Potential Energy of a Pendulum Explained

Thread starter fridsatu
Start date Nov 2, 2005
Tags

Pendulum

AI Thread Summary

The potential energy (PE) of a pendulum can be expressed as mg(L - Lcos(theta/2)), where m is mass, g is gravitational acceleration, L is the length of the pendulum, and theta is the angle from the vertical. The formula indicates that the height (h) in the context of potential energy is related to the vertical displacement of the pendulum bob. The discussion clarifies that the common formula for potential energy is not simply mgcos(theta/2)/2l, as initially questioned. Understanding the derivation of this formula involves recognizing the relationship between the pendulum's angle and its height. This breakdown helps clarify the concept of potential energy in pendulum motion.

Nov 2, 2005

fridsatu

I am sure this is simple to most people...but i don't understand how to show potential energy of a pendulum to be mgcostheta/2l. is there anybody that could break that down for me??

Physics news on Phys.org

Nov 2, 2005

Chi Meson

Science Advisor

Homework Helper

First of all, it's not.

PE of a pendulum is mg [L-Lcos(theta/2)].

What's the usual formula for PE? What of the above formula is equal to the "h"?

Thread 'I've come across another fluid pressure problem I don't understand'

Neglect gravity other than that of water; neglect friction. F1=ρgsh=1000*10*1*(1+4)=50000 N; F1=ρgsh=1000*10*0.1*4=4000 N; F3=50000-4000=46000 N?

View full post »

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...

View full post »

Thread 'Understanding Forces and their Vector Components'

I was told forces are vectors so they can be divided into x and y components, but what about the normal force or the force of static friction? do you divide those into x and y components if say you had a block that was lying on an angled surface?

View full post »