Pendulum velocity using energy forumula

Click For Summary

Homework Help Overview

This problem involves a pendulum and seeks to determine the velocity at the lowest point of its swing after being released from a height. The context is rooted in energy conservation principles, specifically relating kinetic and potential energy.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply the conservation of energy principle but questions the differences in their calculations compared to their professor's method. They express confusion over the use of subtraction in the energy equations.

Discussion Status

Participants are exploring the reasoning behind the energy conservation equations used in the problem. Some guidance has been offered regarding the interpretation of potential energy changes and their impact on kinetic energy, but no consensus has been reached on the correct approach.

Contextual Notes

The discussion highlights potential misunderstandings regarding the signs and definitions of energy changes in the context of the pendulum's motion. There is also a mention of the need for clarity in visual representations of the problem.

Garrit
Messages
6
Reaction score
0

Homework Statement



This problem deals with a pendulum. Imagine you're just letting it dangle (so perpendicular to ground) and you lift it to the side by some angle theta. This point your holding it at will be Point A. You release the pendulum from your grip and want to find out at Point B (where it was originally at, just dangling in a straight line) what the velocity is there.
upload_2017-3-7_22-42-36.png


Homework Equations


[/B]
Since only the work of gravity is being done, I am using Ea = Eb. So the kinetic energy at Point A plus the potential energy at point A equals the kinetic energy at Point B plus the potential energy at point B.
Kinetic = (0.5)(m)v^2 Potential = mgl

The Attempt at a Solution


[/B]
I did (0.5)(m)(0) + mglcos(theta) = (0.5)m(Vb)^2 + mgl
Vb = sqrt(2gl(cos(theta)-1)))

The problem is, I don't think this is right. My professor did this in class and got a different answer.
He did this 0 - mglcos(theta) = (0.5)m(Vb)^2 - mgl -------> Vb = sqrt(2gl(1-cos(theta)))

Why did he subtract? I thought Eb= Ea was kinetic PLUS potential
 
Last edited by a moderator:
Physics news on Phys.org
I don't know why my picture doesn't show. I drew it (its real simple) and uploaded it. If my wording doesn't make enough sense here it is https://imgur.com/YqxEwFH
 
Garrit said:
I don't know why my picture doesn't show. I drew it (its real simple) and uploaded it. If my wording doesn't make enough sense here it is https://imgur.com/YqxEwFH
Instead of posting a url to some offsite-stored image which can sometimes be "fragile", it's better to upload your image to the PF server. Use the UPLOAD feature (button at the bottom right of the edit window).

This time I'll insert a copy of your image for you.
 
Garrit said:
Why did he subtract? I thought Eb= Ea was kinetic PLUS potential
The gravitational potential energy decreases with height above the Earth's surface. You've chosen coordinates such that the distance increases downwards, so that you should have ΔPE ∝ -Δh.

It's often worthwhile to look at the change in elevation that occurs and ask yourself whether you should be gaining KE from the change in PE or losing KE to PE. Then make sure that your equation reflects this gain or loss for the given change in elevation.
 
Ohh ok. That makes sense. Thanks!
 

Similar threads

  • · Replies 20 ·
Replies
20
Views
2K
Replies
2
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
55
Views
6K
Replies
6
Views
1K
Replies
1
Views
1K