System of Particles and Rotational Motion Test

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System of Particles and Rotational Motion Test - 49

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Question 1
1 / -0

The position of axis of rotation of a body is changed so that its moment of inerted decreases by 36%. The % change in the radius of gyration is

A
decreases by 18%

B
increases by 18%

C
decreases by 20%

D
increases by 20%

Solution

$$K = \sqrt {\frac{I}{m}} $$
MOI is decreases by $$36\% $$
so new MOI $$=0.641$$
$${K^1} = \sqrt {\frac{{0.641}}{m}} = 0.8K$$
$$\% \,change = \frac{{K - {K^1}}}{K} \times 100$$
$$ = 20\% $$
decreases by $$20\% $$
Hence,
option $$(C)$$ is correct answer.
Question 2
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Four rings each of mass $$M$$ and radius $$R$$ are arranged as shown in the figure. The moment of inertia of the system about the axis $$yy'$$ is

A
$$2 M R ^ { 2 }$$

B
$$3 M R ^ { 2 }$$

C
$$4 M R ^ { 2 }$$

D
$$5 M R ^ { 2 }$$

Solution

We have,

$$I_4=I_3=\dfrac{MR^2}{2}$$

$$I_1=I_2=\dfrac{MR^2}{2}+MR^2$$

$$I_{system}=I_1+I_2+I_3+I_4$$

$$=2\times \dfrac 32 MR^2+2\times \dfrac{MR^2}{2}=4MR^2$$
Question 3
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Three identical thin rods each of length l and mass M are joined together to from a letter H. What is the moment of intertia of the system about one of the sides of H

A
$$\dfrac{Ml^{2}}{3}$$

B
$$\dfrac{Ml^{2}}{4}$$

C
$$\dfrac{2Ml^{2}}{3}$$

D
$$\dfrac{4Ml^{2}}{3}$$

Solution

The three rods form H, So let $$ H = I_1(AB) + I_2(EF) + I_3(CD)$$
Here we will find about left vertical axis i.e. $$I_1$$ AB

Moment of inertia of rod AB about the axis is $$0$$

Moment of inertia of rod CD(parallel axis theorem) about the axis is $$M{{L}^{2}}$$

Moment of inertia of rod EF (perpendicualr axis theorem) about the axis is $$\dfrac{1}{3}M{{L}^{2}}$$

So, Moment of inertia of the rod about the axis is

$$ =0+M{{L}^{2}}+\dfrac{1}{3}M{{L}^{2}} $$

$$ =\dfrac{4}{3}M{{L}^{2}} $$
Question 4
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A particle of mass M kg describes a circle of radius 1 m. The centripetal acceleration of the particle is $$ 4m/s^2 $$. What will be the momentum of the particle ?

A
4 M

B
2 M

C
8 M

D
M

Solution

Centripetal acc. =$$ v^2/r$$
4 =$$ v^2$$ => |v|=2
Now, momemntum =$$ Mv$$= 2M
Question 5
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A circular disc of M.I. 10 $$kg-m^2$$ rotates about its own axis at a constant speed of 60 r.p.m. under the action of an electric motor of power 31.4 W. If the motor is switched off, how many rotations will it cover before coming to rest?

A
$$3.14$$

B
$$31.4$$

C
$$314$$

D
$$6.28$$

Solution

The moment of inetia of the disc is $$10\ kg-m^2$$
The speedof the disc is $$60\ rpm$$ i.e. $$60\times\dfrac{2\pi}{60}\ rad/s$$
The power supplied by the motor is $$31.4\ W$$

The relation between the power and the torque developed is:
$$P=\tau\omega$$

$$\tau=\dfrac{P}{\omega}=\dfrac{31.4}{2\pi}$$
$$\tau=5\ Nm$$

The angular acceleration of motor due to torque is :
$$\alpha=\dfrac{\tau}{I}$$

$$\alpha=\dfrac{5}{10}=0.5\ rad/s^2$$

Now, when the motor is switched off, the angular displacement of the disc to come to rest is given as:
$$\omega^2_f=\omega^2_0+2\alpha\theta$$

$$0-(2\pi)^2=2\times0.5\times\theta$$
$$\theta=4\pi^2$$

In one rotation, displacement is $$2\pi$$. SO, total number of rotations is:
$$n=\dfrac{\theta}{2\pi}=\dfrac{4\pi^2}{2\pi}$$

$$n=2\pi=6.28\ rotations$$
Question 6
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Two particles of equal mass 'm' go around a circle of radius R under the action of their mutual gravitation attraction. The speed of each particle with respect to their centre of mass is :

