- #1
Garrit
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Homework Statement
This problem deals with a pendulum. Imagine you're just letting it dangle (so perpendicular to ground) and you lift it to the side by some angle theta. This point your holding it at will be Point A. You release the pendulum from your grip and want to find out at Point B (where it was originally at, just dangling in a straight line) what the velocity is there.
Homework Equations
[/B]
Since only the work of gravity is being done, I am using Ea = Eb. So the kinetic energy at Point A plus the potential energy at point A equals the kinetic energy at Point B plus the potential energy at point B.
Kinetic = (0.5)(m)v^2 Potential = mgl
The Attempt at a Solution
[/B]
I did (0.5)(m)(0) + mglcos(theta) = (0.5)m(Vb)^2 + mgl
Vb = sqrt(2gl(cos(theta)-1)))
The problem is, I don't think this is right. My professor did this in class and got a different answer.
He did this 0 - mglcos(theta) = (0.5)m(Vb)^2 - mgl -------> Vb = sqrt(2gl(1-cos(theta)))
Why did he subtract? I thought Eb= Ea was kinetic PLUS potential
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