Torques of equal magnitude are applied to a hollow cylinder and a solid sphere,
Question:

Torques of equal magnitude 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 centre. Which of the two will acquire a greater angular speed after a given time?

Solution:

Let m and r be the respective masses of the hollow cylinder and the solid sphere.

The moment of inertia of the hollow cylinder about its standard axis, $I_{1}=m r^{2}$

The moment of inertia of the solid sphere about an axis passing through its centre, $I_{11}=\frac{2}{5} m r^{2}$

We have the relation:

$\tau=I \alpha$

Where,

$\alpha=$ Angular acceleration

$\mathrm{T}=$ Torque

$I=$ Moment of inertia

For the hollow cylinder, $\tau_{1}=I_{1} \alpha_{1}$

For the solid sphere, $\tau_{\|}=I_{\|} \alpha_{\|}$

As an equal torque is applied to both the bodies, $\tau_{1}=\tau_{2}$

$\therefore \frac{\alpha_{\mathrm{II}}}{\alpha_{\mathrm{I}}}=\frac{I_{1}}{I_{\mathrm{II}}}=\frac{m r^{2}}{\frac{2}{5} m r^{2}}=\frac{2}{5}$

$\alpha_{\mathrm{n}}>\alpha_{1}$      $\ldots(i)$

Now, using the relation:

$\omega=\omega_{0}+\alpha t$

Where,

$\omega_{0}=$ Initial angular velocity

$t=$ Time of rotation

$\omega=$ Final angular velocity

For equal $\omega_{0}$ and $t$, we have:

$\omega \propto \alpha \ldots(i i)$

From equations (i) and (ii), we can write:

$\omega_{\|}>\omega_{1}$

Hence, the angular velocity of the solid sphere will be greater than that of the hollow cylinder.

 

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