Question

A track is mounted on a large wheel that is free to turn with negligible 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.)

• A. $\frac{v}{(M/m+1)R}$
• B. $\frac{v}{(m/M+1)R}$
• C. $\frac{v}{(M/m+2)R}$
• D. $\frac{v}{(m/M+2)R}$

Answer 1

Answer: (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 $I=m{R}^2$ be the rotational inertia of the wheel (which we treat as a hoop). Its angular momentum is
${\bar{L}}_{\text{wheel }}=(I\omega )\hat{k}=-m{R}^2|\omega |\hat{k}$
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 value for $\omega .$ The linear speed of a point on the track is $-|\omega |R$ and the speed of the train (going counterclockwise in figure with speed $v$ 'relative to an outside observer) is therefore $v^{\prime }=v-|\omega |R$ where $v$ is its speed relative to the tracks. Consequently, the angular momentum of the train is ${\vec{L}}_{\text{train }}=M(v-|\omega |R)R\hat{k}$. Conservation of angular momentum yields
$0={\vec{L}}_{\text{wheel }}+{\vec{L}}_{\text{train }}=-m{R}^2|\omega |\hat{k}+M(v-|\omega |R)R\hat{k}$
which we can use to solve for $|\omega |$. Solving for the angular speed, the result is
$|\omega |=\frac{MvR}{(M+m)R^2}=\frac{v}{(m/M+1)R}$