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Homework Help: Angular momentum

  1. Jun 21, 2005 #1
    A uniform solid cylindrical drum of mass 1.5kg and radius 0.5m is free to rotate about a fixed, smooth, horizontal axis which coincides with the axis of the cylinder. The axis is at a height of 2m above a horizontal table, and a light string of AB of length 4m has one end attached to the heighest point of the cylinder. A block of mass 0.3kg is attached to end B of the string and rests on the table. The drum begins to rotate at a constant angular speed of 4 rad s-1 in a clockwise direction. Calculate the angular speed of the drum immediately after the block is jerked into motion.

    I have no idea how to approach this question! The answer is supposed to be 2.86 rad s-1.

    Any help is greatly appreciated. :smile:

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    Last edited: Jun 21, 2005
  2. jcsd
  3. Jun 21, 2005 #2
    I'm looking at this and the only thing i can think of is to find the angular acceleration of the drum due to the block pulling it down.
  4. Jun 21, 2005 #3
    I think the way I'm supposed to approach this question is by equating the sums of initial and final angular momentum for both the drum and the block. I can do the drum just fine, but all my attemps of finding the angular momentum of the block didn't produce the required answer.

    Here's what I get for the drum:
    initial angular momentum = moment of inertia * angular speed = 3/16 * 4 = 0.75 Nms
    final angular momentum = 3/16 w, where w is the new angular speed
  5. Jun 21, 2005 #4
    Ohhh right. Ofcourse.

    The final moment of inertia has changed, because its now pulling the block. You can simplify the block as a point mass at point A, and find the new MOI.
  6. Jun 21, 2005 #5
    I tried that:
    final angular momentum = [3/16 + 0.3*(2^2 + 0.5^2)] w = 1.4625w
    conservation of angular momentum => w = 0.75/1.4625 = 20/39

    Which isn't the right answer. :frown: Am I doing something wrong?
  7. Jun 21, 2005 #6
    final angular momentum = [3/16 + 0.3*(2^2 + 0.5^2)] w = 1.4625w

    Your radius isnt right. How far is the point A from the axis of rotation?
  8. Jun 21, 2005 #7
    If I take it to be at point A then I get the right answer! But.. Why should it be at point A?
  9. Jun 21, 2005 #8
    I thought about it a bit and this is what I concluded:
    Initial jerk turns the drum 4 rad, and so the block is pulled up 4*0.5 m, because A turns an arc length [itex]r \theta[/itex]. So the block adheres to the circumference of the circle and can be considered as a point mass 0.5m away from the axis of rotation.

    Is this correct?
  10. Jun 21, 2005 #9
    Exactly, the block is 'pulling' or resisting motion AT the radius 0.5m, thus you can treat it as a particle in motion at the edge of the drum. Good thinking.
  11. Jun 21, 2005 #10
    Thanks for the help -- and good night!
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