A solid sphere is set into motion on a rough horizontal surface with a linear speed ν in the forward direction and an angular speed ν/R in the anticlockwise directions as shown in the following figure. Find the linear speed of the sphere (a) when it stops rotating and (b) when slipping finally ceases and pure rolling starts.

#### Solution

(a)

Initial angular momentum about point A,

\[L = m\left( v \times R \right) - I\omega\]

\[ = mvR - \frac{2}{5}m R^2 \left( \frac{v}{R} \right)\]

\[ = \frac{3}{5}mvR\]

Angular momentum about point A' after it stops rotating,

\[L' = mv'R\]

As no external torque is applied, angular momentum will be conserved.

Therefore, we have

`L = L'`

\[\Rightarrow mvR = \frac{5}{3}mv'R\]

\[ \Rightarrow v' = \frac{3v}{5}\]

(b)

Angular momentum about point A after it stops rotating,

\[L' = mv'R\]

Angular momentum about point A' after it starts pure rolling.

\[L'' = I\omega'' + m\left( v'' \times R \right)\]

\[ = \frac{2}{5}m R^2 \left( \frac{v''}{R} \right) + mv'' R\]

\[ = \frac{7}{5}mVR\]

As no external torque is applied, angular momentum will be conserved.

Therefore, we have

\[L' = L''\]

\[\Rightarrow m\frac{3V}{5}R = \frac{7}{5}mv'' R\]

\[ \Rightarrow v'' = \frac{3v}{7}\]