- #36
Delzac
- 389
- 0
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.
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.