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Kinetic energy of a rotating wheel?

  1. Jul 6, 2005 #1
    A simple wheel has the form of a solid cylinder of radius r with a mass m uniformly distributed throughout its volume. The wheel is pivoted on a stationary axle through the axis of the cylinder and rotates about the axle at a constant angular speed. The wheel rotates n full revolutions in a time interval .
    Express your answer in terms of m, r , n ,t and, pi .

    does anyone have any pointers for me?

    i do know that this formula, 1/2mr^2 can help me, but i dont know how create the right equation using the other arts given to me.
     
  2. jcsd
  3. Jul 7, 2005 #2
    Rotational kinetic energy is given by

    [tex]KE_{rot} = \frac{1}{2}I\omega^2[/tex].

    Just write everything in terms of the variables you've been given.
     
    Last edited: Jul 7, 2005
  4. Jul 7, 2005 #3
    this is my answer so far, but im having a problem with the angular of velocity
    1/2*m*r^2*((2*PI)/n*t)^2. can anyone point me in the right direction
     
  5. Jul 7, 2005 #4
    It should have [tex]\frac{1}{4}[/tex] at the front, because you have [tex]\frac{1}{2}\frac{1}{2}mr^2[/tex].

    In your angular velocity, your n should be in the numerator of the fraction.
     
  6. Jul 7, 2005 #5

    BobG

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    Two problems with your answer.

    You're using moment of inertia for a ring, not a solid wheel of uniform density. Technically, moment of inertia is:

    [tex]I = \int_0^m r^2 dm[/tex]

    Unless you have to solve the integrals, it's usually easier to look up the solution. Moment of inertia of several shapes are at Eric Weisstein's World of Physics (you need to scroll down a little to see the formulas)

    Your angular velocity is measured in radians per second. You were given n revolutions in t seconds. Convert the revolutions per second:

    [tex]\frac{n_- revs}{t_- sec} * \frac{2 \pi_- rad}{1_- rev} = \omega[/tex]
     
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