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  4. A solid cylinder on moving with constant speed v0 reaches the bottom of an incline of 30° . A hollow cylinder of same mass and radius moving with the same constant speed v0 reaches the bottom of a different incline of inclination θ. There is no slipping and both of them go through the same distance in the same time; θ is then equal to

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Q. A solid cylinder on moving with constant speed $v_{0}$ reaches the bottom of an incline of $30^{\circ} .$ A hollow cylinder of same mass and radius moving with the same constant speed $v_{0}$ reaches the bottom of a different incline of inclination $\theta$. There is no slipping and both of them go through the same distance in the same time; $\theta$ is then equal to

System of Particles and Rotational Motion

Solution:

For solid cylinder, $\theta=30^{\circ},\, K^{2}=\frac{1}{2} R^{2}$
For hollow cylinder, $\theta=?,\, K^{2}=R^{2}$
Hence,
$\frac{\left(1+\frac{1}{2}\right)}{\sin 30^{\circ}}=\frac{1+1}{\sin \theta}$
$\therefore \sin \theta =\frac{2}{3}=0.6667$
$\theta =42^{\circ}$