Question

A force $F=2\text{N}$ acts at a distance $x=0.5\text{m}$ above the centre of a thin spherical shell of mass $m=3\text{kg}$ and radius $R=2\text{m}$. Find out the linear acceleration $\left(a\right)$ and angular acceleration $\left(\alpha \right)$ of the shell.

• A. $a=\frac{1}{3}{\text{m/s}}^2;\alpha =\frac{3}{8}{\text{rad/s}}^2$
• B. $a=\frac{2}{3}{\text{m/s}}^2;\alpha =\frac{1}{16}{\text{rad/s}}^2$
• C. $a=\frac{2}{3}{\text{m/s}}^2;\alpha =\frac{1}{8}{\text{rad/s}}^2$
• D. $a=\frac{1}{4}{\text{m/s}}^2;\alpha =\frac{2}{3}{\text{rad/s}}^2$

Answer 1

The correct option is C $a=\frac{2}{3}{\text{m/s}}^2;\alpha =\frac{1}{8}{\text{rad/s}}^2$


Given, $F=2\text{N},x=0.5\text{m},m=3\text{kg}$ and $R=2\text{m}$
From F.B.D.
Along $y-$ axis, $N=mg$
Along $x-$ axis, $F=m{a}_{COM}$
$\therefore a_{COM}=a=\frac{F}{m}{\text{m/s}}^2=\frac{2}{3}{\text{m/s}}^2$

Now, net torque about $O$ is;
${\tau }_o=F.x$
$\Rightarrow {\tau }_o=2\times 0.5\text{N-m}=1\text{N-m}$

We know that,
$\tau =I\alpha$ or $\alpha =\frac{\tau }{I}$
and $I$ for spherical shell is $\frac{2m{R}^2}{3}$
$\therefore \alpha =\frac{{\tau }_o}{I}=\frac{1}{\frac{2\times 3\times 2^2}{3}}=\frac{1}{8}{\text{rad/s}}^2$
$\therefore$ option (c) is correct.