A bicycle wheel is mounted on a fixed, frictionless axle, with a light string wound around its rim. The wheel has moment of inertia I=kmr2, where m is its mass, r is its radius, and k is a dimensionless constant between zero and one. The wheel is rotating counterclockwise with angular speed ω_0, when at time t=0 someone starts pulling the string with a force of magnitude F. Assume that the string does not slip on the wheel.
Here’s a picture: http://www.chegg.com/homeworkhelp/questionsanda…
1) What is the instantaneous power P delivered to the wheel via the force F⃗ at time t=0?
Express the power in terms of some or all of the variables given in the problem introduction.
Ps: The answer is not 0 nor FrW_0, believe me I’ve tried. I am so confused. On the hints it said the answer should be torque * d theta/dt, but what is d theta/dt. So, far I know the answer is F*r d theta/dt, but what is d theta/dt in terms of this question on mastering physics and how would I express and write that? This is a mastering physics question by the way for classical mechanics. If anyone could give me the correct answer, that would be really helpful in not only completing the homework but a good gain of knowledge for my upcoming physics exam. Thank You all!
5 Answers

The correct answer *is* Frω₀.
(Technically we should write F. But since this wasn’t necessary for the first part of the question in the link, this is probably unnecessary.)
___________________________________________
θ is the angle through which the wheel rotates.
So (in calculus notation) dθ/dt is rate of change of θ, which is the angular speed ω.
ω = dθ/dt
__________________________
When the wheel rotates through an angle θ radians, a point on the rim moves through an arc length s:
s = r θ
This means that when F causes an infinitesimal rotation dθ, there is an infinitesimal displacement ds:
ds = r dθ
The element of work, dW, done by the force is then:
dW = F ds = F r dθ
If this ocurs in time dt then dividing by dt gives
dW/dt = F r dθ/dt
(This is just the same as differentiating with respect to time.)
dW/dt is the instantaneous power, P(t).
Fr is the torque,τ.
dθ/dt at t=0 is is the initial angular speed, ω₀.
So:
P = τω₀
In terms of the variables in the question:
P = Frω₀
So it sounds like there is a mistake in the ‘official’ answer. It sometimes happens.

Power = P = Work done(W) / time taken
instantaneous P = dW / dt
W = τθ where τ is the torque provided by force F
W = Frθ
=> dW / dt = d/dt (Frθ) = Fr dθ/dt —————–> [given F and r are constants, θ is the only variable ]
since θ is in radians = L / r => dθ = dL / r
=> Fr dθ/dt = Fr /r dL = FdL
also if ωₔ is ω(final) and ωₒ is initial
=> for a small increment dω, ωₔ = (ωₒ + dω)
ωₔ² = ωₒ² + 2FL/I
or => ωₔ² – ωₒ² = (ωₔ – ωₒ)(ωₔ + ωₒ) = dω(2ωₒ + dω) = 2FdL/I
=> 2ωₒdω + (dw)² = 2FdL/I
=> ωₒdω ~=FdL/I [ignoring the relatively small (dw)² ]
or FdL = Iωₒdω
=>
Ans: dW/dt = Fr dθ/dt watt = FdL/dt = Iωₒdω/dt which at t=0 s translates to
instantaneous P = Frωₒ since ω = angular speed in rad / s = dθ/dt
[please note even Iωₒdω/dt leads to the same Ans at t=0
as dω/dt = α = τ/I = Fr/I => Iωₒdω/dt = (FrI/I)ωₒ = Frωₒ]
hope this helps

I wanted to post this question too this morning

Probably it is Okay

a squrt((2FL/I)+wo)
b frwo