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 it stops rotating and (b) when slipping finally ceases and pure rolling starts.

Answer 1

(a) If we take moment at A then external torque will be zero. Therefore, the initial angular momentum = the angular momentum after rotation stop (i.e. only linear velocity exist.)

$MV\times R-\left(\frac{2}{5}\right)I\times w=M{V}_0\times R$

$\Rightarrow MVR-\frac{2}{5}\times M{R}^2\frac{V}{R}=M{V}_{0}R$

$\Rightarrow V_0=\frac{3v}{5}$

(b) Again, after some time pure rolling starts.

$\Rightarrow Mx{V}_0\times R=\left(\frac{2}{5}\right)M{R}^2\times \left(VR\right)+M{v}^{\prime }R$

$\Rightarrow m\times \frac{3V}{5}\times R=\left(\frac{2}{5}\right)MVR+MvR$

$\Rightarrow V=\frac{3V}{7}$