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.

Answer 1

Consider, m and r be the masses and radius of both the hollow cylinder and solid sphere respectively.

The expression for the moment of the inertia of the hollow cylinder I 1 about its principle axis is,

I 1 =m r 2

The expression for the moment of the inertia of the solid sphere I 2 about axis passing through centre is,

I 2 = 2 5 m r 2

The expression of the torque τ is,

τ=Iα

Here, I is the moment of inertia and α is the angular acceleration.

The expression of the torque for the hollow cylinder τ 1 is,

τ 1 = I 1 α 1

Here, I 1 is the moment of inertia of hollow cylinder and α 1 is the angular acceleration.

The expression of the torque for the solid sphere τ 2 is,

τ 2 = I 2 α 2

Here, I 2 is the moment of inertia of the solid sphere and α 2 is the angular acceleration.

Equal torque is applied to both the bodies,

τ 1 = τ 2 α 2 α 1 = I 1 I 2 = m r 2 2 5 m r 2 = 2 5

Above relation shows that α 2 > α 1 .

The expression for the angular velocity ω is,

ω= ω 0 +αt

Here, ω 0 is the initial angular velocity and tis the time of the rotation.

For same angular velocities and time of rotation above equation shows that

ω∝α ω 1 ω 2 = α 1 α 2 ω 1 ω 2 = 5 2

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