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Angular Velocity Help

  1. Oct 14, 2016 #1
    1. The problem statement, all variables and given/known data
    A speed skater increases his speed from 10 m/s to 12.5 m/s over a period of 3 seconds while coming out of a curve of 20 m radius.
    a. What is the magnitude of his angular velocity as he leaves the curve? Be sure to include proper units for your answer.
    b. What is his average linear acceleration during this curve?

    2. Relevant equations
    ω=Δθ/Δt
    vt=rω

    3. The attempt at a solution
    I am not sure how to start this problem because there is not a given angle. I am wondering if there is a different way to solve for angular velocity or if there is a way to calculate the angle based on the given information.

    Thank you in advance.
     
  2. jcsd
  3. Oct 14, 2016 #2

    andrewkirk

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    What is the equation relating linear velocity to angular velocity?
    We know the radius and we know the linear velocity at the instant the skater leaves the curve, so we can use that formula to calculate the angular velocity at that instant.
     
  4. Oct 14, 2016 #3
    tangential velocity: vt = rω
    so you could rearrange to ω = vt/r
    ω = 12/20
    ω = 0.6 rad/sec

    would that be correct?

    And for the second part is it just a simple a=Δv/Δt question because it is asking for linear acceleration?
     
  5. Oct 14, 2016 #4

    andrewkirk

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    The method is right, but you appear to have used 12m/s instead of 12.5m/s for the linear velocity.

    I find the second question ambiguous. To calculate the skater's tangential linear acceleration you can do what you suggest, and my best guess is that that's what they're after. But the actual linear acceleration of the skater at any point in time while in the curve is the vector sum of her tangential and radial accelerations, where the latter is the centripetal acceleration required to maintain the curve. So the true linear acceleration is the average of that vector sum, which is more complex and also depends on info that we don't have, which is the pattern of acceleration while in the curve (eg constant accel, vs accelerate hard then ease off, vs build up then shut off suddenly).

    I suggest you do it the way you indicated, but reply that that is the average linear tangential acceleration.
     
  6. Oct 14, 2016 #5

    haruspex

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    Not sure that resolves the difficulty.

    [Edit: next lines are incorrect. See post #8.
    The average tangential acceleration should be (final tangential velocity - initial tangential velocity)/time.
    But the velocity is tangential throughout, so this is the same as the average acceleration. ]

    As you say, to find that we need to assume e.g. that the rate of change of speed is constant.
    I think what you mean is the average rate of change of speed.
     
    Last edited: Oct 14, 2016
  7. Oct 14, 2016 #6

    andrewkirk

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    Perhaps we're at cross purposes. What I meant was this:
    I get an average tangential acceleration of (12.5-10)/3 which is about 0.8 m/s^2.
    The centripetal acceleration will be 10^2 / 20= 5m/s^2 at the beginning of the curve and about 7.8m/s^2 at the end. If we assume the magnitude of tangential acceleration was constant throughout the curve, the magnitude of total acceleration just before the end is ##\sqrt{7.8^2+0.8^2}=7.9ms^{-2}## and just after the beginning it is ##\sqrt{5^2+0.8^2}=5.1ms^{-2}##. The tangential acceleration is dominated by the much larger centripetal acceleration.

    The average acceleration is going to be somewhere between 5.1 and 7.9##ms^{-2}##, depending on the pattern of the tangential acceleration over the three seconds.

    Unless 'linear acceleration' is used in this course with a specific meaning that somehow excludes centripetal acceleration, eg if it means 'magnitude of tangential acceleration expressed as a linear acceleration'. I'm not familiar with such a convention though.
     
  8. Oct 14, 2016 #7

    haruspex

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    I am saying that is wrong. That's the average magnitude of tangential acceleration.
    The average tangential acceleration would be ##\frac{\int \vec a_T.dt}{\Delta t}##, where ##\vec a_T## is the instantaneous tangential acceleration. We can write ##\vec a_T = \vec r \times \vec\alpha##. If ##\vec\alpha## is constant, we have ##\frac{\int \vec r.dt\times \vec \alpha}{\Delta t}##, but ##\int \vec r.dt## still looks messy.
     
  9. Oct 14, 2016 #8

    andrewkirk

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    Fair point. So I'm guessing that what the question is after is the average magnitude of tangential acceleration. Calculating anything else would be too complex for the level of question this appears to be.

    Do you think there's any terminology convention under which average linear acceleration means that?

    I'm thinking that (b) is a badly worded question.
     
  10. Oct 15, 2016 #9

    haruspex

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    Agreed.
    Not that I'm aware of, but that does not mean much.
     
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