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Pendulum speed with only height from equilibrium

  • Thread starter Serpentia
  • Start date
  • #1
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Homework Statement


A pendulum is dropped from 0.50 metres above the equilibrium position. What is the speed of the pendulum as it passes through the equilibrium position? (The pendulum is in a isolated system.)

h1= 0.05m
h2= 0.00m
g= 9.81 m/s^2
Length of string (L)= 1.0m.

Homework Equations


Em=Ek+Ep
Em=[(1/2)mv^2]+[mgh]
v=√(2Ek/m)


The Attempt at a Solution



I'm not entirely sure what to try. I'm not given mass, or any value of any form of Energy, be it Kinetic or Gravitational Potential. I've tried to find the value of mass, but I don't have the energy or velocity factor to figure it out;

v=√(2Ek/m)
m=(2Ek/v^2), or
m=(Ep/gh)

Just don't work because I don't have that information. I'm totally lost... And that's all the formulas I can extrapolate from what information they've given me. All I need is a formula to follow, and I can get it from there, thanks muchly.
 

Answers and Replies

  • #2
rock.freak667
Homework Helper
6,230
31
At 0.05 metres what is the potential energy?

This same energy is converted entirely to ke as it passes through the equilibrium position.

1/2 mv2=mgh
 
  • #3
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At 0.05 metres what is the potential energy?

This same energy is converted entirely to ke as it passes through the equilibrium position.

1/2 mv2=mgh
I'm not supplied potential, kinetic, or mechanical energy.
 
  • #4
rock.freak667
Homework Helper
6,230
31
I'm not supplied potential, kinetic, or mechanical energy.
Using the formula, m cancels out. you have g and h. find v.
 
  • #5
3
0
Using the formula, m cancels out. you have g and h. find v.
Oh, doy, thanks so much. Got it now.
 

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