System of Particles and Rotational Motion Test

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System of Particles and Rotational Motion Test - 63

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Question 1
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The moment of inertia of a thin uniform rod of mass M and length L about an axis passing through its midpoint and perpendicular to its length is $$0_0$$. Its moment of inertia about an axis passing through one of its ends and perpendicular to its length is :

A
$$0_0 + \dfrac{ML^2}{2}$$

B
$$0_0 + \dfrac{ML^2}{4}$$

C
$$0_0 + 2ML^2$$

D
$$0_0 + ML^2$$

Solution

Let the end points of the rod are designated as A and B.
Applying parallel axis theorem,
Given, $$I_{cm}=O_o$$
$$I_A= I_{cm} + M(\frac{L}{2})^2= O_o + M\frac{L^2}{4}$$
Question 2
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The ratio of the radii of gyration of a circular
disc to that of a circular ring, each of same mass
and the radius around their respective axes is -

A
$$\dfrac {\sqrt2}{1}$$

B
$$\dfrac{\sqrt2}{\sqrt3}$$

C
$$\dfrac{\sqrt3}{\sqrt2}$$

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

Solution

$$MK_{1}^{2}=MR^{2}$$
$$K_{1} = R$$
$$MK_{2}^{2}=\frac{MR^{2}}{2}$$
$$K_{2}=\frac{R}{\sqrt{2}}$$
$$\frac{K_{2}}{K_{1}}=\frac{\sqrt{2}}{R}=\frac{1}{\sqrt{2}}$$
Question 3
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Two small balls $$A$$ and $$B$$ are made of same material, $$A$$ is solid but $$B$$ is hollow. These are connected by an ideal spring and this system is kept over a smooth plank such that center of mass of two balls coincide with mid point of the plank. Initially the spring is compressed by a certain amount and the plank is in horizontal position. Now we release the system then,

A
Plank will rotate clock wise

B
Plank will rotate anti clock wise

C
Plank will not rotate

D
It will depend upon string constant.

Solution

according to the given question ,
in the question only it is given that center of mass of body are coinciding with mid point of plank,therefore with respect to the condition of equilibrium,these bodies will be at constant point only.
plank will not rotate
Question 4
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Directions For Questions

A closed system consists of two particles of masses $$m_1$$ and $$m_2$$, which move at right angles to each other with velocities $$v_1$$ and $$v_2$$. Given $$\mu= \dfrac{m_1m_2}{m_1+m_2}$$

...view full instructions

Find the momentum of each particle.

A
$$p=2\mu\sqrt{v_1^2+v_2^2}$$

B
$$p=\mu\sqrt{v_1^2+v_2^2}$$

C
$$p=3\mu\sqrt{v_1^2+v_2^2}$$

D
$$p=4\mu\sqrt{v_1^2+v_2^2}$$

Solution

The particles $$m_1$$ and $$m_2$$ move with velocities $$\vec{v_1}$$ and $$\vec{v_2}$$ respectively.

$$\vec{v_{cm}} = \dfrac{m_1\vec{v_1} + m_2\vec{v_2}}{m_1 + m_2}$$

In the CM frame,

$$\vec{v'_1} = \vec{v_1} - \vec{v_{cm}}$$

$$\vec{v'_2} = \vec{v_2} - \vec{v_{cm}}$$

$$ \vec{v_{rel}} = \vec{v_1} - \vec{v_2}$$

$$\vec{p'_1}=m_1( \vec{v_1} - \vec{v_{cm}})=\dfrac{m_1m_2}{m_1 + m_2}(\vec{v_1} -\vec{v_2})=\dfrac{m_1m_2}{m_1 + m_2}\vec{v_{rel}}$$

$$\vec{p'_2}=m_2( \vec{v_2} - \vec{v_{cm}})=\dfrac{m_1m_2}{m_1 + m_2}(\vec{v_2} -\vec{v_1})=-\dfrac{m_1m_2}{m_1 + m_2}\vec{v_{rel}}$$

$$|\vec{p'_1}| = |\vec{p'_2}| =|\dfrac{m_1m_2}{m_1 + m_2}\vec{v_{rel}}|=\dfrac{m_1m_2}{m_1 + m_2}|\vec{v_{rel}}|= \dfrac{m_1m_2}{m_1 + m_2}\sqrt{(v_1^2 + v_2^2)}= \mu \sqrt{(v_1^2 + v_2^2)} $$

since
$$\vec{v_1} \perp \vec{v_2}$$ and $$|\vec{v_1} -\vec{v_2}|=\sqrt{v_1^2 + v_2^2 -2v_1v_2 \cos \dfrac{\pi}{2}}=\sqrt{v_1^2 + v_2^2} $$

where $$\mu =\dfrac{m_1m_2}{m_1 + m_2}$$ is defined as the reduced mass.
Question 5
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A chain $$AB$$ of mass $$m$$ and length $$L$$ is hanging on a smooth horizontal table as shown in the figure. If it is released from the position shown then the displacement of centre of mass of chain in magnitude, when end $$A$$ moves a distance $$\dfrac{L}{2}\ is\ X\sqrt{2}m $$.
Find $$X$$. ($$L = 32\ m$$)

A
$$8$$

B
$$9$$

C
$$10$$

D
$$16$$

Solution

In the first case, we consider two parts one of mass $$3m/4$$ and length $$3L/4$$ and other of mass $$m/4$$ and length $$L/4$$.
The center of mass of these masses is calculated as

$$x_1=\displaystyle\dfrac{(3m/4)(-3L/8)+(m/4)(0)}{m}=-9L/32=-9$$
and
$$y_1=\displaystyle\dfrac{(3m/4)(0)+(m/4)(-L/8)}{m}=-L/32=-1$$

