Pendulum speed with only height from equilibrium

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Homework Help Overview

The problem involves a pendulum dropped from a height of 0.50 meters above its equilibrium position, with the goal of determining its speed as it passes through that position. The context includes gravitational effects and energy conservation principles.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between potential and kinetic energy, questioning how to express energy without mass. There are attempts to derive formulas and clarify the conversion of energy types as the pendulum moves through equilibrium.

Discussion Status

Some participants have offered insights into energy conservation, noting that mass cancels out in the equations. There appears to be a shift towards understanding how to calculate speed using gravitational potential energy and kinetic energy principles.

Contextual Notes

Participants express concern over the lack of specific values for mass and energy types, which complicates their calculations and reasoning. The discussion reflects a focus on the theoretical aspects of energy conversion in a pendulum system.

Serpentia
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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.
 
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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
 
rock.freak667 said:
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.
 
Serpentia said:
I'm not supplied potential, kinetic, or mechanical energy.

Using the formula, m cancels out. you have g and h. find v.
 
rock.freak667 said:
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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