Question regarding tension in a string.

1. Aug 9, 2010

p75213

I have a cylinder wound with string. The free end of the string has a weight dangling from it. Is the tension in the string simply mass of the weight*gravity?

2. Aug 9, 2010

rcgldr

Ignoring the weight of the string, then the tension in the free part of the string is equal to the weight (mass x gravity).

3. Aug 9, 2010

p75213

A cylindrical spool with mass 0.5kg and radius 0.5m is wrapped with 4m of string. A 5kg mass is hung from the end of the string. How fast will the spool be rotating when all of the string has been pulled off?

angular velocity^2= (initial angular velocity)^2 +2*(angular acceleration)*theta.
omega^2=initial_omega^2+2*alpha*theta
initial angular velocity=0
theta=8
omega^2=2*alpha*8
Find angular acceleration (alpha).

Let T be the tension in the string.
T-mg=ma
T-5*9.8=5a
T-49=5a
T-49=5r*alpha (tangental accel = r*alpha)
T-49=2.5*alpha (a)
----------------------------------
Torque=I*alpha
Torque=rT
I*alpha=rT
0.0625alpha=0.5T
T=0.125alpha
----------------------------------
Substitute T=0.125alpha in (a)
0.125alpha-49=2.5*alpha
-2.375alpha=49
------------------------------------
omega=sqr(2*20.63*8)
------------------------------------
That's the worked solution. Now lets look at T - equation (a)
T-49=2.5*alpha
T=2.5*alpha+49
T=100.6N
I would have thought T was equal to mg=5*9.8=49N

4. Aug 9, 2010

Staff: Mentor

No, not if the mass is accelerating.

5. Aug 9, 2010

Staff: Mentor

Some points:
(1) Redo your solution as you've made an error. You have the tension greater than the weight of the hanging mass. That would mean the mass rises instead of falls!
(2) If the tension equaled mg, then the hanging mass would be in equilibrium.
(3) You can check your work by solving the problem using energy methods.

6. Aug 9, 2010

p75213

I see. That makes sense. That's what I was missing. The tension includes the acceleration of the string.

7. Aug 9, 2010

p75213

Good points. Yes 100N didn't make sense to me either. (-2.575N)