Pendulum Conservation of Energy

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  • #1
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Hi, I have a general question to pendulums. I hope it is ok to post it in this format.
Please accept my apologies for my poor English.

Homework Statement

:
As a general Example:
I have a Pendulum of length L with Angle Theta as maximum displacement.


I know how to solve these problems. Find the period, the speed etc.
But,
I think I'm missing something fundamental here.
I would be very glad if you could help me understand.

So when the Pendulum is at maximum displacement it is also at maximum height in it's oscillation.
Which means it has the most potential energy in mgh.

So when it's in the middle shouldn't the total energy be 1/2Iw2 + 1/2mv2.

Why isn't mg(change in)h = 1/2Iw2 + 1/2mv2?

Am I missing some term?


 

Answers and Replies

  • #2
PeroK
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Hi, I have a general question to pendulums. I hope it is ok to post it in this format.
Please accept my apologies for my poor English.

Homework Statement

:
As a general Example:
I have a Pendulum of length L with Angle Theta as maximum displacement.


I know how to solve these problems. Find the period, the speed etc.
But,
I think I'm missing something fundamental here.
I would be very glad if you could help me understand.

So when the Pendulum is at maximum displacement it is also at maximum height in it's oscillation.
Which means it has the most potential energy in mgh.

So when it's in the middle shouldn't the total energy be 1/2Iw2 + 1/2mv2.

Why isn't mg(change in)h = 1/2Iw2 + 1/2mv2?

Am I missing some term?


What do you mean by ##\omega##? Is that the angular velocity about the fulcrum or the rotation of the bob about its centre of mass?
 
  • #3
111
21
Hi, sorry, the angular velocity about the fulcrum.
What do you mean by ##\omega##? Is that the angular velocity about the fulcrum or the rotation of the bob about its centre of mass?
 
  • #4
429
15
So when it's in the middle shouldn't the total energy be 1/2Iw2 + 1/2mv2.
If you consider the plane perpendicular to the string for θ=0,θ is the angle made by the string with the vertical.,as the reference for potential, at any angle θ the total energy is E=KE+mgl(1-cosθ),and for max. height E=mgl(1-cosθmax.) and for θ=0 ,E is simply KEmax.
 
  • #5
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To maybe add something, for the longest time I thought it should just be mgΔh = 1/2mv2 when the bob is in the middle.
I played around for a few hours and then come to 1/2Iω2 + 1/2mv2

now I realize that the bob is not rotating.

So to better formulate my question.

What is the total energy when theta is at it's max and what is it when the bob is in the middle at maximum velocity?
 
  • #6
PeroK
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Hi, sorry, the angular velocity about the fulcrum.

Is that not just an alternative way to express the kinetic energy? Try calculating ##I\omega^2## and ##mv^2## and see whether they are always the same for a pendulum bob.
 
  • #7
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21
If you consider the plane perpendicular to the string for θ=0,θ is the angle made by the string with the vertical.,as the reference for potential, at any angle θ the total energy is E=KE+mgl(1-cosθ),and for max. height E=mgl(1-cosθmax.) and for θ=0 ,E is simply KEmax.
Hi, thank you.

that is what I thought, but it doesn't work out numerically if you solve for v when θ=0.
 
  • #8
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Is that not just an alternative way to express the kinetic energy? Try calculating ##I\omega^2## and ##mv^2## and see whether they are always the same for a pendulum bob.
Hi, I don't think so. Consider a ball rolling down a ramp. mgh = 1/2Iω2 + 1/2mv2
the potential energy got converted to rotational and kinetic energy
 
  • #9
PeroK
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Hi, I don't think so. Consider a ball rolling down a ramp. mgh = 1/2Iω2 + 1/2mv2
the potential energy got converted to rotational and kinetic energy

I thought we were talking about a pendulum, not a ball rolling down a ramp.

Note that if you consider the pendulum bob as something greater than a point mass, then the kinetic energy of the pendulum becomes more complicated.
 
  • #10
PeroK
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Note that for a point mass, you have:

##I = ml^2## and ##v = \omega l##

Hence

##I\omega^2 = ml^2\omega^2 = mv^2##

And so the two expressions give the same KE of the point mass. Hence, you could use either. But not both added together.
 
  • #11
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Hi, thank you.

that is what I thought, but it doesn't work out numerically if you solve for v when θ=0.
v is max. for θ=0.,it has to be
Hi, I don't think so. Consider a ball rolling down a ramp. mgh = 1/2Iω2 + 1/2mv2
the potential energy got converted to rotational and kinetic energy
And an angle θ the tangential component of velocity is v=rθdot.=rω

And putting mv2=mr2ω2=Iω2,so they are the same.
 
  • #12
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I thought we were talking about a pendulum, not a ball rolling down a ramp.

Note that if you consider the pendulum bob as something greater than a point mass, then the kinetic energy of the pendulum becomes more complicated.
Ah, I see what you meant. Yes, I misunderstood you. For the point mass, yes I is mr2 so Iω2 is mv2 since v = ωr.

