Two solid cylinders P and Q of same mass and | vedclass

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(D) The moment of inertia $I$ of a body depends on the distribution of mass. Since cylinder $P$ has mass concentrated near its surface,it has a larger

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Cylinder $Q$ reaches the ground with larger angular speed.

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(D) The moment of inertia $I$ of a body depends on the distribution of mass. Since cylinder $P$ has mass concentrated near its surface,it has a larger moment of inertia $(I_P > I_Q)$.The linear acceleration $a$ of a rolling body on an inclined plane is given by $a = \frac{g \sin 	heta}{1 + \frac{I}{mR^2}}$.Since $I_P > I_Q$,it follows that $a_P Using the kinematic equation $s = \frac{1}{2}at^2$,the time taken to reach the bottom is $t = \sqrt{\frac{2s}{a}}$. Since $a_P  t_Q$,meaning $Q$ reaches the ground first.The final velocity $v$ is given by $v^2 = 2as$. Since $a_Q > a_P$,$v_Q > v_P$. The translational kinetic energy is $\frac{1}{2}mv^2$,so $Q$ has higher translational kinetic energy.Finally,since $v = \omega R$,the angular speed $\omega = \frac{v}{R}$. Since $v_Q > v_P$,it follows that $\omega_Q > \omega_P$. Thus,cylinder $Q$ reaches the ground with a larger angular speed.

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