|May11-03, 02:35 PM||#1|
I'm not getting the same answer as that which is posted in the back of my textbook for the following problem.
Can someone please tell me where I'm going wrong?
A rod of mass m and length L is pivoted at its center. A constant vertical force, F = mg/2, is exerted by a cord attached to the top end of the rod. The rod is released from rest when it is in the vertical position.
a) find the linear velocity of the end of the rod at the instant when the rod is horizontal.
I use the idea that work done by non-conservative forces = the change in mechanical energy =>
mg/2 = 1/2Iw^2 where I = rotational Inertia of rod
w = angular velocity of the rod
mg/2 = 1/2(1/12)mL^2(tangential velocity/radius)^2
mg/2 = (1/24)mL^2(vtan/(L/2)^2)
vtan = (3g)^1/2
The answer in the back of the text is (3gL/2)^1/2.
Any help would be greatly appreciated.
|May11-03, 02:47 PM||#2|
Sorry about this post.
I figured out my mistake. Forgot to multiply Force times distance on the left side of the eq.
I tried to delete the post, but the software wouldn't let me.
Sorry for wasting your time.
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