What Is the Final Velocity at the Highest Point of a Pendulum?

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SUMMARY

The final velocity of a pendulum at its highest point, after being released from its lowest point with a velocity of \(\sqrt{2gL}\), is calculated using the conservation of energy principle. The kinetic energy at the bottom is converted into potential energy and kinetic energy at the top. The correct calculation shows that the height at the top is \(0.6L\), leading to the final velocity being \(\sqrt{1.2gL}\), as confirmed by the discussion participants.

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Edit: sorry, i meant a problem concerning kinetic and potential energy :)

Homework Statement



http://img136.imageshack.us/img136/8787/diagramln.jpg

A mass on a pendulum with length L has a velocity of [tex]\sqrt{2gL}[/tex] at it's lowest point. At it's lowest point, the string hits a pole at a distance of .8L sticking out of the wall. What is it's velocity when it reaches its highest point?

Homework Equations



[tex]KE_{bottom} = PE_{top}+KE_{top}[/tex]

The Attempt at a Solution



The distance between the pole and the bottom is .2L, so the height at the top is .6L.

Since it has a velocity and a height at the top, it has both kinetic and potential energy. The potential energy at the bottom of the pendulum is 0. So I thought:

[tex]\frac{1}{2}m{v^2}_{bottom} = mg(.6L)+\frac{1}{2}m{v^2}_{top}[/tex]
[tex]\frac{1}{2}2mgL = m(.6gL+\frac{v^2}{2})[/tex]
[tex]gL = .6gL+\frac{v^2}{2}[/tex]
[tex].4gL = \frac{v^2}{2}[/tex]
[tex].8gL = {v^2}[/tex]
[tex]\sqrt{.8gL} = v[/tex]

but the book says the answer is [tex]\sqrt{1.2gL}[/tex]
 
Last edited by a moderator:
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In the first equation replace mg(0.6L) by mg( 0.4L).
 

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