A solid sphere is pushed on a horizontal surface such that it slides with a speed $3.5 \mathrm{~ms}^{-1}$…
A solid sphere is pushed on a horizontal surface such that it slides with a speed $3.5 \mathrm{~ms}^{-1}$ initially without rolling. The sphere will start rolling without slipping when its velocity becomes
$2.5 \mathrm{~ms}^{-1}$
$5 \mathrm{~ms}^{-1}$
$3.5 \mathrm{~ms}^{-1}$
$7 \mathrm{~ms}^{-1}$
Solution
Solid sphere with velocity, $v_1=3.5 \mathrm{~m} / \mathrm{s}$ By conservation of angular momentum
$\begin{aligned}
& L_i=L_f \\
& m_1 R=I \omega+m_2 R
\end{aligned}$
after some time sphere starts pure rolling
$\mathrm{v}_2=\mathrm{R} \omega$
Moment of inertia of solid sphere, $\mathrm{I}=\frac{2}{5} \mathrm{mR}^2$
$\begin{aligned}
& \mathrm{mv}_1 \mathrm{R}=\frac{2}{5} \mathrm{mR}^2 \times \frac{\mathrm{v}_2}{\mathrm{R}}+\mathrm{mv}_2 \mathrm{R} \\
& \mathrm{v}_1=\frac{2}{5} \mathrm{v}_2+\mathrm{v}_2 \\
& \Rightarrow \mathrm{v}_2=\frac{5 \mathrm{v}_1}{7}=\frac{5}{7} \times 3.5=2.5 \mathrm{~m} / \mathrm{s}
\end{aligned}$