Class 11 Physics - Chapter System of Particles and Rotational Motion NCERT Solutions | Torques of equal magnitude are applied t

Welcome to the NCERT Solutions for Class 11th Physics - Chapter System of Particles and Rotational Motion. This page offers a step-by-step solution to the specific question from Exercise 1, Question 11: torques of equal magnitude are applied to a hollow....
Question 11

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

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,II = mr2

The moment of inertia of the solid sphere about an axis passing through its centre, III  =  2/5 mr2

We have the relation:

τ  =  I α

Where, α = Angular acceleration

τ = Torque

I = Moment of inertia

For the hollow cylinder, τI   =  II αI

For the solid sphere, τII  = III αII

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

∴ αII / αI   I/ III    =  mr2 /  2/5 mr2    =  2/5

αII > αI              ....  (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:

ωII > ωI

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

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