A track is mounted on a large wheel that is free to turn with neigligible friction about a vertical axis (Fig). A toy train of mass `M` is placed on the track and, with the system initially at rest, the train's electrical power is turned on. The train reaches speed `v` with respect to the track. What is the wheel's angular speed if its mass is `m` and its radius is `r` ? (Treat it as a hoop, and neglect the mass of the spokes and hub).
.

Options

A. `(v)/((M//m + 1)R)`
B. `(v)/((m//M + 2)R)`
C. `(v)/((M//m + 2)R)`
D. `(v)/((m//M + 2)R)`

Answer

B

Explanation

(b) No external torque acts on the system consisting of the train and wheel, so the total angular momentum of the system (which is initially zero) remains zero.
Let `I = mR^2` be the rotational inertia of the wheel (which we treat as a hoop). Its angular momentum is
`vec L_(wheel) = (I omega)hat k = - mR^2|omega|hat k`
Where `hat k` is up in figure and that last step (with the minus sign) is done in recognition that the wheel's clockwise rotation implies a negative vlaue for `omega`. The linear speed of point on the track is `-|omega| R` and the speed of the train (going counterclockwise in figure with the speed `v'` relative to an outside observer) is therefore `v' = v - |omega|R` where `v` is its speed relative to the tracks. Consequently, the angular momentum of the train is `vec L_(train) = M(v - |omega|R) R hat k`. Conservation of angular momentum yields.
`0 = vec L_("wheel") + vec L_("train") = - mR^2|omega|hat k + M (v-|omega|R) R hat k`
which we can use to solve for `|omega|`
Solving for the angular speed, the result is
`|omega| = (mvR)/((M + m)R^2) = (v)/((m//M + 1)R)`.]