×200=30×103
⇒P=36kW
⇒P=36kW
Therefore, the power required by the engine is 36kW 36kW

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16. From a uniform disk of radius R R , a circular hole of radius R2
R2 is cut out. The centre of the hole is at R2 R2
from the centre of the original disc. Locate the centre of gravity of the resulting flat body.

Ans: Given that,
Mass per unit area of the original disc can be given as σ
σ . Radius of the original disc =R =R
Mass of the original disc,
M=πR2σ M=πR2σ
The disc with the cut portion is shown in the given figure:
Radius of the smaller disc is given =R2 =R2
Mass of the smaller disc is given as M′=π(R2)2σ=14πR2σ=M4 M′=π(R2)2σ=14πR2σ=M4
Let us suppose that O O and O′ O′
be the respective centres of the original disc and the disc cut off from the original. As per definition of the centre of mass, the centre of mass of the original disc is assumed to be concentrated at O O , while that of the smaller disc is assumed to be concentrated at O′ O′
. It is provided that:
OO′=R2
OO′=R2
After the smaller disc has been cut from the original disc, the remaining portion left over after cutting is considered to be a system of two masses. The two masses can be expressed as:
M(concentratedatO)−M′=(M4)concentratedatO′ M(concentratedatO)−M′=(M4)concentratedatO′
(The negative sign in the above statement indicates that this portion has been removed from the original disc.) Let us suppose that x
x be the distance through which the centre of mass of the remaining portion shifts from point O O
. The relation between the centres of masses of two masses is given as:
x=m1r1+m2r2m1+m2 x=m1r1+m2r2m1+m2
For the given system, it can be written as:
⇒x=M×0−M′×(R2)M+(−M′) ⇒x=M×0−M′×(R2)M+(−M′)
⇒x=−M4×R2M−M4=−MR8×43M=−R6 ⇒x=−M4×R2M−M4=−MR8×43M=−R6
(The negative sign in the above statement indicates that the centre of mass gets shifted toward the left of point O O .)
The centre of gravity of the resulting flat body can be located from the original centre of the body and opposite to the centre of the cut portion.

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17. A meter stick is balanced on a knife edge at its centre. When two coins, each of mass 5g 5g
are put one on top of the other at the 12cm 12cm
mark, the stick is found to be balanced at 45cm 45cm
. What is the mass of the meter stick?

Ans: Let us assume that W W and W′ W′
be the respective weights of the meter stick and the coin.
The mass of the meter stick is supposed to be concentrated at its mid-point, i.e., at the 50cm 50cm mark.
Mass of the meter stick is m′ m′
Mass of each coin is m=5g m=5g
When the coins are placed 12cm 12cm away from the end P P
, the centre of mass gets shifted by 5cm 5cm from point R R
toward the end P P
. The centre of mass is located at a distance of 45cm 45cm
from point P P
. The net torque will be thus, conserved for rotational equilibrium about point R R
. This can be expressed by the equation,
10×g(45−12)−m′g(50−45)=0 10×g(45−12)−m′g(50−45)=0
⇒m′=10×335=66g
⇒m′=10×335=66g
Therefore, the mass of the meter stick is 66g

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18. A solid sphere rolls down two different inclined planes of the same heights but different angles of inclination.

