Question

A solid sphere is at rest on a horizontal surface. A horizontal impulse $J$ is applied at a height $h$ from the centre. The sphere starts rolling just after the application of impulse. The ratio $\frac{h}{R}$ will be

• A. $\frac{2}{5}$
• B. $\frac{1}{3}$
• C. $\frac{3}{5}$
• D. $\frac{5}{2}$

Answer 1

The correct option is A $\frac{2}{5}$

We can observe Rolling motion as pure rotational motion about Instantaneous axis of rotation passing through point $\text{P}$.

Applying Impulse - momentum theorem about $\text{P}$:
$J=\Delta P=mv.....\left(1\right)$
The Angular momentum due to Impulse $J$ about point $\text{P}$
$J\times (R+h)=I_p\omega .......\left(2\right)$
Friction cannot provide the impulsive force.
From the conditon of pure rolling : $v=R\omega .......\left(3\right)$
Moment of inertia of solid sphere about tangential axis of rotation $I_p=\frac{2}{5}m{R}^2+m{R}^2=\frac{7}{5}m{R}^2$
Substituting this in $\left(2\right)$ we get,
$J\times (R+h)=\frac{7}{5}m{R}^2\omega$
Using $\left(3\right)$ in the above equation we get,
$J\times (R+h)=\frac{7}{5}mvR.....\left(4\right)$
From $\left(2\right)$ and $\left(4\right)$ we can write that,
$R+h=\frac{\frac{7}{5}mvR}{mv}\Rightarrow R+h=\frac{7R}{5}$ $\Rightarrow \frac{h}{R}=\frac{2}{5}$
Thus, option (a) is the correct answer.