A solid sphere and hollow sphere of the same mass and radius are given a spin about their centre of mass and then, they are placed on a rough horizontal surface. The spin angular velocity is the same for both the spheres and it is equal to ${\mathrm{\omega }}_0$. Once the pure rolling starts; let $v_1$ and $v_2$ be the linear speeds of their centres of mass, then

• A

  $v_1=v_2$
• B

  $v_1>v_2$
• Correct answer
  C

  $v_1<v_2$
• D

  Data is insufficient

Solution

From conservation of angular momentum about point of contact :



              $\mathit{\text{I}}{\mathrm{\omega }}_0=\mathit{\text{I}}\mathrm{\omega }+\mathit{\text{mRv}}$
or           $\mathit{\text{I}}{\mathrm{\omega }}_0=\mathit{\text{I}}\frac{\mathit{\text{v}}}{\mathit{\text{R}}}+\mathit{\text{mRv}}$
or               $\mathit{\text{v}}=\frac{\mathit{\text{I}}{\mathrm{\omega }}_0}{\frac{\mathit{\text{I}}}{\mathit{\text{R}}}+\mathit{\text{mR}}}$
or 
or               $\mathit{\text{v}}=\frac{{\mathrm{\omega }}_0}{\frac{1}{\text{R}}+\frac{\mathit{\text{mR}}}{\mathit{\text{I}}}}$
Now ${\mathit{\text{I}}}_{\text{solid}\text{sphere <}}{\mathit{\text{I}}}_{\text{hollow}}$
∴             ${\mathit{\text{v}}}_{\text{solid}<}{\mathit{\text{v}}}_{\text{hollow}}$
∴                  ${\mathit{\text{v}}}_1<{\mathit{\text{v}}}_2$

Hence, the correct answer is option: 'C'.