But what I meant to ask. What is the total kinetic energy at the middle? is it not mgΔh?
 
  • #13
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Ah, I see what you meant. Yes, I misunderstood you. For the point mass, yes I is mr2 so Iω2 is mv2 since v = ωr.

But what I meant to ask. What is the total kinetic energy at the middle? is it not mgΔh?

Yes, it must be.
 
  • #14
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Ah, I see what you meant. Yes, I misunderstood you. For the point mass, yes I is mr2 so Iω2 is mv2 since v = ωr.

But what I meant to ask. What is the total kinetic energy at the middle? is it not mgΔh?
Yes if Δh being the heighest displacement of the Bob from the reference potential level.
 
  • #15
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Hi, thank you both. I'm a slow typer. I'm sorry. I appreciate that for a point mass r2w2 turns out to be just v2

my first thought has been just mgΔh = 1/2mv2

then after a few hours and countless problems I thought I might be missing the rotational energy.
For neither I get the correct speed.
Which was wrong.

I realized that after Perok's first comment.

So my question is when theta is at max what is the total energy? is there anything else but mgh?

and when theta is zero what is the total energy is there anything else but 1/2mv2
 
  • #16
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Yes, it must be.
Thank you. So is there anything else but 1/2mv2 because if you solve for v I always get the wrong answer.
 
  • #17
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Thank you. So is there anything else but 1/2mv2 because if you solve for v I always get the wrong answer.

You should post what you're doing. There is nothing more for a point mass. If the bob is larger, then it's not all travelling at the same speed. But, that's irrrelevant here.
 
  • #18
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Thank you. So is there anything else but 1/2mv2 because if you solve for v I always get the wrong answer.

Just a thought. If you are using ##\omega = \dot{\theta}##, you may be confusing that the angular frequency of pendulum's motion - which is something different.
 
  • #19
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Note that for a point mass, you have:

##I = ml^2## and ##v = \omega l##

Hence

##I\omega^2 = ml^2\omega^2 = mv^2##

And so the two expressions give the same KE of the point mass. Hence, you could use either. But not both added together.
Since they are equal, he can take half of each and add them to get the same thing. That is a strange way of doing it, but it will give him the correct answer every time.
CORRECTION: I forgot that the sum will be two times two great.
 
Last edited:
  • #20
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21
I'm sorry. Also for being so slow, I have real trouble with science notation on my laptop.

I have done about 30 problems yesterday and today and some a few days ago.
I get v usually by x = Lsin(theta)cos(wt)
so the derivative is v = Lsin(theta)ωsin(wt) hence v is Lsin(theta)ω (sin(90) = 1)

This would always give me the correct answer.

but mgΔh = 1/2mv2

squareroot(2gΔh) = v is never right. In every single problem.

Now I feel dumb and like I may be making algebra mistakes
 
  • #21
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Hi, I have a general question to pendulums. I hope it is ok to post it in this format.
Please accept my apologies for my poor English.

Homework Statement

:
As a general Example:
I have a Pendulum of length L with Angle Theta as maximum displacement.


I know how to solve these problems. Find the period, the speed etc.
But,
I think I'm missing something fundamental here.
I would be very glad if you could help me understand.

So when the Pendulum is at maximum displacement it is also at maximum height in it's oscillation.
Which means it has the most potential energy in mgh.

So when it's in the middle shouldn't the total energy be 1/2Iw2 + 1/2mv2.

Why isn't mg(change in)h = 1/2Iw2 + 1/2mv2?

Am I missing some term?

The two haves are equal. ##L\omega^2 = mv^2## So you can write the expression in either term only or partition the total between them as you have.
CORRECTION: I forgot that the sum will be two times two great.
 
Last edited:
  • #22
111
21
Since they are equal, he can take half of each and add them to get the same thing. That is a strange way of doing it, but it will give him the correct answer every time.
Thanks, I try to always check my answers by getting to them in two alternative ways. For motion problems I usually use conservation of energy for my alternative. Works fine in all problems except pendulums. So I feel I'm not seeing something.

Now i realize I maybe messing up algebra. I'll have to try again.
 
  • #23
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21
The two haves are equal. ##L\omega^2 = mv^2## So you can write the expression in either term only or partition the total between them as you have.
Hi, I got to this too at one of my last attempts but v was still wrong.
 
  • #24
PeroK
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I'm sorry. Also for being so slow, I have real trouble with science notation on my laptop.

I have done about 30 problems yesterday and today and some a few days ago.
I get v usually by x = Lsin(theta)cos(wt)
so the derivative is v = Lsin(theta)ωsin(wt) hence v is Lsin(theta)ω (sin(90) = 1)

This would always give me the correct answer.

but mgΔh = 1/2mv2

squareroot(2gΔh) = v is never right. In every single problem.

Now I feel dumb and like I may be making algebra mistakes

How are you calculating ##\Delta h##?
 
  • #25
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21
How are you calculating ##\Delta h##?
L minus cos(theta)L
 

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