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

Answer:

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, I2 = (2/5)mr2

We have the relation:

τ = Iα

Where,

α = Angular acceleration

τ = Torque

I = Moment of inertia

For the hollow cylinder, τ1 = I1α1

For the solid sphere, τn = Inαn

As an equal torque is applied to both the bodies, τ1 = τ2

∴ α2 / α1 = I1 / I2 = mr2 / (2/5)mr2

α2 > α1 ...(i)

Now, using the relation:

ω = ω0 + αt

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:

ω2 > ω1

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

Answer 2

Answer:

Explanation:

Let angular accelerations produced by hollow cylinder and solid sphere are a1 and a2 respectively.

We know,

Torque = moment of inertia × angular acceleration

So,

Torque of hollow cylinder = torque of solid sphere

I1 × a1 = I2 × a2

Moment of inertia of hollow cylinder = mR²

Moment of inertia of solid sphere = (2/5)mR²

mR² × a1 = (2/5)mR² × a2

a2 = 2.5a1

After a given, time w1 and w2 be the angular speed of hollow cylinder and solid sphere respectively.

W1 = Wo + a1t

W2 = Wo + 2.5a1t

hence, w2 > w1

e.g solid sphere will acquire gretar angular speed after a given time .