System of Particles and Rotational Motion Test

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System of Particles and Rotational Motion Test - 30

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following is not correct about centre of mass ?

A
It depends on the choice of frame of reference

B
In centre of mass frame, momentum of a system is always zero

C
Internal forces may affect the motion of centre of mass

D
Centre of mass and centre of gravity coincide in uniform gravitational field

Solution

The internal forces all balance each other so that there is no change in the mass distribution of the body, thus center of mass remains unchanged. Rest all options are correct.
Question 2
1 / -0

Four identical rods each of mass $$M$$ are joined to form as square frame. The moment of inertia of the system about an axis passing through the point of intersection of diagonals and perpendicular to the plane of the square is

A
$$\displaystyle\ \frac{4Ml^{2}}{3}$$

B
$$\displaystyle\ \frac{13Ml^{2}}{3}$$

C
$$\displaystyle\ \frac{Ml^{2}}{6}$$

D
$$\displaystyle\ \frac{13Ml^{2}}{6}$$

Solution

Moment of Inertia of a rod about it Center of Mass= $$\dfrac { M{ L }^{ 2 } }{ 12 } $$
Moment of Inertia of a rod about any other point at a distance D from its COM= $$\dfrac { M{ L }^{ 2 } }{ 12 } +M{ D }^{ 2 }$$
The distance of rod COM from point of intersection of diagonals is $$D=L/2$$
$$\therefore $$ Moment of Inertia of a rod about an axis passing through the point of intersection of diagonals and perpendicular to the plane of the square is ($$D=L/2$$)
$${ I }_{ 1 }=\dfrac { M{ L }^{ 2 } }{ 12 } +MD^{ 2 }=\dfrac { M{ L }^{ 2 } }{ 12 } +\dfrac { M{ L }^{ 2 } }{ 4 } =\dfrac { M{ L }^{ 2 } }{ 3 } $$

As the distance of each rod COM from point of intersection, mass & length of each rod is constant.
$$\therefore $$ Moment of Inertia of 4 rod about an axis passing through the point of intersection of diagonals and perpendicular to the plane of the square is
$$4{ I }_{ 1 }=\dfrac { 4M{ L }^{ 2 } }{ 3 } $$
Question 3
1 / -0

A man stands at one end of the open truck which can run on frictionless horizontal rails. Initially, the man and the truck are at rest. Man now walks to the other end and stops. Then which of the following is true?

A
The truck moves opposite to direction of motion of the man even after the man ceases to walk.

B
The centre of mass of the man and the truck remains at the same point throughout the man's walk.

C
The kinetic energy of the man and the truck are exactly equal throughout the man's walk.

D
The truck does not move at all during the man's walk.

Solution

since there is no external force on the system the COM of man-truck system will not change.
hence the truck will move in opposite direction until the man stops.
it will also stop if man stops.
also momentum will be conserved but no kinetic energy as the man is doing work.
Question 4
1 / -0

For a rigid body made up of N particles each of mass m, the radius of gyration, about a given axis of rotation, equals to the

A
Distance of any constituent particles from the axis of rotation

B
Mean of the distance of the constituent particles from axis of rotation

C
Root mean square value of the distance of the constituent particle from axis if rotation

D
Harmonic mean of the distance of the constituent particles from the axis of rotation

Solution

As $$k=\sqrt{\displaystyle\frac{I}{m}}$$. The moment of inertia of each N particle is given as $$I=m(r_1^2+r_2^2+r_3^2+....+r_N^2)$$. Thus $$k=\sqrt{\displaystyle\frac{I}{m}}=\sqrt{r_1^2+r_2^2+r_3^2+....+r_N^2}$$ which is the root mean square value of the distance of the constituent particle from axis if rotation.
Question 5
1 / -0

Directions For Questions

Two persons, A of mass 60 kg and B of mass 40 kg, are standing of a horizontal platform of mass 50 kg. The platform is supported on wheels on a horizontal frictionless surface and is initially at rest. Consider the following situations.
(i) Both A and B just from the platform simultaneously and in the same horizontal direction.
(ii) A jumps first in a horizontal direction and after a few seconds B also jumps in the same direction.
In both the situations above, just after the jump, the person (A or B) moves away from the platform with a speed of 3 m/s relative to the platform and along the horizontal.
Answer these questions.

...view full instructions

In situation (i), just after both A and B jump from the platform, velocity of centre of mass of the system (A, B and the platform) is :

A
2 m/s

B
6 m/s

C
5 m/s

D
none of these

Solution

Since in situation (i) both of them jump simultaneously, there is no external force acting on the system, so velocity of COM will be zero.
Question 6
1 / -0

Two particles A and B initially at rest, move towards each other by mutual force of attraction. At the instant when the speed of A is n and the speed of B is 3n, the speed of the centre of mass of the system is :

A
3n

B
2n

C
1.5n

D
0

Solution

$$\vec{F}_{external} = M \vec{a_{CM}}$$
Mutual force of attraction is an internal force i.e $$\vec{F}_{external} = 0$$
$$\implies \vec{a_{CM}} = 0$$
$$\therefore$$ $$ \vec{V_{CM}}_f = \vec{V_{CM}}_i$$
But initially, the system is at rest i.e $$\vec{V_{CM}}_i = 0$$
$$\implies \vec{V_{CM}}_f = 0$$
Hence the speed of centre of mass is zero at any instant.
Question 7
1 / -0