A
$$\sqrt { \dfrac { Gm }{ 4R } } $$

B
$$\sqrt { \dfrac { Gm }{ 3R } } $$

C
$$\sqrt { \dfrac { Gm }{ 2R } } $$

D
$$\sqrt { \dfrac { Gm }{ R } } $$

Solution

Lets consider $$m=$$ mass of particle

$$R=$$ radius of a circle

Centripetal force is equal to the gravitational force,

$$\dfrac{mv^2}{R}=\dfrac{Gm.m}{(2R)^2}$$

$$\dfrac{v^2}{R}=\dfrac{Gm}{4R^2}$$

$$v^2=\dfrac{Gm}{4R}$$

$$v=\sqrt{\dfrac{Gm}{4R}}$$

The correct option is A.
Question 7
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A football roots through the ground. The path followed by centre of mass of football is

A
Linear

B
Circular

C
Rotational

D
All the above

Solution
Question 8
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Two bodies of masses $$2kg$$ and $$4kg$$ are moving with velocities $$2m/s$$ and $$10m/s$$ respectively towards each other due to mutual gravitational attraction. What is the velocity of their centre of mass (Bodies are at rest initially) :

A
$$5.3\ m/s$$

B
$$6.4\ m/s$$

C
$$zero$$

D
$$8.1\ m/s$$

Solution
Question 9
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A thin circular ring first slip down a smooth incline then rolls down a rough incline of identical geometry from the same height. The ratio of time taken in the two motion is :

A
$$\dfrac{1}{2}$$

B
$$1$$

C
$$\dfrac{1}{{\sqrt 2 }}$$

D
$$\dfrac{1}{4}$$

Solution

case (i) Transitional motion
acceleration=$$g\sin \theta $$
case (ii) acceleration =$$\dfrac{{g\sin \theta }}{{1 + \dfrac{1}{{M{R^2}}}}}$$
$$\begin{array}{l} I=M{ R^{ 2 } } \\ a=\dfrac { { g\sin \theta } }{ 2 } \end{array}$$
Now using $$\rho = ut + \frac{1}{2}a{t^2}$$
$${a_1}{t_1}^2 = {a_2}{t_2}^2\,\,\,\,\,\theta \,\& \rho \,\,\,same\,for\,both\,\,are\& \,\,v = 0$$
$$\dfrac{{{t_1}^2}}{{{t_2}^2}} = \dfrac{{{a_2}}}{{{a_1}}} = \dfrac{{g\sin \theta }}{{2g\sin \theta }}$$
$$\dfrac{{{t_1}}}{{{t_2}}} = \dfrac{1}{{\sqrt 2 }}$$
$$\therefore \dfrac{{{t_1}}}{{{t_2}}} = \dfrac{1}{{\sqrt 2 }}$$
Option $$C$$ is correct answer.
Question 10
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Two spherical bodies of mass $$M$$ and $$5M$$ and radii $$R$$ and $$2R$$ respectively are released in free space with initial separation between their centers equal to $$12R$$. If they attract each other due to gravitational force only, then the distance covered by the smaller body just before the collision is

A
$$4.5\ R$$

B
$$7.5\ R$$

C
$$1.5\ R$$

D
$$2.5\ R$$

Solution

We know that the center of mass will remain constant, taking origin at the center of sphere of mass $$M$$ and $$x-axis$$ to the left towards the sphere $$5M$$
$$x=\dfrac{0\times M+5M\times 12R}{M+5M}$$
$$x=10R$$
After collision
$$\dfrac{mx+5M(x+3R)}{6m}=10R$$
$$6x+15R=60R$$
$$6x=45R$$
$$x=7.5R$$