Falling solid cylinder with string

[Suyash Singh]

No, I mean using the equation you wrote in post #21 what is T's torque about the axis of the cylinder?

Torque=F R sin(theta)

 

[haruspex]

Torque=F R sin(theta)

Yes, but that's a generic equation. You need to write it in terms of the variables in this question: T, r etc. What are F, R and θ here?

 

[Suyash Singh]

Yes, but that's a generic equation. You need to write it in terms of the variables in this question: T, r etc. What are F, R and θ here?

F= tension
R= radius of cylinder
theta = 90 degrees

Torque=T R

 

[haruspex]

F= tension
R= radius of cylinder
theta = 90 degrees

Torque=T R

Right, so correct your first equation in post #15 and go from there.

 

[Suyash Singh]

Torque= T R
=m(g-a)R

 

[haruspex]

Torque= T R
=m(g-a)R

This is the first equation in your post #15:

T= 1/(2) (mr^2) a/r

As I wrote, the left hand side is wrong because it is a force, not a torque. Replace that T by the torque it exerts (as in post #33).

 

[Suyash Singh]

Torque=F R
Torue=I alpha
=1/2 MR^2 a/R

1/2 MR^2 a/R = F R
1/2 Ma=F
a=2F/m
a=2(g-a)
3a=2g

a=2g/3

Tension=m(g-a)
=m(g-2g/3)
T=mg/3

 

[haruspex]

Torque=F R
Torue=I alpha
=1/2 MR^2 a/R

1/2 MR^2 a/R = F R
1/2 Ma=F
a=2F/m
a=2(g-a)
3a=2g

a=2g/3

Tension=m(g-a)
=m(g-2g/3)
T=mg/3

That's it.