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?
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 = II / 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.
NCERT questions are designed to test your understanding of the concepts and theories discussed in the chapter. Here are some tips to help you answer NCERT questions effectively:
Welcome to the NCERT Solutions for Class 11 Physics - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 11: Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same....
Comments