warfreak131
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Edit: sorry, i meant a problem concerning kinetic and potential energy :)
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?
[tex]KE_{bottom} = PE_{top}+KE_{top}[/tex]
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]
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]
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