a.Will it reach the bottom with the same speed in each case?
Ans: Mass of the sphere =m =m
Height of the plane =h =h
Velocity of the sphere at the bottom of the plane is given as =v =v
At the top of the plane, the total energy of the sphere i.e., Potential energy (P.E.)=mgh (P.E.)=mgh
At the bottom of the plane, the sphere has both translational and rotational kinetic energies which can be expressed as,
Therefore, total energy T.E.=12mv2+12Iω2 T.E.=12mv2+12Iω2
Using the law of conservation of energy, we can state that: 12mv2+12Iω2=mgh 12mv2+12Iω2=mgh
…… (1) For a solid sphere, the moment of inertia about its centre can be given as,
I=25mr2
I=25mr2
Therefore, equation (1) becomes:
12mv2+12(25mr2)ω2=mgh
12mv2+12(25mr2)ω2=mgh
⇒12v2+(15r2)ω2=gh
⇒12v2+(15r2)ω2=gh
But we have the formula,
v=rω
v=rω
12v2+15v2=gh 12v2+15v2=gh
⇒v2(710)=gh ⇒v2(710)=gh
⇒v=107gh−−−−−√ ⇒v=107gh
Therefore, the velocity of the sphere at the bottom depends only on height (h) and acceleration due to gravity (g). Both values are constants and do not change. Therefore, the velocity at the bottom remains the same from whichever inclined plane the sphere is rolled.
b.Will it take longer to roll down one plane than the other?
Ans: Let us consider two inclined planes with inclinations θ1 θ1 and θ2 θ2 respectively related as:
θ1<θ2
θ1<θ2
The acceleration generated in the sphere when it rolls down the plane inclined at θ1 θ1 is:
gsinθ1 gsin⁡θ1 The different forces acting on the sphere are shown in the given figure.
R1 R1 is the normal reaction to the sphere as shown in the above figure.
Similarly, the acceleration generated in the sphere when it rolls down the plane inclined at θ2 θ2 is:
gsinθ2
gsin⁡θ2
The different forces that act on the sphere are shown in the given figure.
R2 R2 is the normal reaction to the sphere as given in the figure.
θ1<θ2 θ1<θ2 , sinθ2>sinθ1
sin⁡θ2>sin⁡θ1
…… (1) a2>a1
a2>a1
…… (2) Initial velocity of sphere, u=0 u=0
Final velocity of sphere, v= v=
constant
Now, by using the first equation of motion, we can obtain the time of roll as:
v=u+at
v=u+at
t∝1a t∝1a
For inclination of angle θ1 θ1
: t1∝1a1
t1∝1a1
For inclination of angle θ2 θ2
: t2∝1a2 t2∝1a2
…… (3) c.If so, which one and why?
Ans: From equations (2) and (3), we obtain:
t2 Therefore, conclude that the sphere will take a longer time to reach the bottom of the inclined plane having the smaller inclination.

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19. A hoop of radius 2m weighs 100kg 100kg . It rolls along a horizontal floor so that its centre of mass has a speed of 20cm/s 20cm/s . How much work has to be done to stop it?

Ans: Radius of the hoop is given as, r=2m r=2m
Mass of the hoop is, m=100kg m=100kg
Velocity of the hoop is,
v=20cm/s=0.2m/s
v=20cm/s=0.2m/s
Total energy of the hoop = =
Translational KE + +
Rotational KE
Er=12mv2+12Iω2
Er=12mv2+12Iω2
Moment of inertia of the hoop about its centre can be given as I=mr2 I=mr2
Er=12mv2+12(mr2)ω2
Er=12mv2+12(mr2)ω2
But we have the formula, v=rω v=rω
⇒E1=12mv2+12mr2ω2
⇒E1=12mv2+12mr2ω2
⇒E1=12mv2+12mv2=mv2
⇒E1=12mv2+12mv2=mv2
The work needed to be done for halting the hoop is the same as the total energy of the hoop.
Hence, required work to be done can be given as,
W=mv2=100×(0.2)2=4J

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20. The oxygen molecule has a mass of 5.30×1026kg 5.30×1026kg and a moment of inertia of 1.94×10−46kgm2 1.94×10−46kgm2 about an axis through its centre perpendicular to the lines joining the two atoms. Suppose the mean speed of such a molecule in a gas is 500m/s 500m/s and that its kinetic energy of rotation is two thirds of its kinetic energy of translation. Find the average angular velocity of the molecule.

