Q.

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.)

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a

v(Mm+1)R

b

v(M+mM)R

c

v(Mm+2)R

d

v(mM+2)R

answer is B.

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Detailed 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 = mR2 be the rotational inertia of the wheel (which we treat as a hoop). Its angular momentum is Lwheel→ = (Iω)k^ = -mR2|ω|k^where 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 ofthe train is Ltrain→ = M(v-|ω|R)Rk^. Conservation of angular momentum yields0 = Lwheel→+Ltrain→ = -mR2|ω|k^+ M(v-|ω|R)Rk^which we can use to solve for |ω|.Solving for the angular speed, the result is|ω| = MvR(M+m)R2= v(mM+1)R
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