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Two solid cylinders $P$ and $Q$ of same mass and same radius start rolling down a fixed inclined plane from the same height at the same time. Cylinder $P$ has most of its mass concentrated near its surface, while $Q$ has most of its mass concentrated near the axis. Which statement(s) is(are) correct?
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The correct answers are:
Cylinder $Q$ reaches the ground with larger angular speed.
As we know, acceleration of the center of mass of cylinder rolling down an inclined plane
$a_{c}=\frac{g \sin \theta}{1+\frac{I}{M R^{2}}}$
In case of $P$ the mass is concentrated away from the axis, So $I_{P}>I_{Q}$
$\therefore \quad a_{P} < a_{Q} \Rightarrow v_{\mathrm{P}} < v_{\mathrm{Q}} \Rightarrow \omega_{\mathrm{P}} < \omega_{\mathrm{Q}}$
$a_{c}=\frac{g \sin \theta}{1+\frac{I}{M R^{2}}}$
In case of $P$ the mass is concentrated away from the axis, So $I_{P}>I_{Q}$
$\therefore \quad a_{P} < a_{Q} \Rightarrow v_{\mathrm{P}} < v_{\mathrm{Q}} \Rightarrow \omega_{\mathrm{P}} < \omega_{\mathrm{Q}}$
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