Ans: Given that,
Mass of an oxygen molecule is given as, m=5.30×1026kg m=5.30×1026kg
Moment of inertia of oxygen molecule is given as, I=1.94×10−46kgm2 I=1.94×10−46kgm2
Velocity of the oxygen molecule is given as, v=500m/s v=500m/s
Let the separation between the two atoms of the oxygen molecule be 2r 2r
Mass of each oxygen atom in the oxygen molecule =m2 =m2
Therefore, moment of inertia I I
, can be calculated as:
(m2)r2+(m2)r2=mr2
(m2)r2+(m2)r2=mr2
r=lm−−−√ r=lm
⇒1.94×10−465.36×1026−−−−−−−−−−−√=0.60×10−10m
⇒1.94×10−465.36×1026=0.60×10−10m
It is provided that:
KErot=23KEtrans
KErot=23KEtrans
⇒12Iω2=23×12mv2
⇒12Iω2=23×12mv2
⇒mr2ω2=23mv2
⇒mr2ω2=23mv2
⇒ω=23−−√×vr ⇒ω=23×vr
⇒ω=23−−√×5000.6×10−10 ⇒ω=23×5000.6×10−10
⇒ω=6.80×1012rad/s ⇒ω=6.80×1012rad/s
, which is the required average angular velocity.

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1. A solid cylinder rolls up an inclined plane of angle of inclination 30∘ 30∘ . At the bottom of the inclined plane the centre of mass of the cylinder has a speed of 5m/s 5m/s . How far will the cylinder go up the plane?
How long will it take to return to the bottom?

Ans: A solid cylinder rolling up an inclination is pictured in the following figure.
Initial velocity of the solid cylinder on the inclined plane, v=5m/s v=5m/s
Angle of inclination is 30∘ 30∘
Height reached by the cylinder on the inclined plane =h =h
Energy of the cylinder on the inclined plane at point A A
: KErot=KEtrans KErot=KEtrans
⇒12Iω2=12mv2 ⇒12Iω2=12mv2
Energy of the cylinder at point B B =mgh =mgh
Let us use the law of conservation of energy, we can express:
12Iω2=12mv2=mgh 12Iω2=12mv2=mgh
Moment of inertia of the solid cylinder is I=12mr2 I=12mr2
⇒12(12mr2)ω2+12mv2=mgh ⇒12(12mr2)ω2+12mv2=mgh
⇒14mr2ω2+12mv2=mgh ⇒14mr2ω2+12mv2=mgh
⇒14r2ω2+12v2=gh ⇒14r2ω2+12v2=gh
But we have the expression, v=rω v=rω
⇒14v2+12v2=gh ⇒14v2+12v2=gh
⇒34v2=gh ⇒34v2=gh
⇒h=34×v2g ⇒h=34×v2g
⇒h=34×5×59.8=1.91m ⇒h=34×5×59.8=1.91m
In ΔABC ΔABC
, sinθ=BCAB sin⁡θ=BCAB
⇒sin30∘=hAB ⇒sin⁡30∘=hAB
AB=1.910.5=3.82m AB=1.910.5=3.82m
Therefore, the cylinder will move 3.82m 3.82m
up the inclined plane.
For radius of gyration K K
, the velocity of the cylinder at the instance when it rolls back to the bottom is given by the formula: v=⎛⎝2gh1+K2R2⎞⎠12 v=(2gh1+K2R2)12
⇒v=⎛⎝2gABsinθ1+K2R2⎞⎠12 ⇒v=(2gABsin⁡θ1+K2R2)12
For the solid cylinder we can write K2=R22 K2=R22
⇒v=(2gABsinθ1+12)12 ⇒v=(2gABsin⁡θ1+12)12
⇒v=(43gABsinθ)12 ⇒v=(43gABsin⁡θ)12
The time taken to return to the bottom can be given as:
t=ABv t=ABv
⇒t=AB(43gABsinθ)12 ⇒t=AB(43gABsin⁡θ)12
⇒t=(3AB4gsinθ)12 ⇒t=(3AB4gsin⁡θ)12
⇒t=(11.4619.6)12=0.764s ⇒t=(11.4619.6)12=0.764s
Therefore, the total time taken by the cylinder to return to the bottom is (2×0.764)=1.53s

