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## Homework Statement

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.

Find:

(a) the energy supplied by the cyclist in one turn of the pedal shaft and

(b) the cyclist's average power output.

## Homework Equations

__Work:__

W = τθ, with θ the angular displacement;

__Work-energy theorem:__

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:

E=τ0 sin(ωt)θ

with:

- θ=2π rad=one revolution

- t=θ/ω

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