Torque and energy conservation of a yoyo

AI Thread Summary
The discussion revolves around calculating the velocity of a giant yo-yo dropped from a height of 57 meters, using energy conservation principles. The equation .5mv² + .5Iw² = mgh is highlighted, with a focus on determining angular velocity (ω) and its relationship to linear velocity (v) through the equation v = Rω. Participants express confusion about how to find angular velocity without time, but it is noted that this relationship does not require time for calculations. The conversation emphasizes the importance of understanding torque and energy conservation in solving the problem. Ultimately, the key takeaway is to incorporate the relationship between linear and angular motion into the energy conservation equation for a solution.
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In 1993, a giant yo-yo of mass 480 kg and measuring about 1.9 m in radius was dropped from a crane 57 m high. Assuming that the axle of the yo-yo had a radius of r=0.1 m, find the velocity of the descent v at the end of the fall.

I know that .5mv2+.5Iw2=mgh, but I don't have a clue how to find w (the angular velocity).

I really don't even know where to start. The only thing I can think of is the force due to gravity creating a torque, which would allow me to solve for the angular acceleration, but even then I would need the amount of time it takes to travel 57m in order to get the angular velocity.

If someone could point me in the right direction I would appreciate it.
 
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It is easy to show that \frac{dz}{dt}=v=R\frac{d\theta}{dt}=R\omega.
(To show this, simply draw a diagram - If the yo-yo rotates through an angle d\theta what is the consequent change in z?)
 
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Wouldn't I need the angular velocity, velocity, or time to do anything with that?
 
scavok said:
Wouldn't I need the angular velocity, velocity, or time to do anything with that?

According to my last post, v = R\omega. This relation doesn't involve time. Try putting this into your energy conservation equation.
 
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