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Additional Exercise

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22. As shown in figure, the two sides of a step ladder BA BA and CA CA are 1.6m 1.6m long and hinged at A A . A rope DE DE , 0.5m 0.5m is tied halfway up. A weight 40kg 40kg is suspended from a point F F , 1.2m 1.2m from B B along the ladder BA BA . Assuming the floor to be frictionless and neglecting the weight of the ladder, find the tension in the rope and forces exerted by the floor on the ladder. (Take g=9.8m/s2 g=9.8m/s2 )

Ans: The given situation can be depicted as follows: NB= Force exerted on the ladder by the floor point B NC=
Force exerted on the ladder by the floor point C T=
Tension in the given rope
BA=CA=1.6m
DE=0.5m
BF=1.2m
Mass of the given weight, m=40kg m=40kg
Draw a perpendicular line from A A on the floor BC BC . This line intersects DE DE at mid-point H H.
In ΔABI ΔABI and ΔAIC ΔAIC are similar
BI=IC Therefore, I I is the mid-point of BC BC
. DE∥BC BC=2×DE=1m
AF=BA−BF=0.4m …… (1)
It can be said that D D is the mid-point of AB. Therefore, we can express:
AD=12×BA=0.8m ……. (2) Using equations (1) and (2), we get:
FE=0.4m
Therefore, F F
is the mid-point of AD AD
. FG∥DH and F F is the mid-point of AD
AD . Therefore, G
G will also be the mid-point of AH AH
. ΔAFGDH ΔADH are similar triangles.
FGDH=AFAD
⇒FGDH=0.40.8=12
⇒FG=12DH
⇒FG=12×0.25=0.125m
In ΔADH
ΔADH we can state, AH=AD2−DH2
⇒AH=0.82−0.252=0.75m
For translational equilibrium of the ladder, the upward force should be same as the downward force.
NC+NB=mg=392
…… (3) For rotational equilibrium of the ladder, the net moment about A can be given as:
⇒−NB×0.5+40×9.8×0.125+NC×0.5=0
⇒(NC−NB)×0.5=49
⇒NC−NB=98
…… (4) Solving equations (3) and (4), we can write
NC=245N
NB=147N
For rotational equilibrium of the side AB, let us consider the moment about A.
−NB×BI+mg×FG+T×AG=0
⇒−245×0.5+40+9.8×0.125+T×0.76=0
⇒0.76T=122.5−49
⇒T=96.7N
, which is the required tension.

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23. A man stands on a rotating platform, with his arms stretched horizontally holding a 5kg 5kg weight in each hand. The angular speed of the platform is 30 30 revolutions per minute. The man then brings his arms close to his body with the distance of each weight from the axis changing from 90cm 90cm to 20cm 20cm . The moment of inertia of the man together with the platform may be taken to be constant and equal to 7.6kgm2 7.6kgm2

a.What is his new angular speed? (Neglect friction.)
Ans: Moment of inertia of the man-platform system is given as,
7.6kgm2
Moment of inertia when the man stretches his hands to a distance of 90 cm:
MIHandsStretched=2×mr2
⇒MIHandsStretched=2×5×(0.9)2
⇒MIHandsStretched=8.1kgm2
Initial moment of inertia of the system can be given as,
Ii=7.6+8.1=15.7kgm2
Angular speed can be expressed as,
ω1=300rev/min
Angular momentum can be given as,
Li=Iiωi=15.7×30
…… (1) Moment of inertia when the man folds his hands to a distance of 20 cm becomes:
MIHandsat20cm=2mr2
⇒MIHandsat20cm=2×5(0.2)2=0.4kgm2
Final moment of inertia is given as,
If=7.6+0.4=8kgm2
Final angular speed can be given as, ωf ωf
Final angular momentum can be expressed as,
Lf=Ifωf=0.79ω
…… (2) From the conservation of angular momentum, we can write:
Iiωi=Ifωf
ωf=15.7×308=58.88rev/min , which is the new angular speed.
b. Is kinetic energy conserved in the process? If not, from where does the change come about?
Ans: Kinetic energy is not conserved in the mentioned process. With the decrease in the moment of inertia, there is an increase in kinetic energy. The additional kinetic energy is generated from the work done by the man to fold his hands toward himself.

