Stability of a teeter-toy-classical mechanics

In summary, the conversation discusses a physics homework problem where two identical masses are hanging from a peg and the potential energy is taken at the pivot. The question asks for the correct solution, which is provided by another user. The conversation also addresses the use of the term "pivot" and urges the asker to submit their own attempt at the homework task.
  • #1
Saph
17
9

Homework Statement

:
teeter.totter-expl.c.jpg
[/B]
In the image above, the toy consists of two identical mass hanging from a peg. we take the zero potential energy at the pivot.

Homework Equations

:[/B]
http://www.physicsmynd.com/?p=1172

The Attempt at a Solution


Since the potential energy is zero at the pivot, and the potential energy is defined by U(θ)=mgh, I had:
U(θ)= mg(- l*cos(α+θ)- l*cos(α-θ))
but the solution is: U(θ)= mg(L*cos(θ) - l*cos(α+θ)) +mg(L*cos(θ) - l*cos(α-θ))

So my question is,what is wrong with my answer, and why?
 
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  • #2
The zero of PE is at the pivot point.
l*cos(α-θ) is not the vertical distance between the pivot point and the mass on the right.
 

Attachments

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Likes Saph
  • #3
Thank you very much sir, I think my problem was with English more than it was with physics, I thought what he meant be pivot is the joining point of the arms with the peg, now I got the solution right.
 
  • Like
Likes ERK1
  • #4
OK. Good.
 
  • #5
Hello, could you please publish or send to me the correct solution to that example?
 
  • #6
Hello, could you please publish or send to me the correct solution to that example?
 
  • #7
ERK1 said:
Hello, could you please publish or send to me the correct solution to that example?
That's not how things are done here. If you have been given the same homework task and need help, please post your own attempt; maybe in a new thread.
 
  • #8
Can someone do all my physics homeworks please and post them to me on time and not a minute late ?
 

1. What is the definition of stability in classical mechanics?

Stability in classical mechanics refers to the ability of a system to return to its original position or state after being disturbed by an external force. It is a measure of how easily a system can be disturbed or how much energy is required to destabilize the system.

2. How is stability of a teeter-toy determined?

The stability of a teeter-toy, or any system, is determined by analyzing its potential energy. A system with a lower potential energy is more stable because it requires less energy to return to its original state. In the case of a teeter-toy, this potential energy is influenced by factors such as the shape and weight distribution of the toy.

3. What factors affect the stability of a teeter-toy?

The stability of a teeter-toy is affected by several factors including the shape and weight distribution of the toy, the surface on which it is placed, and any external forces acting on the toy. Additionally, the height of the center of mass and the base of support also play a role in the stability of the toy.

4. How does the center of mass affect the stability of a teeter-toy?

The center of mass is a point where the entire mass of an object can be considered to be concentrated. In the case of a teeter-toy, the height of the center of mass in relation to the base of support affects its stability. A lower center of mass provides more stability as it requires more energy to disturb the system. Conversely, a higher center of mass makes the toy more unstable and easier to tip over.

5. How can the stability of a teeter-toy be increased?

To increase the stability of a teeter-toy, the center of mass can be lowered by adding weight to the base or by changing the shape of the toy. Increasing the base of support, for example by widening the legs of the toy, can also improve stability. Additionally, placing the toy on a surface with more friction can help prevent it from sliding and increasing its stability.

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