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Find the net work required to accelerate a cylinder

  1. Dec 10, 2016 #1
    1. The problem statement, all variables and given/known data
    A merry-go-round has a mass of 1440kg and a radius of 7.50m . How much net work is required to accelerate it from rest to a rotation rate of 1.00 revolution per 7.00s? Assume it is a solid cylinder.

    2. Relevant equations
    I believe my relevant equations are:
    tau=Iα (moment of inertia * alpha)

    3. The attempt at a solution
    First I calculated the moment of inertia as:
    ICYL=½(1440kg)(7.50m)2=40500 kg⋅m2

    Then I calculated ω as 2π/7.00s

    Knowing ω allowed me to calculate α as: (2π/7.00s)/7.00s=2π/49.00s2

    Then I used the tau equation to calculate tau as Iα=(40500 kg⋅m2)*(2π/49.00s2)=5193.244999J

    Then I used the work equation to calculate work as tau*Δθ=5193.244999J*2π=32630.12067J

    With 3 sig figs, this would be 3.26*104J

    The problem is, the book says the answer is 1.63*104J and I can't figure out why.
    Last edited: Dec 10, 2016
  2. jcsd
  3. Dec 10, 2016 #2
    Your answer is twice as much. Look for a rogue 2 somewhere in your calculations.
  4. Dec 10, 2016 #3
    Thanks for the reply. I realize my answer is exactly 2 times as high. (I wouldn't have gone through the trouble of creating an account here and posting a question for something as obvious as that). But after carefully going over the steps multiple times, I cannot account for the fact that the answer should be (my answer / 2). So I am hoping someone can explain what I'm doing wrong which logically explains the problem. I can't just arbitrarily divide my answer by 2 and call it a day.
  5. Dec 10, 2016 #4
    This is correct.
    This is an assumption you made - that it took 7 seconds to reach the angular velocity above. No such information was given in the problem statement. For all we know, it might have taken 3 weeks to get up to that angular velocity. But I think we can still work with your assumption.
    That looks good based on your assumption of 7 seconds to accelerate up to the final angular velocity.
    Now you are multiplying the torque times the angle over which it was applied, which is 2π. But is 2π really the correct number? We know, based on your assumption, that it took 7 seconds to accelerate up to the final velocity. At your calculated (or assumed) angular acceleration, what angle of rotation would it take to reach the final angular velocity of 2π/7??

    Edit: Oh yeah, welcome to Physics Forums.
  6. Dec 10, 2016 #5
    Ah, yes. I see what you are saying. That was sloppy reading on my part. Since I don't know how long it took to reach that angular velocity, I can't calculate angular acceleration, so my entire approach is wrong. The fact that my answer comes out twice as high as it should be was just coincidental. Back to the drawing board.
  7. Dec 10, 2016 #6
    Actually, I think you should find your error in the solution that you have started with. Even though you made that assumption, you can still find the right answer based on that assumption.
  8. Dec 10, 2016 #7


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    Work is closely related to which dynamic quantity?
  9. Dec 10, 2016 #8
    Thanks for the replies. I went back over the problem using the work energy theorem. W≡K-K0=ΔK where K=½Iω2. This allowed me to get the same answer as the book without any assumptions.

    Using the same I and ω that I already found from the original post: K=½(40500 kg⋅m2)([2π/7.00s])2=16315.06034J

    So that takes care of that problem. Thanks again to everyone who replied.
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