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24. A bullet of mass 10g 10g and speed 500m/s 500m/s is fired into a door and gets embedded exactly at the centre of the door. The door is 1.0m 1.0m wide and weighs 12kg 12kg . It is hinged at one end and rotates about a vertical axis practically without friction. Find the angular speed of the door just after
the bullet embeds into it. (Hint: The moment of inertia of the door about the vertical axis at one end is ML23 ML23 .)

Ans: Given that,
Mass of the bullet is given as, m=10g=10×10−3kg m=10g=10×10−3kg
Velocity of the bullet is given as, v=500m/s v=500m/s
Width of the door, L=1.0m L=1.0m
Radius of the door, r=12m r=12m
Mass of the door is given, M=12kg M=12kg
Angular momentum transmitted by the bullet on the door:
α=mvr α=mvr
⇒α=(100×10−3)×(500)×12=2.5kgm2s−1 ⇒α=(100×10−3)×(500)×12=2.5kgm2s−1
Moment of inertia of the door can be given as:
I=ML23
⇒I=13×12×(1)2=4kgm2
But we have the relation,
α=Iω
⇒ω=αI=2.54=0.625rads−1 which is the required angular speed.

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25. Two discs of moments of inertia I1 I1 and I2 I2 about their respective axes (normal to the disc and passing through the centre), and rotating with angular speeds ω1 ω1 and ω2 ω2 are brought into contact face to face with their axes of rotation coincident.
a. What is the angular speed of the two-disc system?

Ans: Given that,
Moment of inertia of disc 1 is I1
Angular speed of disc 1 is ω1
Moment of inertia of disc 2 is I2
Angular speed of disc 2 is ω2
Angular momentum of disc 1 is L1=I1ω1
Angular momentum of disc 2 is L2=I2ω2
Total initial angular momentum is Li=I1ω1+I2ω2
When the two discs are joined together, their moments of inertia get summed up.
Moment of inertia of the system of two discs can be given as,
I=I1+I2
Let ω ω be the angular speed of the system.
Total final angular momentum is given as, Lf=(I1+I2)ω
Let us use the law of conservation of angular momentum,
Li=Lf ⇒I1ω1+I2ω2=(I1+I2)ω
⇒ω=I1ω1+I2ω2I1+I2
, which is the required angular speed.
b.Show that the kinetic energy of the combined system is less than the sum of the initial kinetic energies of the two discs. How do you account for this loss in energy? Take ω1≠ω2
Ans: Kinetic energy of disc 1 is given as, E1=12I1ω21
Kinetic energy of disc 2, E2=12I2ω22
Total initial kinetic energy can be given as, Ei=12(I1ω21+I2ω22)
When the discs are joined, their moments of inertia get summed up.
Moment of inertia of the system can be given as, I=I1+I2
Angular speed of the system can be given as: ω
Final kinetic energy Ef=12(I1+I2)ω2
⇒Ef=12(I1+I2)(I1ω1+I2ω2I1+I2)2=12×(I1ω1+I2ω2)2I1+I2
And Ei=Ef
⇒Ei=12(I1ω21+I2ω22)−(I1ω1+I2ω2)22(I1+I2)
⇒Ei=12I1ω21+12I2ω22−12I21ω21(I1+I2)−12I22ω22(I1+I2)−122I1I2ω1ω2(I1+I2)
⇒Ei=1(I1+I2)[12I2ω21+12I1I2ω21+12I1I2ω22+12I2ω22−12I1ω21−12I2ω22−I1I2ω1ω1]
⇒Ei=I1I22(I1+I2)(ω12+ω22−2ω1ω2)
⇒Ei=I1I2(ω1−ω2)22(I1+I2)
All the quantities on right hand side are positive
Ei−Ef>0
Ei>Ef
The loss of K.E. can be attributed to the frictional force that comes into play when the two discs come in contact with each other.

