# Q7.11     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.

The moment of inertia of hollow cylinder is given by   $=\ mr^2$

And the moment of inertia of solid cylinder is given by :

$=\ \frac{2}{5}mr^2$

We know that :                $\tau \ =\ I\alpha$

Let the torque for hollow cylinder be $\tau_1$ and for solid cylinder let it be $\tau_2$.

According to the question :                 $\tau_1\ =\ \tau_2$

So we can write the ratio of the angular acceleration of both the objects.

$\frac{\alpha _2}{\alpha _1}\ =\ \frac{I_1}{I_2}\ =\ \frac{mr^2}{\frac{2}{5}mr^2}\ =\ \frac{2}{5}$

Now for angular velocity,

$\omega \ =\ \omega _o\ +\ \alpha t$

Clearly, the angular velocity of the solid sphere is more than the angular velocity of the hollow sphere. (As the angular acceleration of solid sphere is greater).

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