Thus the coordinates of the center of mass is $$(-9,-1)$$
Similarly, when the chain moves by $$L/2$$ distance we get the new position of the center of mass as:

$$x_2=\displaystyle\dfrac{(m/4)(-L/8)+(3m/4)(0)}{m}=-9L/32=-1$$
and
$$y_2=\displaystyle\dfrac{(m/4)(0)+(3m/4)(-3L/8)}{m}=-L/32=-9$$

Thus the coordinates of the center of mass in this case is $$(-1,-9)$$
Thus the distance between the two position of the center of mass is calculated as $$8\sqrt 2$$. Thus the value of $$X$$ is $$8$$.
Question 6
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Two bars of masses $$m_1$$ and $$m_2$$, connected by a weightless spring of stiffness $$k$$, rest on a smooth horizontal plane. Bar 2 is shifted by a small distance $$x_0$$ to the left and released. The velocity of the centre of mass of the system when bar 1 breaks off the wall is :

A
$$x_0\sqrt {\dfrac {km_2}{m_1+m_2}}$$

B
$$\dfrac {x_0}{m_1+m_2}\sqrt {km_2}$$

C
$$x_0k\dfrac {m_1+m_2}{m_2}$$

D
$$x_0\dfrac {\sqrt {km_1}}{(m_1+m_2)}$$

Solution

$$\displaystyle \frac {1}{2}kx_0^2=\frac {1}{2}m_2v_2^2$$
$$\displaystyle \therefore v_2=\sqrt {\frac {k}{m_2}}x_0$$
$$\displaystyle v_{CM}=\frac {m_1v_1+m_2v_2}{m_1+m_2}$$
$$\displaystyle =\frac {\sqrt {km_2}}{(m_1+m_2)}x_0($$ as $$v_1=0$$)
Question 7
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Two identical bricks of length $$L$$ are piled one on top of the other on a table as shown in the figure. The maximum distance $$S$$ the top brick can overhang the table with the system still balanced is :

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

B
$$ \dfrac{2}{3}L $$

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

D
$$ \dfrac{7}{8}L $$

Solution

The center of mass of the set of bricks must be above or to the left of the point of support for the system of be balanced.
The center of mass of the upper brick needs to be above the right end of lower brick. The center of upper brick is at a distance of $$L/2$$ from its right end.
The distance center of mass of both the books from the right edge of upper brick$$=\dfrac{(mL/2)+(mL)}{2m}=\dfrac{3L}{4}=S$$.
Question 8
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Three point like equal masses $$m_1, m_2$$ and $$m_3$$ are connected to the ends of a mass-less rod of length $$L$$ which lies at rest on a smooth horizontal plane. At $$t = 0$$, an explosion occurs between $$m_2$$ and $$m_3$$, and as a result, mass $$m_3$$ is detached from the rod, and moves with a known velocity $$v$$ at an angle of $$30^o$$ with the y-axis. Assume that the masses $$m_2$$ and $$m_3$$ are unchanged during the explosion.
What is the velocity of the centre of mass of the system consisting of three masses after the expulsion?

A
$$\displaystyle \dfrac{v}{4} (\widehat i - 3 \widehat j)$$

B
$$\displaystyle \dfrac{v}{4} (- \widehat i + \sqrt{3} \widehat j)$$

C
$$-v$$

D
none of the above

Solution

considering all the three masses in one system, the COM of mass will not change its position until unless an external force is acted upon the system. Explosion is an impulsive but internal force for the system, so COM will not move as it was already at rest initially. Thus, the velocity of COM is zero after explosion.
Question 9
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The arrangement shown in figure above consists of two identical uniform solid cylinders, each of mass $$m$$, on which two light threads are wound symmetrically. The friction in the axle of the upper cylinder is assumed to be absent. If the tension of each thread in the process of motion is $$\displaystyle T=\frac{mg}{x}$$, then the value of $$x$$ is :

A
$$10$$

B
$$20$$

C
$$15$$

D
$$5$$

Solution

For the cylinder from the equation $$N_z=I\beta_z$$ about its stationary axis of rotation.
$$\displaystyle2Tr=\frac{mr^2}{2}\beta$$ or $$\displaystyle\beta=\frac{4T}{mr}$$......(1)
For the rotation of the lower cylinder from the equation $$N_{cz}=I_c\beta_z$$
$$\displaystyle2Tr=\frac{mr^2}{2}\beta^\prime$$ or, $$\displaystyle\beta^\prime=\frac{4T}{mr}=\beta$$
Now for the translational motion of lower cylinder from the equation $$F_x=mw_{cx}$$,
$$mg-2T=mw_c$$ ......(2)
As there is no slipping of threads on the cylinders,
$$w_c=\beta^\prime r+\beta r=2\beta r$$ (3)
Simultaneous solution of (1), (2) and (3) yields
$$\displaystyle T=\frac{mg}{10}$$
Question 10
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A particle of mass $$'m'$$ is rigidly attached at $$'A'$$ to a ring of mass $$'3m'$$ and radius $$'r.'$$ The system is released from rest and rolls without sliding. The angular acceleration of ring just after release is

A
$$\cfrac{g}{4r}$$

B
$$\cfrac{g}{6r}$$

C
$$\cfrac{g}{8r}$$

D
$$\cfrac{g}{2r}$$