How do you solve this Physics Question that involves power?

NetherCraft 0

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:…

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!

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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.


    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

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