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

[tiny-tim]

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.

 

[Delzac]

[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?

 

[tiny-tim]

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.

 

[Delzac]

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.

 

[tiny-tim]

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.

 

[Delzac]

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.

 

[tiny-tim]

Now I'm totally confused: you had a √5 earlier, not a √3.

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.

 

[Delzac]

The √5 is from a calculation error that i didn't pick up. kk. That cleared things up. Thanks for the help.