Question

A tangential force $F$ acts at the top of a thin spherical shell of mass $m$ and radius $R$. The acceleration of the shell if it rolls without slipping is
$(f=$ rolling friction $)$

• A. $\frac{5F}{6m}$
• B. $\frac{6F}{5m}$
• C. $\frac{3F}{2m}$
• D. $\frac{F}{6m}$

Answer 1

Answer: (B)

Torque due to the force $F$ on a thin spherical shell,
$\tau =r\times F$
$=2RF\sin 90^{\circ }=2RF\left[\because \sin 90^{\circ }=1\right]$
$\because$ Angular acceleration,
$\alpha =\frac{\tau }{I}....(i)$
Where, $I$ is moment of inertia of a thin spherical shell.
From parallel axes's theorem, moment of inertia of spherical shell,
$I=I_{cm}+M{r}^2$
or $I=\frac{2}{3}M{R}^2+M{R}^2(\because r=R)$
From Eqs. (i), we get
$\alpha =\frac{2RF}{\frac{2}{3}M{R}^2+M{R}^2}$
$=\left(\frac{6}{5}\right)\frac{F}{RM}$
Hence, the tangential acceleration, $a_T=R\alpha$
Or $a_T=\left(\frac{6}{5}\right)\frac{F}{RM}\times R=\frac{6}{5}\frac{F}{M}$