Question

A solid sphere is set into motion on a rough horizontal surface with a linear speed $\nu$ in the forward direction and an angular speed $\nu$/R in the anticlockwise directions as shown in figure (10-E16). Find the linear speed of the sphere (a) when it stops rotating and (b) when slipping finally ceases and pure rolling starts.

Figure

Answer 1

(a)

Initial angular momentum about point A,
$$\begin{gathered} L=m\left(v\times R\right)-I\omega \\ =mvR-\frac{2}{5}m{R}^2\left(\frac{v}{R}\right) \\ =\frac{3}{5}mvR \end{gathered}$$
Angular momentum about point A' after it stops rotating,
$L\text{'}=mv\text{'}R$

As no external torque is applied, angular momentum will be conserved.
Therefore, we have:
L = L'
$$\begin{gathered} \Rightarrow mvR=\frac{5}{3}mv\text{'}R \\ \Rightarrow v\text{'}=\frac{3v}{5} \end{gathered}$$

(b)

Angular momentum about point A after it stops rotating,
$L\text{'}=mv\text{'}R$
Angular momentum about point A' after it starts pure rolling.
$$\begin{gathered} L"=I\omega "+m\left(v"\times R\right) \\ =\frac{2}{5}m{R}^2\left(\frac{v"}{R}\right)+mv"R \\ =\frac{7}{5}mVR \end{gathered}$$
As no external torque is applied, angular momentum will be conserved.
Therefore, we have:
L' = L"
$$\begin{gathered} \Rightarrow m\frac{3V}{5}R=\frac{7}{5}mv"R \\ \Rightarrow v"=\frac{3v}{7} \end{gathered}$$