First slide

Angular momentum

Question

A track is mounted on a large wheel that is free to tum with negligible friction about a vertical axis (Fig). A toy train of mass Mis 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.)

Moderate

A

$\frac{v}{(\frac{M}{m + 1}) R}$
B

$\frac{v}{(\frac{M + m}{M}) R}$
C

$\frac{v}{(\frac{M}{m + 2}) R}$
D

$\frac{v}{(\frac{m}{M + 2}) R}$

Solution

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
$\vec{L_{wheel}} = (Iω) \hat{k} = - {mR}^{2} | ω | \hat{k}$
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 $ω$ . The linear speed of a point on the track is -| $ω$ |R and the speed of the train (going counterclockwise in figure with speed v'relative to an outside observer) is therefore $v^{'} = v - | ω | R$ where v is its speed relative to the tracks. Consequently, the angular momentum of
the train is $\vec{L_{train}} = M (v - | ω | R) R \hat{k}$ . Conservation of angular momentum yields
$0 = \vec{L_{wheel}} + \vec{L_{train}} = - {mR}^{2} | ω | \hat{k} + M (v - | ω | R) R \hat{k}$
which we can use to solve for | $ω$ |.
Solving for the angular speed, the result is
$| ω | = \frac{MvR}{(M + m) R^{2}} = \frac{v}{(\frac{m}{M} + 1) R}$

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