JEE MAIN - Physics (2002 - No. 64)

Initial angular velocity of a circular disc of mass $$M$$ is $${\omega _1}.$$ Then two small spheres of mass $$m$$ are attached gently to diametrically opposite points on the edge of the disc. What is the final angular velocity of the disc?
$$\left( {{{M + m} \over M}} \right)\,\,{\omega _1}$$
$$\left( {{{M + m} \over m}} \right)\,\,{\omega _1}$$
$$\left( {{M \over {M + 4m}}} \right)\,\,{\omega _1}$$
$$\left( {{M \over {M + 2m}}} \right)\,\,{\omega _1}$$

Explanation

When two small spheres of mass $$m$$ are attached gently, the external torque, about the axis of rotation, is zero.

So, $${{d\overrightarrow L } \over {dt}} = \overline z $$ = 0

$$\overrightarrow L $$ = conserved

So the angular momentum about the axis of rotation is conserved.

$$\therefore$$ $${I_1}{\omega _1} = {I_2}{\omega _2}$$

$$ \Rightarrow {\omega _2} = {{{I_1}} \over {{I_2}}}{\omega _1}$$

Here Moment of inertia of Disc $${I_1} = {1 \over 2}M{R^2}$$ and

After adding two sphere Moment of Inertia of disc and two sphere,

$${I_2} = {1 \over 2}M{R^2} + $$$$2\left( {{1 \over 2}m{R^2} + {1 \over 2}m{R^2}} \right)$$

$$\therefore$$ $${\omega _2} = {{{1 \over 2}M{R^2}} \over {{1 \over 2}MR + 2m{R^2}}} \times {\omega _1} = {M \over {M + 4m}}{\omega _1}$$

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