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

Let M be the mass and R the radius of the hollow cylinder, and also of the solid sphere. Their moments of inertia about the respective axes are I₁ = MR² and I₂ = 2/5 MR²

Let τ be the magnitude of the torque applied to the cylinder and the sphere, producing angular accelerations α₁and α₂ respectively. Then τ=I₁ α₁ = I₂ α₂

The angular acceleration 04 produced in the sphere is larger. Hence, the sphere will acquire larger angular speed after a given time.

Answer 2

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 = mr^2`

The moment of inertia of the solid sphere about an axis passing through its centre,` I_n = 2/5 mr^2`

We have the relation:

`t = Ialpha`

Where

`alpha`  = Angular acceleration

τ = Torque

I = Moment of inertia

For the hollow cylinder, `t_1 = I_1alpha1`

For the solid sphere, `t_n = I_nalpha_n`

As an equal torque is applied to both the bodies,` t_1 = t_2`

`:.(alpha_II)/(alpha_I) = (I_I)/(I_"II") = (mr^2)/(2/5mr^2) =2/5`

`alpha_(II) > alpha_I` ...(i)

Now, using the relation:

`omega = omega_0 + alpha t`

where

`omega_0` = Initial angular velocity

t = Time of rotation

ω = Final angular velocity

For equal ω₀ and t, we have:

ω ∝ α … (ii)

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

ω_II > ω_I

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