What are the Tensions and Accelerations in a Rotating Triangle Mass System?

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Homework Help Overview

The problem involves a system of two point masses attached to the vertices of an equilateral triangle, which is pivoted and allowed to swing in a vertical plane under the influence of gravity. Participants are tasked with computing the tensions in the rods and the accelerations of the masses immediately after release.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the applicability of angular momentum, torque, and moment of inertia, questioning the type of acceleration required (linear vs. angular). Some express uncertainty about the relationship between the two masses' accelerations and the nature of the forces acting on them.

Discussion Status

There are multiple interpretations of the problem, with some participants suggesting calculations for angular acceleration while others focus on linear acceleration. Various values for acceleration have been proposed, but no consensus has been reached regarding the correct approach or the validity of the assumptions made.

Contextual Notes

Participants note that certain concepts, such as angular momentum, have not been covered in class, which may limit their ability to apply certain methods. There is also a discussion about the implications of using the center of mass for torque calculations and the potential inaccuracies that may arise from this approach.

  • #31
Delzac said:
Isn't the centre of mass in between the 2 mass.

That's right … it's closer than they are (to the pivot), so it doesn't work.
 
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  • #32
[STRIKE]Are they? Isn't the centre of mass \frac{\sqrt5}{2} L away from pivot? Which is more that 1L.[/STRIKE]

My goodness blooper. You are right, the distance is shorter. But still, is there any proof that the formula breaks down when r becomes shorter?
 
Last edited:
  • #33
Delzac said:
My goodness blooper. You are right, the distance is shorter. But still, is there any proof that the formula breaks down when r becomes shorter?

It must do … mr + mr is different from 2m times something less than r. :wink:
 
  • #34
Here is a question from mastering physics.

If you would take a look, they did what you said cannot work.

The distance of centre of mass is smaller than L. But they still equate it nonetheless.
 

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  • #35
Yes, but they're using I as well as d …

the I of the original object.

In our case, a = Lα = Lτ/I = LmgLsin60°/2mL2 = g√3/4.
 
  • #36
Your calculations doesn't use centre of mass.

You either can use centre of mass to calculate or your cannot. How can there be " but they are using I as well as d..".

So, if i were to use centre of mass the calculations is as follows.

Distance of centre of mass from pivot = \frac{\sqrt3}{2}L
Then is follows that torque is = \frac{\sqrt3}{2}L 2mg sin30
Then angular acceleration is = \frac{\sqrt3}{4L}g
To find linear acceleration, you take angular acceleration multiply by L, and you get g√3/4.

Which is identical to the result you calculated using only one mass for torque. (since the other mass doesn't contribute to torque, but contribute to moment of inertia)

So, question here is, can i use centre of mass to do torque calculations regardless of whether the r is smaller or larger.

My calculations above seem to suggest i can.
 
  • #37
Now I'm totally confused: you had a √5 earlier, not a √3. :confused:

Anyway, yes, you can use centre of mass for the torque only, but it's difficult to see why you would bother in a case like this where it involves an extra line of calculation, with the danger that extra lines always carry, of making a mistake.
 
  • #38
The √5 is from a calculation error that i didn't pick up. kk. That cleared things up. Thanks for the help.
 

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