A solid sphere ($I = frac{2}{5}MR^2$) rolls without slipping down a …

Question

A solid sphere ($I = 	frac{2}{5}MR^2$) rolls without slipping down a plane inclined at $	heta$. Its **linear acceleration** is:

Accepted Answer

C. $	frac{5}{7} g\sin	heta$

$a = g\sin	heta /(1 + I/MR^2) = g\sin	heta /(1 + 2/5) = (5/7) g\sin	heta$.

Answer

$g\sin	heta$

Answer

$	frac{2}{7} g\sin	heta$

Answer

$	frac{1}{2} g\sin	heta$

A solid sphere ($I = \tfrac{2}{5}MR^2$) rolls without slipping down a plane inclined at $\theta$. Its linear acceleration is: