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Double pulley system

  1. Feb 14, 2005 #1
    A glider is attached to a bathtub full of cement via a double pulley system, (the bath tub is hanging over a cliff). Assume the glider rests on a freely rolling cart. I wasn't sure of exactly how a double pulley works. The picture shows the one where the Glider is attached through the outer radius, and the other is where the Glider is attached through the inner radius. The outer radius is twice as big as the inner one. I figure that the goal of this problem is to get the maximum speed of the glider after the tub falls 60 ft. (The tub weighs twice as much as the glider). I'm just not sure what equations relate to a double pulley system. I know one system will pull pull the glider 2 times as much and the other 1/2 as much, but I can't think of which is which. Any help on this would be great, thanks.

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  3. Feb 15, 2005 #2

    Doc Al

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    Staff: Mentor

    Here's a hint: The inner and outer radii have the same angular speed and angular acceleration. The linear speed and acceleration relate to the angular quantities like so: [itex]v = \omega r[/itex] and [itex]a = \alpha r[/itex].
  4. Feb 15, 2005 #3
    Thanks, I used work energy and got this for the equation with the outer pulley connected to the glider.

    [tex]m_2*g*h=\frac{1}{2}*m_1*(v_1)^2 + \frac{1}{2}*m_2*(\frac{1}{2}*v_1)^2[/tex]

    where [tex]m_1[/tex] is the mass of the glider and [tex]m_2[/tex] is the mass of the tub of cement hanging over the side. [tex]v_1[/tex] is the velocity of the glider.

    and I got this for the inner pulley connected to the glider.

    [tex]m_2*g*h=\frac{1}{2}*m_1*(v_1)^2 + \frac{1}{2}*m_2*(2*v_1)^2[/tex]

    Does that look right?
    Thanks in advance for any help.
  5. Feb 16, 2005 #4

    Doc Al

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    Staff: Mentor

    Exactly correct. (Assuming we can ignore the KE of the pulley itself--assume it's massless.)
  6. Feb 16, 2005 #5
    Yes, the pulley is massless in this problem. Thanks for the help!
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