System of Particles and Rotational Motion Test

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System of Particles and Rotational Motion Test - 66

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Question 1
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A uniform disc of mass $$m$$, radius $$r$$ and a point mass $$m$$ are arranged as shown in the figure. The acceleration of point mass is : (Assume there is no slipping between pulley and thread and the disc can rotate smoothly about a fixed horizontal axis passing through its centre and perpendicular to its plane).

A
$$\dfrac {g}{2}$$

B
$$\dfrac {g}{3}$$

C
$$\dfrac {2g}{3}$$

D
None of these

Solution

F.B.D of point mass we get,
$$mg-T=ma$$ , where $$T$$ is the tension in thread and $$a$$ is the acceleration of thread and point mass.
Now, Applying Torque from the center of the disc
Therefore, we get
$$T*r=I*\alpha$$, where $$r$$ is the radius of disc, $$I$$ is the moment inertia of Disc and $$\alpha$$ is the angular accleration of the disc.
As there is no slipping between the thread and the pulley therefor, $$a=r\alpha$$
We know, $$I=mr^{2}/2$$ for a disc.
Now by substituting the values of $$I$$ in equation we get,
$$T*r=mr^{2}/2*\alpha$$
Evaluating the equation we get, $$T=ma/2$$ also using the equation $$a=r\alpha$$.
Now substituting $$T=ma/2$$ in equation $$mg-T=ma$$ we get,
$$a=2g/3$$
Question 2
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A uniform thin rod is bent in the form of closed loop $$ABCDEFA$$ as shown in the figure. The ratio of moment of inertia of the loop about x-axis to that about y-axis is

A
$$> 1$$

B
$$< 1$$

C
$$=1$$

D
$$=\cfrac{1}{2}$$

Solution
Question 3
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A square frame ABCD is formed by four identical rods each of mass 'm' and length 'T'. This frame is in X - Y plane such that side AB coincides with X - axis and side AD along Y - axis. The moment of inertia of the frame about X - axis is

A
$$\dfrac{5ml^2}{3}$$

B
$$\dfrac{2ml^2}{3}$$

C
$$\dfrac{4ml^2}{3}$$

D
$$\dfrac{ml^2}{12}$$

Solution

Moment of inertia of rod AD and BC are each $$\dfrac{1}{3}mT^2$$

For rod AB, moment of inertia is zero since it is on X-axis and hence r=0.

For, rod CD, every element is at distance $$T$$ from X-axis and hence its moment of inertia is $$mT^2$$.

Hence, $$I=2\times \dfrac{1}{3}mT^2+mT^2$$

$$\implies I=\dfrac{5}{3}mT^2$$
Question 4
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Two thin discs each of mass $$M$$ and radius $$R$$ are placed at either end of a rod of mass $$m$$, length $$l$$ and radius $$r$$. Moment of inertia of the system about an axis passing through the centre of rod and perpendicular to its length is

A
$$\dfrac {mL^{2}}{12} + \dfrac {1}{4}MR^{2} + \dfrac {1}{4}ML^{2}$$

B
$$\dfrac {ML^{2}}{12} + \dfrac {1}{2}mR^{2} + \dfrac {1}{2}mL^{2}$$

C
$$\dfrac {1}{2}mL^{2} + \dfrac {mR^{2}}{2} + \dfrac {ML^{2}}{12}$$

D
$$\dfrac {mL^{2}}{12} + MR^{2} + \dfrac {1}{2} ML^{2}$$

Solution
Question 5
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In the figure shown, the plank is being pulled to the right with a constant speed $$v$$. If the cylinder does not slip then

A
The speed of the centre of mass of the cylinder is $$2v$$

B
The speed of the centre of mass of the cylinder is zero

C
The angular velocity of the cylinder is $$v/R$$

D
The angular velocity of the cylinder is zero

Solution

$$V=u+\omega R$$ ...(1)
For pure rolley $$u=\omega R$$ then (1) becomes $$2V=u+u$$
therefore $$u=v$$ =velocity of center of mass w.r.t plank
Now velocity of center of mass of cylinder w.r.t ground,
$$u-$$(speed pf plank)
=$$u-(-v)$$
=$$v-(-v)$$
=$$2v$$
Question 6
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Moment of inertia of a straight wire about an axis perpendicular to the wire and passing through one of its end is $$I$$. This wire is now framed into a circle (a ring) of single turn. The moment of inertia of this ring about an axis passing through centre and perpendicular to its plane would be:

