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nineeyes
Feb14-05, 10:18 PM
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.

Doc Al
Feb15-05, 05:56 AM
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: v = \omega r and a = \alpha r.

nineeyes
Feb15-05, 09:58 PM
Thanks, I used work energy and got this for the equation with the outer pulley connected to the glider.

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

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

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

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

Does that look right?
Thanks in advance for any help.

Doc Al
Feb16-05, 08:45 AM
Exactly correct. (Assuming we can ignore the KE of the pulley itself--assume it's massless.)

nineeyes
Feb16-05, 06:37 PM
Yes, the pulley is massless in this problem. Thanks for the help!