Question

If torques of equal magnitudes are applied to a hollow cylinder and a solid sphere both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry and the sphere is free to rotate about an axis passing through its center. Which of the two will acquire a greater angular speed after a given time?

• A. ${\omega }_1>{\omega }_2$
• B. ${\omega }_1={\omega }_2$
• C. ${\omega }_2>{\omega }_1$
• D. None of these

Answer 1

Answer: (C)

${\omega }_2>{\omega }_1$

Consider $I_1$ and $I_2$ be the moments of inertia of the hollow cylinder and solid sphere about its axis through its center respectively.
Then, $I_1=M{R}^2$ ....(i)
and $I_2=\frac{2}{5}M{R}^2$ ....(ii)
Let $\tau$ be the magnitude of the torque applied on each of them. If ${\alpha }_1$ and ${\alpha }_2$ are the angular accelerations produced in the cylinder and sphere respectively, then
$\tau =I_1{\alpha }_1$
and $\tau =I_2{\alpha }_2$
$\therefore I_1{\alpha }_1=I_2{\alpha }_2\Rightarrow \frac{{\alpha }_1}{{\alpha }_2}=\frac{I_2}{I_1}$
$=\frac{\frac{2}{5}M{R}^2}{M{R}^2}=\frac{2}{5}\Rightarrow {\alpha }_2=\frac{5}{2}{\alpha }_1$
$\Rightarrow {\alpha }_2=2.5{\alpha }_1$ ....(iii)
If ${\omega }_1$ and ${\omega }_2$ be the angular speed of the cylinder and sphere after time $t$, then
${\omega }_1={\omega }_0+{\alpha }_{1}t$ ....(iv)
and ${\omega }_2={\omega }_0+{\alpha }_{2}t$
$={\omega }_0+2.5{\alpha }_{1}t$ .....(v)
where, ${\omega }_0=$ initial angular speed
$\therefore$ From equation (iv) and (v), it is clear that
${\omega }_2>{\omega }_1$
$\therefore$ The sphere will acquire more angular speed as compared to that of the cylinder after a given time.