Falling solid cylinder with string

In summary: No, you hadbut the T is a force not a torque. Just replace the T with the torque that T exerts about the centre of the cylinder.torque is the tangential forceNo, a torque is not a force. How do you find the torque of a force about an axis?torque=R F sin(theta)where theta is the between R and FRight, so correct your equation in post #15 and go from there.torque=T RF= tensionR= radius of cylindertheta = 90 degreesTorque=T RRight, so correct your first equation in post #15 and go
  • #36
Suyash Singh said:
Torque= T R
=m(g-a)R
This is the first equation in your post #15:
Suyash Singh said:
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).
 
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  • #37
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
 
  • #38
Suyash Singh said:
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.
 
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