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26.Prove the theorem of perpendicular axes. (Hint: Square of the distance of a point (x,y) in the x–y plane from an axis through the origin perpendicular to the plane is x2+y2

Ans: It is stated by the theorem of perpendicular axes states that the moment of inertia of a planar body (lamina) about an axis perpendicular to its plane is equal to the sum of its moments of inertia about two perpendicular axes concurrent with perpendicular axis and lying in the plane of the body.
A physical body with centre O
and a point mass m
, in the x–y plane at (x, y) is shown in the following figure.
Moment of inertia about x-axis can be given as, Ix=mx2
Moment of inertia about y-axis can be given as, Iy=my2
Moment of inertia about z-axis can be given as, Iz=m(x2+y2
Ix+Iy=mx2+my2=m(x2+y2)
⇒Ix+Iy=m(x2+y2−−−−−−√)2
⇒Ix+Iy=Iz
Therefore, the theorem is proved.

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27. Prove the result that the velocity v
of translation of a rolling body (like a ring, disc, cylinder or sphere) at the bottom of an inclined plane of a height h
is given by v2=2gh(1+k2R2)
using dynamical consideration (i.e. by consideration of forces and torques). Note k
is the radius of gyration of the body about its symmetry axis, and R
is the radius of the body. The body starts from rest at the top of the plane.

Ans: A body rolling on an inclined plane of height h, is depicted in the given figure: m=Mass of the body
R=Radius of the body
K=Radius of gyration of the body
v=Translational velocity of the body
h=Height of the inclined plane
g=Acceleration due to gravity
Total energy at the top of the plane is given as, E1=mgh
Total energy at the bottom of the plane can be given as,
Eb=KErot+KEtrans
⇒Eb=12Iω2+12mv2
But I=mk2 I=mk2 and ω=vR ω=vR
⇒Eb=12mk2(v2R2)+12mv2
⇒Eb=12mv2k2R2+12mv2
⇒Eb=12mv2(1+k2R2)
From the law of conservation of energy, we can write:
ET=EB
⇒mgh=12mv2(1+k2R2)
⇒v=2gh(1+k2R2)

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28. A disc rotating about its axis with angular speed ω0 ω0 is placed lightly (without any translational push) on a perfectly frictionless table. The radius of the disc is R R . What are the linear velocities of the points A A , B B and C C on the disc shown in figure? Will the disc roll in the direction indicated?

Ans: From the question we can infer that:
vA=Rω0
vC=(R2)ω0
The rolling of the disc will not take place.
Angular speed of the disc is given =ω0
Radius of the disc is given =R
Let us use the relation for linear velocity, v=ω0R
For point A we can write: vA=Rω0
in the direction tangential to the right
For point B we can write: vB=Rω0
in the direction tangential to the left
For point C it can be written as: vC=(R2)ω0
in the direction same as that of vA The directions of motion of points
on the disc are depicted in the following figure
Because the disc is placed on a frictionless table, the disc will not roll. This is due to the presence of friction is essential for the rolling of a body.

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29. Explain why friction is necessary to make the disc in figure given roll in the direction indicated.
a.Give the direction of frictional force at B, and the sense of frictional torque, before perfect rolling begins.

Ans: To roll the given disc, some torque is necessary. As per the definition of torque, the rotating force must be tangential to the disc. Since the frictional force at point B
is along the tangential force at point A
, a frictional force is necessary for making the disc roll.
Force of friction will act in the opposite direction to the direction of velocity at point B . The direction of linear velocity at point B
can be pointed tangentially leftward. Therefore, frictional force will act tangentially rightward. The frictional torque before the start of perfect rolling is perpendicular to the plane of the disc in the outward direction. b.What is the force of friction after perfect rolling begins?
Ans: Since frictional force will act opposite to the direction of velocity at point B , perfect rolling will start when the velocity at that point becomes equal to zero. This will make the frictional force that acts on the disc as zero.

