Question

A tangential $F$ acts at the top of a thin spherical shell of mass $m$ and radius $R$. Find the acceleration of the shell, if it rolls without slipping.

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

Answer 1

The correct option is A

$\frac{6F}{5m}$
Let $f$ be the force of friction between the shell and the horizontal surface.



For translational motion from figure,

$F+f=ma....\left(1\right)$

For rotational motion, net torque

$F\times R-f\times R=I\alpha =I\frac{a}{R}$

$[\because a=R\alpha ]$ for pure rolling

$\Rightarrow F-f=I\frac{a}{R^2}....\left(2\right)$

Adding equation $\left(1\right)$ and $\left(2\right)$, we get

$2F=(m+\frac{I}{R^2})a$

Substituting $I=\frac{2}{3}m{R}^2$ for spherical shell, in the above equation,

$\Rightarrow 2F=(m+\frac{2}{3}m)a$

$\Rightarrow F=\frac{5}{6}ma$

$\therefore a=\frac{6F}{5m}$

Hence, option (a) is correct answer.

Why this question? This question checks the understanding of rolling, Newton's second law in rotational world and torque due to internal forces.

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