A
$$\left( \cfrac { 3 }{ { \pi }^{ 2 } } \right) I$$

B
$$\left( \cfrac { 3 }{ { 4\pi }^{ 2 } } \right) I$$

C
$$\left( \cfrac { { \pi }^{ 2 } }{ 3 } \right) I$$

D
$$\left( \cfrac { {4 \pi }^{ 2 } }{ 3 } \right) I$$

Solution

For a wire of length l and mass m its moment of inertia about an axis passing through its one end and perpendicular to its length is $$ml^2/3$$ so $$I=ml^2/3$$ or $$ml^2 = 3I$$

Now when this wire is changed in a ring then the circumference of the circle is $$2\pi r =length = l$$ so radius $$r=l/2\pi $$

Now Moment of Inertia of a ring of radius r and mass m about an axis passing through its centre and perpendicular to its plane is $$mr^2$$

putting value of r in terms of l we get new M.O.I as $$(m)l^2/4\pi^2 = ml^2/4\pi^2$$

or $$3I/4\pi^2$$
Option B is correct.
Question 7
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A particle of mass $$m$$ is subjected to an attractive central force of magnitude $$k/r^2$$, $$k$$ being a constant. If at the instant when the particle is at an extreme position in its closed orbit, at a distance $$a$$ from the centre of force, its speed is $$(k/2ma)$$, if the distance of other extreme position is $$b$$. Find $$a/b.$$

A
4

B
3

C
5

D
6

Solution

The attractive force in orbit,

$$k_r=-\dfrac{k}{r^2}$$. . . . .(1)

Potential energy,

$$U=-\dfrac{k}{r}$$. . . . . .(2)

By the conservation of at energy at extreme position $$a$$ and $$b$$

$$K_a+U_a=K_b+U_b$$

$$\dfrac{1}{2}mv_1^2-\dfrac{k}{a}=\dfrac{1}{2}mv_2^2-\dfrac{k}{b}$$. . . . .(3)

At the instant when the particle is at an extreme position $$a$$ in the a closed orbit then it's speed is,

$$v_1=\sqrt{\dfrac{k}{2ma}}$$. . . .(4)

By the conservation of angular momentum,

$$mv_1a=mv_2b$$

$$v_2=\dfrac{a}{b}v_1=\dfrac{a}{b}\sqrt{\dfrac{k}{2ma}}$$. . . . . .(5)

Substitute $$v_1$$ and $$v_2$$ in equation (3), we get

$$a/b=3$$

The correct option is B.
Question 8
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Inside a smooth spherical shell of radius R a ball of the same mass is released from the shown position ( fig. ). Find the distance travelled by the shell on the horizontal floor when the ball comes to the just opposite position of itself with respect to its position in the shell.

A
$$\dfrac{3R}{5}$$

B
$$\dfrac{R}{4}$$

C
$$\dfrac{3R}{4}$$

D
$$\dfrac{5R}{4}$$

Solution

The correct option is C.

Given,

A spherical shell having radius $$R$$. And a ball is placed inside the spherical shell.

So,

Let the shell travel $$x$$ distance on the floor.

The ball travels a distance of $$(x+0)$$ in x-direction with respect to the floor.

So, conserving center of mass is:

$$\Rightarrow m(\dfrac{3R}{4})+mx=0$$

$$\Rightarrow m(\dfrac{3R}{4})=-mx$$

$$\Rightarrow x=(-\dfrac{3R}{4})$$

Thus, the Ball travels $$\dfrac{3R}{4}$$ in the opposite direction.
Question 9
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Find the moment of inertia of the rod AC about an axis BD as shown in figure. Mass of the rod is m and length is L

A
$$\dfrac{3mL^2 sin^2 \alpha}{2}$$

B
$$\dfrac{mL^2 sin^2 \alpha}{3}$$

C
$$\dfrac{3mL^2 sin^2 \alpha}{4}$$

D
$$\dfrac{3mL^2 sin^2 \alpha}{5}$$

Solution
Question 10
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A mass $$m$$ is moving with constant velocity parallel to the $$x$$ axis. Its angular momentum w.r.t the origin

A
Is zero

B
Remains constant

C
Goes on increasing

D
Goes on decreasing

Solution

Consider a particle $$P$$ moving parallel to X-axis.
Its angular momentum with respect to origin.
$$L=\overset{\rightarrow}r\times (m\overset{\rightarrow}v)=mr_{1}\cdot \,v$$
All variables $$m,v $$ and $$r_{1}$$ do not change.
$$\therefore\,L=constant$$