In bicycling, each foot pushes on the pedal for half a rotation of the pedal shaft; that foot then rests and the other foot takes over. During each half-cycle, the torque resulting from the force of the active foot is given approximately by τ = τ0 sin ωt, where τ0 is the maximum torque and ω is the angular speed of the pedal shaft (in s-1 as usual). A particular cyclist is turning the pedal shaft at ω=70.0 rpm, and at the same time τ0 is measured at 38.5 N•m.
(a) the energy supplied by the cyclist in one turn of the pedal shaft and
(b) the cyclist's average power output.
W = τθ, with θ the angular displacement;
W= ΔKE = Energy supplied by the cyclist
The Attempt at a Solution
For (a) I thought of using the work-energy theorem the following way:
- θ=2π rad=one revolution
So I get E = 38.5*sin(ω*2π/ω)*2π = 0 !
This is not really what I expect....
For (b) there may be a rotational version of P=F*v that is P=τω.
Thanks for your help