Question

A cylinder is rolling down a inclined plane of inclination $60^{\circ }$ . What is its acceleration?

• A. $g/\sqrt{3}$
• B. $g\sqrt{3}$
• C. $\sqrt{\frac{2g}{3}}$
• D. None of these

Answer 1

Answer: (A)

For cylinder, $I=\frac{1}{2}M{R}^2=1MR2$
So, $\frac{K^2}{R^2}=\frac{1}{2}$
So. acceleration $a=\frac{g\sin \theta }{1+\frac{K^2}{R^2}}$
$=\frac{g\sin 60^{\circ }}{1+\frac{1}{2}}=\frac{g}{\sqrt{3}}$