Energy and power supplied by a bicyclist

  • #1
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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
 

Answers and Replies

  • #2
TSny
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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.
 
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  • #3
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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?

Thanks again for your answer
 
  • #4
TSny
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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.
 
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  • #5
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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.
 
  • #6
TSny
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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.
 
  • #7
TSny
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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).
 
  • #8
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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.
 
  • #9
TSny
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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).
 

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