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30. A solid disc and a ring, both of radius 10cm 10cm are placed on a horizontal table simultaneously, with initial angular speed equal to 10πrads−1 10πrads−1 . Which of the two will start to roll earlier? The co-efficient of kinetic friction is

Ans: Given that,
Radii of the ring and the disc are given as, r = 10 cm = 0.1 m
Initial angular speed is given, u=0
Coefficient of kinetic friction is, μk=0.2
Initial velocity of both the objects, u=0
Motion of the two objects is a result of frictional force. As per Newton’s second law of motion,
we have frictional force, f=ma
μkmg=ma
Where,
a=Acceleration produced in the objects
m=Mass
a=μkg
From the first equation of motion, the final velocity of the objects can be obtained as:
v=u+at
⇒v=0+μkgt
v=μkgt …... (2)
The torque applied by the frictional force will act in perpendicularly outward direction and cause reduction in the initial angular speed.
Torque, τ=−Iα
Angular acceleration
μkmgr=−Iα
⇒α=−μkmgrI
…… (3) Let us use the first equation of rotational motion to obtain the final angular speed:
ω=ω0+αt
⇒ω=ω0+−μkmgrIt …… (4)
Rolling starts when linear velocity, v=rω
v=r(ω0−μkmgrtI) …… (5)
Equating equations (2) and (5), we can write: ⇒μkgt=r(ω0−μkmgrtI) …… (6)
For the ring: I=mr2
μkgt=r(ω0−μkmgrtmr2)
μkgt=rω0−μkgt
⇒2μkgt=rω0
⇒tr=rω02μkg
⇒tr=0.1×10×3.142×0.2×9.8=0.80s
For the disc:
I=12mr2
μkgtd=rω0−μkgmr2td12mr2
⇒μkgtd=rω0−μkgtd12
⇒μkgtd=rω0−2μkgtd
⇒3μkgtd=rω0
td=rω03μkg
⇒td=0.1×10×3.143×0.2×9.8=0.53s
Since, td>tr the disc will start rolling before the ring.

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31. A cylinder of mass 10kg 10kg and radius 15cm 15cm is rolling perfectly on a plane of inclination 30∘ 30∘ . The coefficient of static friction μk=0.25 μk=0.25
a.How much is the force of friction acting on the cylinder?

Ans: Given that,
Mass of the cylinder is given as, m=10kg m=10kg
Radius of the cylinder is given as, r=15cm=0.15m r=15cm=0.15m
Co-efficient of kinetic friction μk=0.25 μk=0.25
Angle of inclination is given as, θ=30∘ θ=30∘
Moment of inertia of a solid cylinder about its geometric axis is, I=12mr2
The various forces acting on the cylinder are depicted in the given figure:
The acceleration of the cylinder is given as:
a=mgsinθm+Ir2
⇒a=mgsinθm+mr22r2=23gsin30∘
⇒a=23×9.8×0.5=3.27m/s2
Let us use Newton’s second law of motion, we can express net force as:
fnet=ma
mgsin30∘−f=ma
⇒f=mgsin30∘−ma
⇒f=10×9.8×0.5−10×3.27
⇒f=49−32.7=16.3N
, which is the frictional force.
What is the work done against friction during rolling?
Ans: During rolling, the instantaneous point of contact with the plane will come to rest. Therefore, the work done against frictional force will be zero.
If the inclination of the plane is increased, at what value of angle does the cylinder begin to skid, and not roll perfectly?
Ans: For rolling without skidding, we have the formula:
μ=13tanθ
⇒tanθ=3μ=3×0.25=0.75
⇒tan−10.75=36.87∘
, which is the required value of angle.

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32. Read each statement below carefully, and state, with reasons, if it is true or false;
a.During rolling, the force of friction acts in the same direction as the direction of motion of the CM of the body.

