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 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, I₁ = mr²
The moment of inertia of the solid sphere about an axis passing through its centre, I_II = \(\frac{2 }{ 5}\) mr²
We have the relation: π = Ia
Where, α = Angular acceleration
τ = Torque
I = Moment of inertia
For the hollow cylinder, t_I = I₁a_I
For the solid sphere, t_II = I_II a_II
As an equal torque is applied to both the bodies, π₁ = π₂

\(∴ \frac{a_{Il}}{ a_I} = \frac{I_I}{I_{II}} = \frac{mr^2 }{ \frac{2 }{ 5} mr^2 }= \frac{2 }{ 5}\)

a_II > a_I ....(i)

Now, using the relation : 

ω = ω₀ + at

Where, 

ω0 = 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.