A solid sphere rolls down without slipping on an inclined plane at angle $60^{{^\circ}}$ over a distance of 10 m. The acceleration (in ${ms}^{- 2}$) is

Question

Dr. Abubacker · Accepted Answer

(c) Here $	heta = 60{^\circ},\ l = 10\ m,\ a = ?$
$a = \frac{g\sin	heta}{1 + \frac{K^{2}}{R^{2}}}$
For solid sphere $K^{2} = \frac{2}{5}R^{2}$
$a = \frac{9.8\sin{60{^\circ}}}{1 + \frac{2}{5}} = \frac{5}{7} 	imes 9.8 	imes \frac{\sqrt{3}}{2} = 6.00\ ms^{- 2}$

Q. A solid sphere rolls down without slipping on an inclined plane at angle $60^{{^\circ}}$ over a distance of 10 m. The acceleration (in ${ms}^{- 2}$) is

Explanation

Reviewed by Dr. Abubacker

Want more practice?