Q. A solid cylinder rolls down an inclined plane of height $3\ m$ and reaches the bottom of plane with angular velocity of $2\sqrt{2}\ rad.s^{- 1}$. The radius of cylinder must be [Take $g = 10ms^{- 2}$]
A) $5\ cm$
B) $0.5\ cm$
C) $\sqrt{10}\ cm$
D) $\sqrt{5}\ m$
Explanation
(d) $v = \sqrt{\frac{2gh}{1 + \frac{I}{mr^{2}}}} = \sqrt{\frac{2 \times 10 \times 3}{1 + \frac{mr^{2}}{2 \times mr^{2}}}} = \sqrt{\frac{2 \times 10 \times 3}{\frac{3}{2}}} = \sqrt{40}$
$\Rightarrow v = r\omega$
$\Rightarrow r = \frac{v}{\omega} = \frac{\sqrt{40}}{2\sqrt{2}} = \sqrt{\frac{40}{8}} = \sqrt{5}\ m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.