Question 45: Centre of Mass and Rotational Motion

Question

A small bob $A$ of mass $m$ is attached to a massless rigid rod of length 1 m pivoted at point $P$ and kept at an angle of $60^{\circ}$ with vertical as shown in figure. At distance of 1 m below point $P$, an identical bob $B$ is kept at rest on a smooth horizontal surface that extends to a circular track of radius $R$ as shown in figure. If bob $B$ just manages to complete the circular path of radius $R$ upto a point $Q$ after being hit elastically by $\operatorname{bob} A$, then radius $R$ is $\_\_\_\_$ m.

Accepted Answer

Bob A is released from rest at an angle $	heta=60^{\circ}$. It swings down to the lowest point (vertical position) where it hits Bob B.
Length of rod is $\mathrm{l}=1 \mathrm{~m}$.
Height lost by $\mathrm{A}(\Delta \mathrm{h})$ :
$\Delta \mathrm{h}=\mathrm{l}-\mathrm{l} \cos 	heta=\mathrm{l}\left(1-\cos 60^{\circ}ight)$
$$\Rightarrow $$ $$ \Delta \mathrm{h}=1(1-0.5)=0.5 \mathrm{~m} $$
Using conservation of energy for Bob A :
$$ \mathrm{mg} \Delta \mathrm{~h}=\frac{1}{2} \mathrm{~m} \mathrm{v}_{\mathrm{A}}^2 $$
$$\Rightarrow $$ $\mathrm{v}_{\mathrm{A}}=\sqrt{2 \mathrm{~g} \Delta \mathrm{~h}}$
$$\Rightarrow $$ $$ \mathrm{v}_{\mathrm{A}}=\sqrt{2 \mathrm{~g} 	imes 0.5}=\sqrt{\mathrm{g}} $$
It is given that the collision is elastic. Both bobs are identical (mass m ).
Initial velocities: $\operatorname{Bob} A$ has $v_A=\sqrt{g}$, $\operatorname{Bob} B$ is at rest $\left(v_B=0ight)$.
In a perfectly elastic collision between two identical masses where one is initially at rest, their velocities are exchanged.
So, after collision, bob A comes to rest and bob B acquires the velocity of Bob A .
$$ \mathrm{v}_{\mathrm{B}}=\mathrm{v}_{\mathrm{A}}=\sqrt{\mathrm{g}} $$
Bob B moves along the smooth horizontal surface and enters the vertical circular track of radius R.
The bob B just manages to complete the circular path up to a point Q .
For a particle to just complete a full vertical circle (reach the top), the minimum velocity at the bottom $(\sqrt{5 \mathrm{gR}})$ is required.
$$ \mathrm{v}_{	ext {bottom }} \geq \sqrt{5 \mathrm{gR}} $$
$\Rightarrow \sqrt{g}=\sqrt{5 g R}$
$$\Rightarrow $$ $\Rightarrow \mathrm{g}=5 \mathrm{gR}$
$$ \Rightarrow \mathrm{R}=\frac{1}{5} \mathrm{~m} $$
Therefore, the required radius $R$ is $\frac{1}{5} \mathrm{~m}$. Hence, the correct option is (A).

Answer

$\frac{1}{5}$

Answer

$\frac{2-\sqrt{3}}{5}$

Answer

$\frac{3}{5}$

Answer

$\frac{2+\sqrt{3}}{5}$

Centre of Mass and Rotational Motion

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