# Energy and power supplied by a bicyclist

## 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

## Answers and Replies

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TSny
Homework Helper
Gold Member
Welcome to PF!
For (a) I thought of using the work-energy theorem the following way:
E=τ0 sin(ωt)θ
Since the torque depends on time, you will need to use calculus.

For (b) there may be a rotational version of P=F*v that is P=τω.
Yes, but this would give the instantaneous power. You want the average power.

welssen
Thank you!
Since the torque depends on time, you will need to use calculus.
So I apply the following integral from 0 to 2π:
W=∫τ0 sin(ωt)dθ
Then :
dθ/2π = dt/T (T:=total time=2π/ω) ⇒ dθ=ω dt
⇒ W = ∫τ0 sin(ωt)/ω dt = -τ0 cos(ωt)/(ω^2)
Evaluating the integral from 0 to 2π gives 0, again. Or should I integrate from 0 to π for the up-down movement?

TSny
Homework Helper
Gold Member
So I apply the following integral from 0 to :
W=∫τ0 sin(ωt)dθ
Note that the expression τ0 sin(ωt) only holds for half a cycle. (This expression would give a negative torque for the second half of the cycle. But this doesn't happen. Instead, the other foot takes over.)
Then :
dθ/2π = dt/T (T:=total time=2π/ω) ⇒ dθ=ω dt
⇒ W = ∫τ0 sin(ωt)/ω dt
How did ω get into the denominator?
Or should I integrate from 0 to π for the up-down movement?
Yes, integrate θ from 0 to π; or integrate t from 0 to the time corresponding to θ = π.

welssen
Very nice, thanks!

How did ω get into the denominator?
This was a mistake indeed, it's supposed to be in the nominator.

So just a recap for (a):
W = ∫τ0 sin(ωt)dθ = -τ0 cos(ωt) → from 0 to π gives → Supplied Energy = W = τ0 cos(ωt)
We must consider that ω in the cos term has unit s^(-1) so we multiply it by 2π to have it in rad/s.
Numerically,
ω=70 rpm = 7.33 rad/s
t=π/ω=0.429 s
So E = 38.5*cos(2π 7.33⋅0.429) = 23.68 J.

So the energy supplied for one turn is:
E = 47.36 J.

(b) The average power is the work difference per unit time which leaves:
P = ΔW/(2*Δt) = 47.36/(2*0.429) = 55.20 W.
(And here Δt is multiplied by 2 for both feet movements)

Hopefully this is correct.

TSny
Homework Helper
Gold Member
So just a recap for (a):
W = ∫τ0 sin(ωt)dθ = -τ0 cos(ωt) → from 0 to π
OK
gives → Supplied Energy = W = τ0 cos(ωt)
I don't think you evaluated at the limits correctly. Otherwise, your work for the rest of parts (a) and (b) looks correct to me regarding method.

TSny
Homework Helper
Gold Member
Just caught something else.
ω=70 rpm = 7.33 rad/s
t=π/ω=0.429 s
So E = 38.5*cos( 7.33⋅0.429) = 23.68 J.
.
Why the factor of 2π? Note that t=π/ω tells you that ωt = π. That's all you need to evaluate cos(ωt).

Thanks.
Just caught something else.
Why the factor of 2π? Note that t=π/ω tells you that ωt = π. That's all you need to evaluate cos(ωt).
Oh I see... I thought that, as ω is specified to be given in s^(-1), we were expected to convert it into rad/s (so multiplying it by 2π)... But OK I understand it now.

So actually
(a) E = 2*38.5 = 77 J.
(b) P = 89.7 W.

TSny
Homework Helper
Gold Member
So actually
(a) E = 2*38.5 = 77 J.
(b) P = 89.7 W.
I think these answers are off by a simple factor due to the fact that you didn't evaluate W = ∫τ0 sin(ωt)dθ = -τ0 cos(ωt) correctly at the lower limit (t = 0).