Question

A tangential force 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.

Answer 1

Given: a tangential force F acts at the top of a thin spherical shell of mass m and radius R.

To find the acceleration of the shell if it rolls without slipping.

Solution:

Let the distance of point of application of force from center is equal to Radius (R)

If sphere rolls without slipping:

$\alpha =\frac{a}{R}$

where $\alpha$ is angular acceleration, $a$ is acceleration of the center of the sphere.

Newton's law of motion:

$(F+F_{fr})R=I\alpha .......\left(i\right)$

and $F-F_{fr}=ma..........\left(ii\right)$

where $F_{fr}$ is force of friction, $I$ is moment of inertia (for solid sphere, $I=\frac{2}{5}m{R}^2$)

$⟹F-F_{fr}=\frac{I}{R}\times \frac{a}{R}$

Substituting the values of $a$ and $I$ in eqn(i), we get

$F+F_{fr}=\frac{2}{5}m{R}^2\times \frac{a}{R^2}⟹F+F_{fr}=\frac{2}{5}ma.......\left(iii\right)$

Adding eqn(ii) and eqn(iii), we get

$F+F_{fr}=\frac{2}{5}ma$

$\underset{–}{F-F_{fr}=ma}$

$\therefore ,2F=\frac{2ma}{5}+ma⟹2F=\frac{7}{5}ma⟹F=\frac{7}{10}ma⟹a=\frac{10F}{7m}$

is the acceleration of the shell if it rolls without slipping.