Question

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

Answer 1

(a)

Initial angular momentum about point A,

$L=m(v\times R)-I\omega$

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

$=\frac{3}{5}mvR$

Angular momentum about point A' after it stops rotating,

$L^{\prime }=m{v}^{\prime }R$

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

Therefore, we have:

$L=L^{\prime }$

$⟹mvR=\frac{5}{3}m{v}^{\prime }R$

$v^{\prime }=\frac{3v}{5}$

(b)

Angular momentum about point A after it stops rotating,

$L^{\prime }=m{v}^{\prime }R$

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

$L"=I\omega "+m(v"\times R)$

$=\frac{2}{5}m{R}^2(v"R)+mv"R$

$=\frac{7}{5}mvR$

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

Therefore, we have:

$L^{\prime }=L"$

$⟹m\frac{3v}{5}R=\frac{7}{5}mv"R$

$v"=\frac{3v}{7}$