The Student Room Group

Potential energy of a pendulum..

If we have a pendulum, we know the potential energy is at its maximum at the end(s) of the oscillations..if you follow..

and potential energy equations are typically expressed as

V=mgh

I have the fact that y=-L(cos(theta)). L is the length of the massless rod holding the bob of weight m, this then follows that..

V=-mgL(cos(theta)). This is shown in loads of books and websites.

The problem is when the pendulum is at 90 degrees (ie maximum potential energy for the pendulum), cos(90) is 0 and so making V=0 so no potential energy.

very confusing and pecking my swede. any help is much thanken.
Reply 1
The potential energy will surely be mgh where h is the length of the pendulum? As it will be this length above it's lowest point
Reply 2
well no, my bad, obviously h is the height of the pendulum above its lowest 'sweep' of its path, ie where theres maximum kinetic energy and lowest potential.

ive seen other sources say that V=-mgL(1-cos(theta)) which can explain the anomaly at pi/2, but why say that the length of the pendulum is in turn its height in relation to potential energy in some books?

basically would it not be correct to have V=-mgLcos(theta)?
Reply 3
would you not have
V=MgL(cos(theta)-1)??

ive seen both this and V=-MgLcos(theta) in books would someone please clarify for what types of pendulum the potentials are

(in my case i am looking at an elliptic pendulum with a mass A able to move horizontally and a mass B attached by light inextensible rod with (theta) measured against the vertical)

thanks
Reply 4
Skaterhaz
If we have a pendulum, we know the potential energy is at its maximum at the end(s) of the oscillations..if you follow..

and potential energy equations are typically expressed as

V=mgh

I have the fact that y=-L(cos(theta)). L is the length of the massless rod holding the bob of weight m, this then follows that..

V=-mgL(cos(theta)). This is shown in loads of books and websites.

The problem is when the pendulum is at 90 degrees (ie maximum potential energy for the pendulum), cos(90) is 0 and so making V=0 so no potential energy.When considering potential energy, all that matters is the change in potential energy as theta changes. So V = 0 when theta = 90, but that's OK, because V = -mgL when theta = 0, so you lose mgL in potential energy as you go from theta = 90 to theta = 0.

Or you could write V = mgL(1-cos(theta)) and V = mgL when theta = 90 and 0 when theta = 0. Either way, you get exactly the same equations of motion, velocities, etc.

P.S. the P.E. is actually highest when theta = 180, not 90.
Reply 5
spot on franklin, cheers, never thought about it like that:biggrin:
Reply 6
Original post by DFranklin
When considering potential energy, all that matters is the change in potential energy as theta changes. So V = 0 when theta = 90, but that's OK, because V = -mgL when theta = 0, so you lose mgL in potential energy as you go from theta = 90 to theta = 0.

Or you could write V = mgL(1-cos(theta)) and V = mgL when theta = 90 and 0 when theta = 0. Either way, you get exactly the same equations of motion, velocities, etc.

P.S. the P.E. is actually highest when theta = 180, not 90.




Can i know the proofing of this equation P.E= mgL(1-cos(theta))