Question

A solid sphere of mass $1\text{kg}$ and radius $2\text{m}$ slips on a rough horizontal plane. At some instant, it has translational velocity $7\text{m/s}$ and angular velocity $\frac{7}{4}\text{rad/s}$ about its centre. The translational velocity after the sphere starts pure rolling is

• A. $4\text{m/s}$
• B. $4.52\text{m/s}$
• C. $6\text{m/s}$
• D. $5.35\text{m/s}$

Answer 1

The correct option is C $6\text{m/s}$


About point of contact $O$, there is no torque on the system as torque due to kinetic friction ${\tau }_{f_k}=0$, hence angular momentum will be conserved about point of contact $O$.
$L_i=L_f$
Taking anticlockwise sense of rotation as $+ve$,
$-\left(m{V}_{0}R\right)-\left(I{\omega }_0\right)=-mVR-\left(I\omega \right)$
$\Rightarrow m{V}_{0}R+I{\omega }_0=mVR+I\omega ...\left(i\right)$
When sphere starts pure rolling,
$V=\omega R$
$\Rightarrow \omega =\frac{V}{R}...\left(ii\right)$
Substituting Eq $\left(ii\right)$ in $\left(i\right)$,
$m{V}_{0}R+\left(\frac{2}{5}m{R}^2\right){\omega }_0=mVR+\left(\frac{2}{5}m{R}^2\right)\times \left(\frac{V}{R}\right)$
$\Rightarrow 1\times 7\times 2+(\frac{2}{5}\times 1\times 2^2)\times \frac{7}{4}=1\times V\times 2+(\frac{2}{5}\times 1\times 2^2)\times \frac{V}{2}$
$\Rightarrow \frac{84}{5}=\frac{14V}{5}$
$\therefore V=6\text{m/s}$