Question

A steel bicycle wheel (without the rubber tire) is rotating freely with an angular speed of 11.2 rad/s. The temperature of the wheel changes from -122 to 243 °C. No net external torque acts on the wheel, and the mass of the spokes is negligible. What is the angular speed of the wheel at the higher temperature?

Answer 1

here,

For thermal expansion consider the wheel as band of length

L0 = 2 * pi * r0

Assuming constant expansion coefficient (alpha) its length changes to

L = L0 * (1 + alpha * dT)

<=>

2 * pi * r = 2 * pi * r0 * (1 + alpha * dT)

Hence the radius changes to:

r = r0 * (1 + alpha * dT)

Moment of inertia of the hot wheel is:

J = m * r^2

= m * r0^2 * (1 + alpha * dT)^2

= J0 * (1 + alpha * dT)^2

Angular momentum of hot and cold wheel are equal

w * J = w0 * J0

hence:

w = w0 * J0/J

w = w0 / (1 + alpha * dT)^2

w = 11.2 rad/s / (1 + 1.2 * 10^-6 * (243 - (-122)))^2

w = 11.19 rad/s

the new angular speed is 11.19 rad/s