Calculating Kinetic and Potential Energy for Simple Pendulum - Homework Problem

[deedsy]

Homework Statement

A simple pendulum consists of a point mass m suspended from a fixed point by a weightless rigid rod of lengh l.

The kinetic energy = 1/2*m*l^2*(dθ/dt)^2

The potential energy = -m*g*l*cosθ

Calculate the time averages of the kinetic and potential energies over one cycle, AND compare them with the total energy of the system.

Homework Equations

(d^2θ/dt^2) + (g/l)sinθ = 0 ∴ w = √(g/l)
τ = 2∏/ω

The Attempt at a Solution

T = 1/2ml^2 (ω)^2
T = 1/2mlg

U = -mglcos(wt)
U = -mglcos(√(g/l) * (2∏√l/g))
U = -mglcos(2∏)
U = -mgl

So 2T + U = 0

Is this the correct answer? Because I thought T should = U?

 

[Simon Bridge]

Is this the correct answer? Because I thought T should = U?

Depends what you are calculating and what sort of approximations you are making.

You appear to be giving the pendulum a constant kinetic energy?
You appear to be saying that $\theta = \omega t$ ... is $\omega$ a constant?
You appear to have put $t=2\pi\sqrt{l/g}$ ... is the time always that value?
Is this all for the time average?

Without some idea of what it is for and how you are thinking about it, I cannot follow your math.

 

[deedsy]

Yes, I'm trying to find the time averages first. Then, I need to compare it to the total E of the system.

Yes, I assumed w is constant because g and l are constant. I plugged in T for t, because I wanted the potential energy for one whole cycle.

Is this one way to go about it? Or is it necessary to do an integral of U from 0 to T, and divide the answer by T?

 

[Hollumber]

If you want the "potential energy for one whole cycle", then it means you want, essentially, a value in units of Joule-seconds. Then divide this by T to get the average.

To get units of J*s from an equation giving J, an integral over time can be performed.
(Hint: Should the integral go from 0 to T, or to some fraction of T?)

 

[deedsy]

The integral should be over one whole cycle - so 0 to T

So for potential energy...
integral from 0 to T of (-m*g*l*cosθ) dt

I'd then divide this answer by T to get the time average potential energy.

 

[haruspex]

The question is not well defined. The potential energy depends on your reference point. -m*g*l*cosθ is taking the reference point to be the top of the pendulum, which cannot give a sensible answer. For small oscillations, the average will be close to m*g*l, while the KE average will be close to 0.
I would take the bottom of the swing as the reference point.
Since a simple pendulum undergoing small oscillations approximates SHM, you know that the altitude and velocity follow sine or cosine functions. The time average will therefore involve integrating those functions.

 

[deedsy]

I agree, I think I need to change the way U and T are defined for this part. The T and U I listed are from previous parts that were required for the correct answers.

Assuming small oscillations the system follows SHM...
In this case, the SHM equation is (d^2θ/dt^2) + (g/l)θ = 0

My book carries out this process for a spring and ends up with E = T+U = .5kA^2

So, should the last part of my final answer also be E = T + U ? Since it is also SHM.
I'll try to define T and U using the bottom of the swing as my reference point, as you suggested.

 

[haruspex]

I agree, I think I need to change the way U and T are defined for this part. The T and U I listed are from previous parts that were required for the correct answers.

It will be helpful to avoid using T to mean both potential energy and oscillation period.

Assuming small oscillations the system follows SHM...
In this case, the SHM equation is (d^2θ/dt^2) + (g/l)θ = 0

Right, so what is the solution for θ as a function of t? What does that give for the PE and KE as functions of t?

So, should the last part of my final answer also be E = T + U ? Since it is also SHM.

E = PE+KE just states conservation of work. It'll be true whether or not it's SHM.

 

[Simon Bridge]

Not to take anything away from the other advise you are getting...

Yes, I'm trying to find the time averages first. Then, I need to compare it to the total E of the system.

Excellent.

Yes, I assumed w is constant because g and l are constant.

$\omega = \frac{d}{dt}\theta$ right?
At each end of the arc, what is the angular speed of the pendulum bob?
Compared with the angular speed in the middle of the arc?

I plugged in T for t, because I wanted the potential energy for one whole cycle.

If you plug in t=T (one period after t=0), then aren't you getting the potential and kinetic energies at time t=T?

Is this one way to go about it? Or is it necessary to do an integral of U from 0 to T, and divide the answer by T?

That would be the definition of an average wouldn't it?
(In this discussion I am taking Haruspex's advise by calling kinetic energy something other than T ... K, say, but you are free to do it the other way around and call the period P or something.)

As Haruspex is suggesting, it will help you to find how θ varies with time.
$$\frac{d^2\theta}{dt^2}+\frac{g}{l}\theta = 0$$... which should be an initial value problem, so you need to work out:
$$\theta (0)=? \\ \left. \frac{d\theta}{dt}\right|_{t=0}=?$$ ... I suspect you already know the solution but, if you have not worked it out yourself before, you should do it from scratch.
Provided of course you know how to solve differential equations :)

 

[deedsy]

Ahh guys thank you so much for the help! I finally got it.
Alrite, θ(t) = Asin(wt) + Bcos(wt)
derivative of this = Awcos(wt) - Bwsin(wt)
Applying initial condition: θ(0)=θ0 and dθ/dt (0) = 0
A=0 and B=θ0

I get θ(t) = θ0cos(wt)

Now, it turns out I CAN use the K and U formulas from my original post. I just have to approximate the cosine in the U equation through Taylor expansion. It turns out to be U = -mglcosθ = (mglθ^2)/2

Plugging in θ to my K and U equations, integrating from 0 to T, and finally dividing this answer by T... K = U = .5E

:)
I really appreciate the help guys

 

[Simon Bridge]

θ(t) = Asin(wt) + Bcos(wt)
derivative of this = Awcos(wt) - Bwsin(wt)

In that case, $\omega = \frac{d}{dt}\theta \neq w$ ... well done though.