A particle having mass $$0.5\ kg$$ is projected under gravity with a speed of $$98\ m/sec$$ at an angle of $$60^{\circ} $$.The magnitude of the change in momentum $$($$in N sec$$)$$ of particle after $$10\ seconds$$ is :

A
$$0.5$$

B
$$49$$

C
$$98$$

D
$$490$$

Solution

There is no change in horizontal velocity , hence no change in momentum in horizontal direction.The vertical velocity at $$t=10\ sec$$ is
$$v=98\times \sin 60^{\circ}-(9.8)\times 10=113.13\ m/sec $$
So change in momentum in vertical direction is
$$(0.5\times 98\times \dfrac{\sqrt{3}}{2})-[(0.5\times13.13)] $$

$$=42.434+6.56=48.997=49 $$
Question 8
1 / -0

A uniform disc of mass $$M$$ and radius $$R$$ is mounted on an axle supported in frictionless bearings. A light cord is wrapped around the rim of the disc and a steady downward pull $$T$$ is exerted on the cord, the tangential acceleration of a point on the rim is

A
$$\displaystyle\ \frac{T}{M}$$

B
$$\displaystyle\ \frac{M}{T}$$

C
$$\displaystyle\ \frac{2T}{M}$$

D
$$\displaystyle\ \frac{M}{2T}$$

Solution

$$TR=I\alpha \\ where,\quad I=\dfrac { M{ r }^{ 2 } }{ 2 } \\ \alpha =\dfrac { TR }{ \dfrac { M{ R }^{ 2 } }{ 2 } } =\dfrac { 2T }{ MR } $$
$$a=\dfrac { 2T }{ MR } R=\dfrac { 2T }{ M } $$
Question 9
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Two bodies of different masses of $$2 kg$$ and $$4 kg$$ are moving with velocities with $$2 {m}/{s}$$ and $$10 {m}/{s}$$ towards each other due to mutual gravitational attraction. What is the velocity of their centre of mass?

A
$$5 {m}/{s}$$

B
$$6 {m}/{s}$$

C
$$8 {m}/{s}$$

D
zero

Solution

$$\vec{V}_{CM} = \dfrac{2 (\vec{v} ) + 4 (\vec{u})}{M}$$
$$\vec{V}_{CM} = \dfrac{-2 (v ) +4 (u)}{M}$$
$$\vec{V}_{CM} = \dfrac{-2 (2 ) + 4 (10)}{2+4} = 6 m/s$$
Question 10
1 / -0

For the same total mass, which of the following will have the largest moment of inertia about an axis passing through the centre of gravity and perpendicular to the plane of body?

A
a ring of radius $$l$$

B
a disc of radius $$l$$

C
a square lamina of side $$2l$$

D
four rods forming square of side $$2l$$

Solution

Moment of inertia of ring of radius, $$l= ml^{2}$$
Moment of inertia of disc of radius, $$l=\dfrac{ml^{2}}{2}$$
Moment of inertia of square lamina of side length, $$2l= \dfrac{2ml^{2}}{3}$$
Moment of inertia of square foremed by four rods of length, $$2l= \dfrac{4ml^{2}}{3}$$.
Therefore, moment of inertia of four rods forming square of length $$2l$$ is larger.

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System of Particles and Rotational Motion Test - 30

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System of Particles and Rotational Motion Test - 30

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SHARING IS CARING

Question 1

It depends on the choice of frame of reference

In centre of mass frame, momentum of a system is always zero

Internal forces may affect the motion of centre of mass

Centre of mass and centre of gravity coincide in uniform gravitational field

Solution

Question 2

$$\displaystyle\ \frac{4Ml^{2}}{3}$$

$$\displaystyle\ \frac{13Ml^{2}}{3}$$

$$\displaystyle\ \frac{Ml^{2}}{6}$$

$$\displaystyle\ \frac{13Ml^{2}}{6}$$

Solution

Question 3

The truck moves opposite to direction of motion of the man even after the man ceases to walk.

The centre of mass of the man and the truck remains at the same point throughout the man's walk.

The kinetic energy of the man and the truck are exactly equal throughout the man's walk.

The truck does not move at all during the man's walk.

Solution

Question 4

Distance of any constituent particles from the axis of rotation

Mean of the distance of the constituent particles from axis of rotation

Root mean square value of the distance of the constituent particle from axis if rotation

Harmonic mean of the distance of the constituent particles from the axis of rotation

Solution

Question 5

2 m/s

6 m/s

5 m/s

none of these

Solution

Question 6

3n

2n

1.5n

0

Solution

Question 7

$$0.5$$

$$49$$

$$98$$

$$490$$

Solution

Question 8

$$\displaystyle\ \frac{T}{M}$$

$$\displaystyle\ \frac{M}{T}$$

$$\displaystyle\ \frac{2T}{M}$$

$$\displaystyle\ \frac{M}{2T}$$

Solution

Question 9

$$5 {m}/{s}$$

$$6 {m}/{s}$$

$$8 {m}/{s}$$

zero

Solution

Question 10

a ring of radius $$l$$

a disc of radius $$l$$

a square lamina of side $$2l$$

four rods forming square of side $$2l$$

Solution

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