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Calculating Kinetic Energy with Rotational Motion

  1. Oct 13, 2011 #1
    1. The problem statement, all variables and given/known data
    A student sits at rest on a piano stool that can rotate without friction. The moment of inertia of the student-stool system is 3.6 kg·m2. A second student tosses a 1.7 kg mass with a speed of 3.0 m/s to the student on the stool, who catches it at a distance of 0.38 m from the axis of rotation. The final angular speed is 0.504 rad/s.
    (a) Does the kinetic energy of the mass-student-stool system increase, decrease, or stay the same as the mass is caught?

    (b) Calculate the initial and final kinetic energy of the system.


    2. Relevant equations
    W=Kf-Ki
    ω=v/r

    3. The attempt at a solution

    All I've gotten so far is Ki=3.6+.5*1.7*3^2=11.25 J. I'm really stuck, please help!
     
  2. jcsd
  3. Oct 13, 2011 #2

    cepheid

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    Welcome to PF embudini!

    How does the moment of inertia of the system change after the person on the stool catches the mass?

    How does the rotational kinetic energy of a system depend on its moment of inertia and on its angular velocity ω?
     
  4. Oct 13, 2011 #3
    The moment of inertia would increase because the system has a greater mass.

    The kinetic energy is .5mv^2+.5Iω^2, the combination of its linear and rotational kinetic energies. But I don't know what to do this these equations in this scenario. I have a lot of problems similar in concept to this one, but I just don't understand how to solve for them!
     
  5. Oct 13, 2011 #4

    cepheid

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    Yes, but how, specifically does it change? Hint: ignoring the mass of the person's arm, you can think of the mass that gets caught as being a point mass that is moving in a circle around the centre of rotation whose radius is the arm length. What is the moment of inertia of such a system? How does it combine with the moment of inertia of the person + stool?

    Yes, but after the catch happens, is there linear motion any longer?
     
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