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 be the radius of the solid sphere and also the hollow cylinder.
The moment of inertia of the hollow cylinder about its standard axis, \( I_1 = MR_2 \)

Moment of inertia of the solid sphere about an axis passing through its centre, \( I_2 = (2MR_2 ) / 5 \)

Let T be the magnitude of the torque being exerted on the two structures, producing angular accelerations of   α2 and α1  in the sphere and the cylinder, respectively.

Thus we have , \( T = I_1 \alpha _1= I_2\alpha_2 \)

∴ \( α_2 / α_1 = I_1 / I_2 = \frac{MR^{2}}{\frac{2}{5}MR^{2}} = 5/2 \)

\( α_2 > α_1 \)         . . . . . .( 1 )

Now, using the relation :

\( ω = ω_0 + αt \)

Where,

\( ω_0 \) = Initial angular velocity

t = Time of rotation

ω = Final angular velocity

For equal ω0 and t, we have:

ω ∝ α             . . . . . . .  . ( 2 )

From equations ( 1 ) and ( 2 ), we can write:

\( ω_2 > ω_1 \)

Thus, from the above relation, it is clear that the angular velocity of the solid sphere will be greater than that of the hollow cylinder.

Thus, from the above relation, it is clear that the angular velocity of the solid sphere will be greater than that of the hollow cylinder.