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

In summary: And finally equate the torque = - I * alpha.So in this case, don't we equate torque = mgdsin(theta) = - I * alpha, and take the distance from point of suspension to cm?In this case, the problem also states that the rods are massless, so the distances should be the same.My apologies if i can't comprehend this concept well, but i am trying to learn from all of you.Hi Delzac! :smile:We equate torque = mgd sin(theta) = - I... and we take the distance from point of suspension to cm. And finally equate the torque = - I * alpha.So
  • #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 = [tex] \frac{\sqrt3}{2}L [/tex]
Then is follows that torque is = [tex] \frac{\sqrt3}{2}L 2mg sin30 [/tex]
Then angular acceleration is = [tex] \frac{\sqrt3}{4L}g [/tex]
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.
 
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  • #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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