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Rotational Dynamics — Physics STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 SciencePhysicsRotational Dynamics4 Marks
Question
Two circular discs A and B, having the same mass, have four identical small circular discs placed on them, as shown in the diagram. They are simultaneously released from rest at the top of an inclined plane. If the discs roll down without slipping, which disc will reach the bottom first?

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Answer

The disc A has the smaller discs closer to the centre than disc B. Hence, the moment of inertia of disc $A\left(I_A\right)$ is less than that of disc $B\left(I_B\right)$.
[Suppose the larger discs have radius $R$, the smaller discs have mass $m$ and radius $r$, and the centre of each smaller disc on disc $\mathrm{A}$ is at a distance $\mathrm{x}$ from the centre. Then, $\mathrm{x}=\sqrt{2} \mathrm{rr}$ and, it can be shown that, $\left.I_B-I_A=4 m\left[R^2-(x-r)^2\right]>0.\right]$
Each composite disc is equivalent to a disc of the same radius $R$ and mass $M^{\prime}=M+4 m$, where $m$ is the mass of each smaller disc, but of different thicknesses.
Suppose, starting from rest, the composite discs roll down the same distance $L$ along a plane inclined at an angle $\theta$, their respective accelerations will be
$
a_{\mathrm{A}}=\frac{g \sin \theta}{1+\left(I_{\mathrm{A}} / M^{\prime} R^2\right)}, \quad a_{\mathrm{B}}=\frac{g \sin \theta}{1+\left(I_{\mathrm{B}} / M^{\prime} R^2\right)}
$
so that, the respective times taken to travel the distance $L$ are
$
\begin{aligned}
t_{\mathrm{A}} & =\sqrt{\frac{2 L}{a_{\mathrm{A}}}}=\sqrt{\frac{2 L}{g \sin \theta} \cdot\left(1+\frac{I_{\mathrm{A}}}{M^{\prime} R_2}\right)} \text { and } \\
t_{\mathrm{B}} & =\sqrt{\frac{2 L}{a_{\mathrm{B}}}}=\sqrt{\frac{2 L}{g \sin \theta} \cdot\left(1+\frac{I_{\mathrm{B}}}{M^{\prime} R_2}\right)} \\
\therefore t_{\mathrm{A}} & <t_{\mathrm{B}},
\end{aligned}
$
i.e., the disc A will reach the bottom first.

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