System of Particles and Rotational Motion Test

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System of Particles and Rotational Motion Test - 52

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Question 1
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Two blocks of masses $$8$$kg are connected by a spring of negligible mass and placed on a frictions less horizontal surface. An impulse gives a velocity of $$12$$m/s to the heavier block in the direction of lighter block. The velocity of the center of mass is:-

A
$$12$$m/s

B
$$10$$m/s

C
$$8$$m/s

D
$$6$$m/s

Solution

Velocity of center of mass is $$=\dfrac{m_1v_1+m_2v_2}{m_1+m_2}$$

$$=\dfrac{8\times 12+8\times 0}{8+8}$$

$$=\dfrac{96+0}{16}$$

$$=6m/s$$
Hence, the answer is $$6m/s.$$
Question 2
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Two particles having mass ratio n : 1 are interconnected by a light in extensible string that passes over a smooth pulley. If the system is released, then the acceleration of the centre of mass of the system is

A
$$( n - 1 ) ^ { 2 } g$$

B
$$\left( \frac { n + 1 } { n - 1 } \right) ^ { 2 } g$$

C
$$\left( \frac { n - 1 } { n + 1 } \right) ^ { 2 } g$$

D
$$\left( \frac { n + 1 } { n - 1 } \right) 9$$

Solution

Given$$,$$
$$\frac{{{m_1}}}{{{m_2}}} = \frac{n}{1} = n$$
Each mass have the acceleration $$a = \frac{{\left( {{m_1} - {m_2}} \right)}}{{{m_1} + {m_2}}}$$
however $${{m_1}}$$ which is heavier will have the will have acceleration $${{a_1}}$$ vertically down while the lighter mass $${{m_2}}$$ will have acceleration $${{a_2}}$$ vertically up $$ \to {a_2} = - {a_1}$$
The acceleration or the centre of mass of the system$$,$$ $${a_{cm}} = \frac{{{m_1}{a_1} + {m_2}{a_2}}}{{{m_1} + {m_2}}}$$
given that $${a_2} = - {a_1} \to {a_{cm}} = \frac{{\left( {{m_1} - {m_2}} \right){a_1}}}{{{m_1} + {m_2}}} = \frac{{{m_1} - {m_2}}}{{{m_1} + {m_2}}} \times \frac{{\left( {{m_1} - {m_2}} \right)g}}{{{m_1} + {m_2}}} = \frac{{{{\left( {{m_1} - {m_2}} \right)}^2}g}}{{{m_1} + {m_2}}}$$
Since $$\frac{{{m_1}}}{{{m_2}}} = n$$ diving by $${{m_2}}$$ and simplifying
$$ \Rightarrow {a_{cm}} = {\left( {\frac{{n - 1}}{{n + 1}}} \right)^2}g$$
Hence,
option $$(C)$$ is correct answer.
Question 3
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A thin uniform rod of length l and m is swinging freely about a horizontal axis passing through its end. Its maximum angular speed is $$\omega$$. Its centre of mass rises to a maximum height of:

A
$$\dfrac{1}{6} \dfrac{l\omega}{g}$$

B
$$\dfrac{1}{2} \dfrac{l^2\omega^2}{g}$$

C
$$\dfrac{1}{6} \dfrac{l^2\omega^2}{g}$$

D
$$\dfrac{1}{3} \dfrac{l^2\omega^2}{g}$$

Solution

At lowest height
$$E=\cfrac{1}{2}I \omega^2$$
At maximum weight $$\omega=0$$
$$E=mgh+0$$ (his height from lower point)
By conservation of energy
$$\cfrac{1}{2}I\omega^2=mgh\\h=\cfrac{I\omega^2}{2mg}\\h=\cfrac{ml^2\omega^2}{6mg}=\cfrac{l^2\omega^2}{6g}$$
Question 4
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Linear mass density of a rod AB ( of length 10 m) varies with distance x from its end A as $$\lambda = {\lambda _o}{x^3}$$$$\left( {{\lambda _o}\;{\text{is}}\;{\text{positive}}\;{\text{constant}}} \right)$$. Distance of the center of mass of the rod, from end B is

A
8 m

B
2 m

C
6 m

D
4 m

Solution
Question 5
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If the moment of inertia of a rigid body is numerically equal to its mass, its radius of gyration is

A
Equal to its radius

B
Equal to diameter

C
Equal to one unit

D
none

Solution
Question 6
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In rotational motion of a rigid body, all particle move with

A
Same linear and angular velocity

B
Same linear and different angular velocity

C
With different linear velocities and same angular velocities

D
With different linear velocities and different angular velocities

Solution
Question 7
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A man spinning in free space changes the shape of his body e.g. by stretching his arms or curling up then, which of the following physical quantity will remain constant

A
Moment of inertia

B
Angular momentum

C
Angular velocity

D
Rotational kinetic energy

Solution

When a man spinning changes his position of body by spreading arms or curing up his angular momentum does not charge as there is no external torque$$,$$ by conservation of angular momentum$$.$$ But his moment of intertia$$,$$ angular velocity changes$$.$$
Hence,
option $$(B)$$ is correct answer.
Question 8
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The moment of inertia of a uniform cylinder of length $$l$$ and radius $$R$$ about its perpendicular bisector is $$1$$. What is the ratio $$l/R$$ such that the moment of inertia is minimum?

A
$$\cfrac{3}{\sqrt{2}}$$

B
$$\sqrt{\cfrac{3}{2}}$$

C
$$\cfrac{\sqrt{2}}{2}$$

D
$$1$$

Solution
Question 9
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Find the coordination of center of mass of a uniform semicircle closed wire frame with respect to the origin which is at its center.The radius of the circular portion is R.

A
$$\left( {\dfrac{{4R}}{{3\pi }},0} \right)$$

B
$$\left( {\dfrac{{2R}}{{\pi }},0} \right)$$

C
$$\left( {\dfrac{R}{{\pi + 2}},0} \right)$$

D
$$\left( {\dfrac{2R}{{\pi + 2}},0} \right)$$

Solution

We know that, $$x+cm=\cfrac{4}{\pi R^2}\int^0_R{-t^2 dt}$$
$$=\cfrac{4}{\pi R^2}|-\cfrac{t^3}{R}|^0_R=\cfrac{4}{\pi R^2}(\cfrac{R^3}{3})=\cfrac{4R}{3\pi}$$
Thus, co-ordinates should be $$=[\cfrac{4}{3\pi},0]$$
Question 10
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Six identical particles each of mass $$m$$ are arranged at the corners of a regular hexagon of side length $$a$$. If the mass of one of the particle is doubled, the shift in the centre of mass is

A
$$a$$

B
$$\dfrac {6a}{7}$$

C
$$\dfrac {a}{7}$$

D
$$\dfrac {a}{\sqrt {3}}$$

Solution

Correct Answer: Option C

Hint: Centre of mass of any regular polygon is at its geometric centre.

Explanation of Correct Option:
Consider $$fig.(i)$$, Since it is given regular hexagon, all the sides are of length $$a$$
The centre of mass in the initial case will be at the centre of the hexagon.ie, at $$O$$.

Consider $$fig.(ii)$$
Here, mass of one particle is doubled,
We can consider the hexagon as a point system with centre of mass at its centre and another mass of m (one mass is doubled) as in $$fig.(iii)$$. Then centre of mass considering the centre of hexagon as origin,

$$C=\dfrac{6m\times 0+m\times a}{6m+m}$$

$$C=\dfrac{a}{7}$$

i.e, centre of mass is at a distance $$\dfrac{a}{7}$$ from the centre.