System of Particles and Rotational Motion Test

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System of Particles and Rotational Motion Test - 29

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Question 1
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Four identical rods, each of mass $$m$$ and length $$l$$, are joined to form a rigid square frame. The frame lies in the xy plane, with its centre at the origin and the sides parallel to the x and y axes. Its moment of inertia about :

A
the x-axis is $$\displaystyle \frac{2}{3} \, ml^2$$

B
the z-axis is $$\displaystyle \frac{4}{3} \, ml^2$$

C
an axis parallel to the z-axis and passing through a corner is $$\displaystyle \frac{10}{3} \, ml^2$$

D
one side is $$\displaystyle \frac{5}{2} \, ml^2$$

Solution

The moment of inertia of one rod about the center of the square frame using parallel axis theorem is $$\displaystyle\frac{1}{12}ml^2+\frac{1}{4}ml^2=\frac{4}{3}ml^2$$
Question 2
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Directions For Questions

A uniform solid cylinder $$A$$ of mass $$m_1$$ freely rotates about a horizontal axis fixed to a mount $$B$$ of mass $$m_2$$ (figure shown above). A constant horizontal force $$F$$ is applied to the end $$K$$ of a light thread tightly wound on the cylinder. The friction between the mount and the supporting horizontal plane is assumed to be absent.

...view full instructions

Find the acceleration of the point $$K$$.

A
$$\displaystyle w=\frac{F(3m_1+2m_2)}{m_1(m_1+m_2)}$$

B
$$\displaystyle w=\frac{F(3m_1+m_2)}{m_1(m_1+m_2)}$$

C
$$\displaystyle w=\frac{F(m_1+m_2)}{m_1(m_1+m_2)}$$

D
$$\displaystyle w=\frac{F(m_1+2m_2)}{m_1(m_1+m_2)}$$

Solution

$$F=(m_1+m_2)a_{cm}$$ ...(1)
torque about center
$$FR=\dfrac{m_1R^2\alpha}{2}\rightarrow \alpha=\dfrac{2F}{m_1R}$$
Accelration of point k $$a_k=a_{cm}+\alpha R$$
$$a_k\dfrac{F(3m_1+2m_2)}{m_1(m_1+m_2)}$$
Question 3
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Two masses $$\displaystyle{M_1}$$ and $$\displaystyle{M_2}$$ are connected with each other with the help of a massless spring of spring-constant $$k$$. Mass $$\displaystyle{M_1}$$ is in contact with a wall and the system is at rest on the frictionless floor. $$\displaystyle{M_2}$$ is displaced by distance $$x$$ to compress the spring and released. The velocity of COM (centre of mass) of the system when $$\displaystyle{M_1}$$ is detached from wall is :

A
$$\displaystyle{\frac{x\sqrt{kM_1}}{M_1+M_2}}$$

B
$$\displaystyle{\frac{x\sqrt{kM_2}}{M_1+M_2}}$$

C
$$\displaystyle{\frac{kx\sqrt{M_2}}{M_1+M_2}}$$

D
$$\displaystyle{\frac{x}{\sqrt{kM_1}(M_1-M_2)}}$$

Solution

Block 1 will leave wall just after the spring is at its natural length.
P.E. will convert into K.E. of 2nd block.
$$\dfrac { 1 }{ 2 } k{ x }^{ 2 }=\dfrac { 1 }{ 2 } { M }_{ 2 }{ v }^{ 2 }\\ v=x\sqrt { \dfrac { k }{ { M }_{ 2 } } } \\ { M }_{ 2 }v=x\sqrt { k{ M }_{ 2 } } $$
velocity of centre of mass $$=\dfrac { { M }_{ 1 }{ v }_{ 1 }+{ M }_{ 2 }{ v }_{ 2 } }{ { M }_{ 1 }+{ M }_{ 2 } } =\dfrac { x\sqrt { k{ M }_{ 2 } } }{ { M }_{ 1 }+{ M }_{ 2 } } $$
Question 4
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A particle of rest mass $$m_0$$ is moving with speed $$c/2$$. What is its momentum?

A
$$\displaystyle \frac{m_0c}{2}$$

B
$$\displaystyle \frac{1}{\sqrt{3}}m_0c$$

C
$$\displaystyle \frac{2}{\sqrt{3}}m_0c$$

D
none of the above

Solution

We know that, mass of moving particle at velocity $$m= \dfrac{m_0}{\sqrt{1-\left(\dfrac{v}{c}\right) ^{2}}}$$
since$$ ,\ v=\dfrac{c}{2}$$
$$\Rightarrow m=\dfrac{2m_0}{\sqrt{3}}$$
$$Momentum=mv =\dfrac{2m_0}{\sqrt{3}} \dfrac{c}{2}= \dfrac{1}{\sqrt{3}} m_0 c$$
Question 5
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A cracker is thrown into air with a velocity of 10 m/s at an angle of 45$$^o$$ with the vertical. When it is at a height of 0.5 m from the ground, it explodes into a number of pieces which follow different parabolic paths. What is the velocity of centre of mass, when it is at a height of 1 m from the ground? $$(g = 10 m/s^2)$$

