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Angular momentum of a merry-go-round

  1. Nov 22, 2004 #1
    1) A merry-go-round is a common piece of playground equipment. A 3.0-m-diameter merry-go-round with a mass of 250 kg is spinning at 20 rpm. John runs tangent to the merry-go-round at 5.0 m/s, in the same direction that it is turning, and jumps onto the outer edge. John's mass is 30 kg. What is the merry-go-round's angular velocity, in rpm, after John jumps on?

    i know how to solve it i could find the moment of inertia for john but i don't think that is easy to do. since the rotation is about an axis of symmetry, the angular momentum L = I*omega is parallel to omega. so all of the angular momentums will be the same. i had L_final = I_disk*omega_f + I_john*omega_f = L_i = I_disk*omega_i.
    but again im not sure how to find the moment of inertia for john or even if this is the correct method to go about this.

    2) A 10 g bullet traveling at 400 m/s strikes a 10 kg, 1.0-m-wide door at the edge opposite the hinge. The bullet embeds itself in the door, causing the door to swing open. What is the angular velocity of the door just after impact?

    i know this is an inelastic collision and i could find the velocity of the door after the impact but am not sure how to find the angular velocity since it is connected to the door.

    any help appreciated.
  2. jcsd
  3. Nov 22, 2004 #2
    Both questions look like conservation of angular momentum.

    For the first one, I think you may have to treat john as a particle and make his moment of inertia mr^2. At the beginning john has angular momentum, so your initial angular momentum would be: I_disk*omega_f + I_john*omega_john. Also, keep in mind, the final angular velocity of both john and the ferris wheel are the same.

    For two, again angular momentum is conserved. Try to find an expression for the initial angular momentum of the door and the bullet and also the final. Btw, angular momentum is conserved because the bullet does not apply a constant torque.
  4. Nov 22, 2004 #3
    ok for the first question, i have the initial angular momentum as (1/2)(m_disk)(radius)^2*(omega_i) + m_john*radius^2*(omega_john){which using v = r*omega, i get v/r}/ (1/2)*m_disk*radius^2 + m_john*radius^2. after doing that, i still get the wrong answer. should the disk spin faster after john gets on or slower? i would think it would be faster but i was getting a slower answer. am i doing this wrong still or not?
    the answer i got was 16.77 rpm but thats wrong.
  5. Nov 23, 2004 #4
    Make sure you are using the right units. The angular momentum must be in radians/sec.
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