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How many revolutions does the merry go round make as it stops?

  1. Nov 26, 2008 #1
    1. The problem statement, all variables and given/known data
    A 5.0m diameter merry go round is initially turning with a 4.0s period. It slows down and stops in 20 s. How many revolutions does the merry go round make as it stops?


    2. Relevant equations
    w=v/r
    v=2(pi)r/T

    3. The attempt at a solution
    v=2(pi)2.5/4
    =3.9 m/s

    w=v/r
    = 3.9/2.5
    =1.56 rad/s

    1.56 rad/s X (1rev/2(pi)rad)=2.48 rev

    how is the answer 24.8??? I don't understand....Please help
     
  2. jcsd
  3. Nov 27, 2008 #2

    Doc Al

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    Staff: Mentor

    Re: Kinematics

    This is the initial angular speed. The final angular speed is zero. What's the average speed as it slows down?

    (Note that ω = 2π/T.)
     
  4. Dec 3, 2008 #3
    Re: Kinematics

    i don't understand...
     
  5. Dec 3, 2008 #4
    Re: Kinematics

    Final velocity (linear and angular) is zero because the problem says the merry-go-round stops. Try thinking of this in terms of linear kinematics, making the appropriate substitutions.
     
  6. Dec 4, 2008 #5

    Doc Al

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    Staff: Mentor

    Re: Kinematics

    You found the initial angular speed ω. Good.

    What you did here was to convert from rad/s to revolutions per second; so that should be 2.48 rev/s.
    To get the number of revolutions, you need to compute the angle traveled by the merry go round as it comes to rest. That's just the rotational analog to Distance = speed X time. Since the speed is not constant, you must use the average speed, not the initial speed. (The average speed is just half the initial speed.)
     
  7. Dec 4, 2008 #6
    Re: Kinematics

    thanks a bunch...solved it!!!
     
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