Q. A solid cylinder $30\ cm$ in diameter at the top of an inclined plane $2.0\ m$ high is released and rolls down the incline without loss of energy due to friction. Its linear speed at the bottom is
A) $5.29\ m/sec$
B) $4.1 \times 10^{3}\ m/sec$
C) $51\ m/sec$
D) $51\ cm/sec$
Explanation
(a) $v = \sqrt{\frac{2gh}{1 + \frac{K^{2}}{R^{2}}}} = \sqrt{\frac{2 \times 10 \times 2}{1 + \frac{1}{2}}} = \sqrt{26.66} = 5.29m/s\ approx$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.