Question

A cord of negligible mass is wound round the rim of a fly wheel of mass 24 kg and radius
24 cm. A steady pull of 30 N is applied on the cord. The flywheel is mounted on a
horizontal axle with frictionless bearings.
(a) Compute the angular acceleration of the wheel.
(b) Find the work done by the pull, when 2m of the cord is unwound.
(c) Find also the kinetic energy of the wheel at this point. Assume that the wheel starts
from rest.
(d) Compare answers to parts (b) and (c).

Answer 1

Answer:

Let us assume the wheel is a ring.

Then its moment of inertia ( I )= MR²

Given mass= 24 kg, radius= 0.24 m

Moment of inertia = 24 × 0.24× 0.24 = 1.38 kg m²

Now force= tension= 30 N

a) Angular acceleration (α)= ?

Torque= Iα

We know that torque is also equal to (force× radius vector)= r×F

Therefore, r×F = Iα

Or           30× 0.24= 1.38 α

Or           α= 5.22 rad/ s²

b) Work done= force × distance covered

                      = 30 × 2= 60 J

c) Kinetic energy = 1/2×I×ω²

   ω²= ω$_{0}$² + 2αФ ( using the major rotatory motion equation )

Here, angular displacement (Ф)= distance covered/ radius= 2/0.24= 8.3 rad

We already know initially angular velocity= 0

Therefore

ω²= 2αФ= 2× 5.2× 8.3= 86.3 rad/s

Now, KE= 1/2 × 1.38 × 86.3 J

              = 59.5 J ≈ 60 J

The answers to (b) and (c) are almost equal therefore we conclude that work done= change in kinetic energy