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Apr 5, 2008

Problem 6/194

How do you relate omega(2) to v(1)? I looked at the example on page 505 of the book which is almost exactly the same as this problem, but I can't see how they did it.
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3 comments:

Steve Musick said...

The book breaks the action down the action into three events: just before the collision, just after the collision, and halfway through the rotation. State 1 and State 2 can be compared with conservation of angular momentum, which will get you omega(2) in terms of v(1). State 2 and State 3 can be looked at with conservation of energy, where you can solve for omega (2) and thus solve for v (1).

April 5, 2008 at 8:29 PM
Keith said...

I can't seem to get how you related the conservation of energy equation. I'm using 1/2mv^2 + 1/2IgW^2 = mg(.76-.88) which is the change in height. Am I missing something?

April 5, 2008 at 9:13 PM
Eric said...

For the conservation of energy from state 2 to 3 (beginning and end of rotation), it should be 0.5*(Ig+mr^2)w^2 = mg(sqrt(.76^2+.88^2)-0.76). The Ig + mr^2 should be the same value as calculated for the angular momentum when in first contact with the curb. The change in height is the distance from the corner to center of gravity minus the original height before rotation.

April 6, 2008 at 10:19 PM

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