Ans: Given that,
Mass of the cylinder is given as, m=10kg
Radius of the cylinder is given as, r=15cm=0.15m
Co-efficient of kinetic friction μk=0.25
Angle of inclination is given as, θ=30∘
Moment of inertia of a solid cylinder about its geometric axis is, I=12mr2
The various forces acting on the cylinder are depicted in the given figure:
The acceleration of the cylinder is given as:
a=mgsinθm+Ir2
⇒a=mgsinθm+mr22r2=23gsin30∘
⇒a=23×9.8×0.5=3.27m/s2
Let us use Newton’s second law of motion, we can express net force as:
fnet=ma
mgsin30∘−f=ma
⇒f=mgsin30∘−ma
⇒f=10×9.8×0.5−10×3.27
⇒f=49−32.7=16.3N
, which is the frictional force.
What is the work done against friction during rolling?
Ans: During rolling, the instantaneous point of contact with the plane will come to rest. Therefore, the work done against frictional force will be zero.
If the inclination of the plane is increased, at what value of angle does the cylinder begin to skid, and not roll perfectly?
Ans: For rolling without skidding, we have the formula:
μ=13tanθ
⇒tan⁡θ=3μ=3×0.25=0.75
⇒tan−10.75=36.87∘ , which is the required value of angle.

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33. Separation of Motion of a system of particles into motion of the centre of mass and motion about the centre of mass:
A.Show pi=pi′+miV pi=pi′+miV Where pi pi is the momentum of the ith ith particle (of massmi mi ) and pi′=mivi′ pi′=mivi′ . Note vi′ vi′ is the velocity of the ith ith particle relative to the centre of mass. Also, prove using the definition of the centre of mass ∑ipi′=0 ∑ipi′=0 .

Ans: Let us take a system of imoving particles.
Mass of the ith particle =mi
Velocity of the ith particle =vi
Therefore, momentum of the ith particle, pi=mivi
Velocity of the centre of mass is V
The velocity of the ith
particle with respect to the centre of mass of the system is given as: vi′=vi−V …… (1)
Multiplying mi throughout equation (1)
we can write:
mivi′=mivi−miV
pi′=pi−miV
Where,
pi′=mivi′
is the Momentum of the ith
particle with respect to the centre of mass of the system.
Hence, pi=pi′+miV
We have the formula: pi′=mivi′
Taking the summation of momentum of all the particles with respect to the centre of mass of the system, we can write:
∑ipr′=∑imivi′=∑imidri′dt
Where
is the position vector of ith particle with respect to the centre of mass vi′=dri′dt
As per the definition of the centre of mass, we have: ∑imivi′=0 ⇒∑imidri′dt=0 ∑ipi′=0
Hence proved.
b.Show dL′dt=∑iri′×ddt(Pi′) dL′dt=∑iri′×ddt(Pi′) . Further, show that dL′dt=τ′ext dL′dt=τ′ext where τ′ext τ′ext is the sum of all external torques acting on the system about the centre of mass. (Hint: Use the definition of centre of mass and Newton’s Third Law. Assume the internal forces between any two particles act along the line joining the particles.)
Ans: We have the following relation:
L′=∑iri′×pi′
dL′dt=ddt(∑iri′×pi′)
⇒dL′dt=ddt(∑iri′)×pi′+∑iri′×ddt(pi′)
⇒dL′dt=ddt(∑imiri′)×vi′+∑iri′×ddt(pi′)
Where,
ri′ is the position vector with respect to the centre of mass of system of particles.
∑imiri′=0
dL′dt=∑iri′×ddt(pi′)
We have the following relation:
⇒dL′dt=∑iri′×middt(vi′)
Where,
ddt(vi′)
is the rate of change of velocity of the ith
particle with respect to the centre of mass of the system.
Therefore, according to Newton’s third law of motion, we can express: middt(vi′)
is the external force acting on the ith particle.
∑i(τi′)ext
i.e., ∑iri′×middt(vi′)=τ′ext
∑iri′×middt(vi′)=τ′ext
is the external torque acting on the system as a whole.
Hence proved that, dL′dt=τ′ext

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Last Updated on: Mar 28, 2024