Question

A 23.4kg solid aluminum cylindrical wheel of radius 0.41 m is rotating about its axle on frictionless bearings with angular velocity ω = 32.8 rad/s. If its temperature is now raised from 20.0oC to 75.0oC, what is the fractional change in ω?

but my friend from the other class said that this question is wrong...idk

Answer 1

To me the question appears in order.

Explanation:

Angular momentum `L`, (considering only the magnitudes of vector quantities)

`L = Iomega`
where `I` is moment of inertia and `omega` angular velocity.

Therefore, for a fixed angular momentum

`omegapropI^-1`

We know that moment of inertia about the central axis of a solid cylinder of mass `m` and of radius `R` is given by

`I=1/2mR^2`

Therefore,

`omegaprop 1/R^2` .....(1)

With the increase of temperature radius of the cylindrical wheel will increase. Using the expression

`R_T = R_0*(1+alpha*ΔT)`
where `alpha=23.1xx10^-6\ K^-1` is Coefficient of linear expansion for aluminum.

`:.R_75=R_20(1+23.1xx10^-6xx55)`
`=>R_75=R_20(1+1.27xx10^-3)` ......(2)

As heating of the wheel is not going to change the angular momentum, therefore, using Law of Conservation of angular momentum; from (1) we have the expression

`omega_75 = omega_20*(R_20/R_75)^2` ......(3)

Fractional change in `omega = (omega_75-omega_20)/omega_20`
`=>`Fractional change in `omega = omega_75/omega_20-1`

Using (3) we get
Fractional change in `omega = (R_20/R_75)^2-1`

Using (2) we get
Fractional change in `omega = (1/(1+1.27xx10^-3))^2-1`
`=>`Fractional change in `omega=-0.0025`