Question

A heavy solid sphere is thrown on a horizontal rough surface with initial velocity u without rolling what will be its speed when it starts pure rolling motion ?

Answer 1

Initial Kinetic energy = $\frac{1}{2}m{u}^2$
Total final Kinetic energy when the sphere starts rolling is
$K_{tot}=\frac{1}{2}I{\omega }^2+\frac{1}{2}m{v}^2$
In pure rolling
$v=r\omega$
Hence
$K_{tot}=\frac{1}{2}m{v}^2\left[1+\frac{k^2}{R^2}\right]$
For solid sphere
$\left[\frac{k^2}{R^2}\right]=\frac{2}{5}$
Hence
$K_{tot}=\frac{7}{10}m{v}^2$
Thus by law of conservation of energy
$$\begin{gathered} \frac{1}{2}m{u}^2=\frac{7}{10}m{v}^2 \\ v=\sqrt{\frac{5}{7}}v \end{gathered}$$=