A
$$4 \sqrt{5} m/s$$

B
$$2 \sqrt{5} m/s$$

C
$$4 \sqrt{4} m/s$$

D
10 m/s

Solution

Since cracker explodes in mid air it does not encounter any external force so velocity of COM will remain unchanged due to explosion.
Horizontal component $$=10\sin { { 45 }^{ o } } =\dfrac { 10 }{ \sqrt { 2 } } =5\sqrt { 2 } m/s$$
From third equation of motion, at a height of 1 m, vertical component of velocity of center of mass $$=\sqrt { { (5\sqrt { 2 } ) }^{ 2 }-2 \times10 \times1 } =\sqrt { 50-20 } =\sqrt { 30 } m/s$$

Hence, net velocity $$=\sqrt { { (5\sqrt { 2 } ) }^{ 2 }+{ (\sqrt { 30 } ) }^{ 2 } } =\sqrt { 80 } =4\sqrt { 5 } m/s$$
Question 6
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The distance of the centre of mass of the T-shaped plate from O is

A
7 m

B
2.7 m

C
4 m

D
1.7 m

Solution

Taking O as Origin for the co-ordinate system, taking horizontal as a X-axis and vertical as a Y-axis.
So
The position of the center of mass is given as $$\displaystyle\frac{A_1y_1+A_2y_2}{A_1+A_2}$$
Substituting the values we get,
$$\displaystyle\frac{16(0)+12(-4)}{16+12}=\frac{-48}{28}=-1.7$$
Negative side indicated $$1.7\ cm$$ downwards.
Question 7
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A particle of mass $$m$$ is acted on by two forces of equal magnitude $$F$$ maintaining their orientation relative to the velocity $$v$$ as shown in figure. The momentum of the particle :

A
increases in (a)

B
decreases in (b)

C
only the direction changes in (c)

D
All of the above

Solution

In figure (a),

The direction of force is same as the direction of momentum is same, so the momentum of particle is increases.

In figure (b),

The direction of force is opposite to the direction of momentum, so the momentum of the particle is decrease.

In figure (c),

The direction of force is perpendicular to the direction of momentum, so only the direction o momentum of particle is change.

The correct option is D.
Question 8
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Two particles are shown in figure. At $$t = 0$$, a constant force $$F = 6$$N starts acting on the 3 kg mass. Find the velocity of the centre of mass of these particles at $$t = 5 s$$.

A
$$5 m/s$$

B
$$4 m/s$$

C
$$6 m/s$$

D
$$3 m/s$$

Solution

We know that the total linear momentum of a system of partilas is equal to linar momentumm of center of man of the system.
$$\Rightarrow (m_{1}+m_{2})\vartheta _{cm}=m_{1}\vartheta _{1}+m_{2}\vartheta _{2}$$
At t= 5 sec,
$$\vartheta _{1}=0 m/s$$
$$\vartheta _{2}=u_{2}+at$$
$$\vartheta _{2}=2\times 5=10 m/s(\because u_{2}=0$$ and $$a=\frac{F}{m_{2}}=2 m/s^{2})$$
$$\Rightarrow 5\times \vartheta _{cm}=0+3\times 10$$
$$\Rightarrow \vartheta _{cm}=6 m/s$$
Question 9
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The moment of inertia of a solid sphere about an axis passing through the centre of gravity is $$2/5 MR^{2}$$, then its radius of gyration about a parallel axis at a distance $$2R$$ from first axis is

A
$$5R$$

B
$$\displaystyle\ \sqrt{\frac{22}{5}} R$$

C
$$\displaystyle\ \frac{5}{2} R$$

D
$$\displaystyle\ \sqrt{\frac{12}{5}}R$$

Solution

By, Parallel axis theorm
$$I$$ at $$2R$$ is $$2MR^2/5 +M(2R)^2=22MR^2/5 $$
Radius of gyration, $$Mk^2=22MR^2/5\Rightarrow k=\sqrt{22/5}\times R$$.
Question 10
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Two blocks of masses 5 kg and 2 kg are placed on a frictionless surface and connected by a spring. An external kick gives a velocity of 14 m/s to the heavier block in the direction of lighter one. The magnitudes of velocities of two blocks in the centre of mass frame after the kick are, respectively :

A
4 m/s, 4 m/s

B
10 m/s, 4 m/s

C
4 m/s, 10 m/s

D
10 m/s, 10 m/s

Solution

Velocities of $$5 kg$$ block and $$2kg$$ block are $$14$$ $$m/s$$ and zero respectively.
Let velocity of its centre of mass be $$V_{CM}$$
$$V_{CM} = \dfrac{m_1v +m_2 v'}{m_1+M_2}$$

$$V_{CM} = \dfrac{5(14) +2 (0) }{5+2} = 10 $$ $$m/s$$

Velocity of $$5kg$$ block in $$COM$$ frame, $$V = v- V_{CM} = 14 - 10 = 4$$ $$m/s$$
Velocity of $$2kg$$ block in $$COM$$ frame, $$V' = v'- V_{CM} = 0 - 10 = -10$$ $$m/s$$
$$\implies |V'| = 10$$ $$m/s$$