Pendulum without oscillation

  • Thread starter amcavoy
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  • #1
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If a mass (m) at the end of a length (L) on a pendulum starts at an angle of θ from the vertical, what is the minimum inital velocity v0 it must have to just barely make it over the top and not oscillate?

This is what I did:

[tex]\Delta K=mgh\implies v_0=\sqrt{v^2-2gh}[/tex]

but at the top, v is zero so it can be written as:

[tex]v_0=\sqrt{-2gh}[/tex]

and h in this case is L+Lcosθ, so using g=-10m/s2 I get:

[tex]v_0=\sqrt{20L\left(1+\cos{\theta}\right)}[/tex]

but whenever I plug in values for θ and L, I get the wrong answer. I can't see what I did wrong.

Any ideas?

Thanks a lot.
 

Answers and Replies

  • #2
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What School do you go to?
 
  • #3
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JeremyM said:
What School do you go to?
I'm in high school.
 
  • #4
HallsofIvy
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At the top the pendulum has 0 velocity (actually, it should have just a tiny non-zero velocity but you can take 0 as "boundary" between going over the top and not making it to the top) and so has no kinetic energy but has potential energy, -mgL (relative to the height of the pivot- the center of the circle the pendulum moves in). Initially, it has velocity v0 and so kinetic energy (1/2)mv0[/sub]2 AND some potential energy. Have you taken into account the fact that the potential energy has to be measured with respect to the center of the circle?
 

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