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Phy 201
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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A simple pendululm of length 3.4 meters and mass .3 kg is pulled back a distance of .222 meters in the horizontal direction from its equilibrium position, which also raises it slightly. If it is released and allowed to fall freely back to its equilibrium position, what will be its velocity at the equilibrium position?
When pulled back a distance less than about 1/4 of its length, a pendulum experiences a very nearly linear restoring force of the form F = - k x, with k = m g / L.
So for this pendulum k = .3 kg * 9.8 m/s^2 / (3.4 m) = .9 N / m, very approximately.
For a pullback of about .2 meters the restoring force is therefore
F = - k x = -.9 N / m * .2 m = -.18 N.
Its potential energy as position x is 1/2 k x^2, so its potential energy at the pullback position is
PE = 1/2 k x^2 = 1/2 * .9 N / m * (.2 m)^2 = .018 Joules, approx..
This potential energy will change to KE as the pendulum swings back to equilibrium. At equilibrium is PE is 0, so it will have lost all of the .018 Joules to KE. Since its KE at release was zero, its KE at equilibrium is therefore .018 J.
Since KE = 1/2 m v^2, it follows that v = sqrt( 2 KE / m) = sqrt( 2 * .018 J / (.3 kg) ) = sqrt(.12 m^2 / s^2) = .35 m/s, approx..
A more rigorous application of the work-energy theorem:
In general `dW_noncons_ON = `dPE + `dKE.
For the ideal pendulum there are no nonconservative forces present, so `dW_noncons = 0 and therefore `dPE + `dKE = 0.
Therefore `dKE = - `dPE.
For this pendulum, from release to equilibrium the PE decreases from .036 J to 0 J, a change of -.036 J.
Thus
`dKE = - `dPE = - ( - .036 J) = .036 J.
sure even where to start or what to do on the problem can you point in the right direction
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Good question. My notes include the full analysis of force and energy for this pendulum.
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