# A light string is wrapped around a solid cylinder

1. Dec 15, 2013

### robbyrandhawa

1. The problem statement, all variables and given/known data

a light string is wrapped around a solid cylinder and a 300 g mass hangs from the free end of the string. When released, the max falls a distance 54 cm in 3.0s.

b) calculate tension in string
c) calculate mass of cylinder

2. Relevant equations

F=ma
I=1/2(mr^2)
a=2y/t^2

3. The attempt at a solution

I got part b, T = 2.9N however for part C i am completely lost.

I know i have to use I=1/2(mr^2) but i am not sure how to go about this.

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Dec 15, 2013

### Kishlay

the question is not complete. the radius of the cylinder should be given.............

3. Dec 15, 2013

### ehild

No, it cancels

ehild

4. Dec 15, 2013

### Kishlay

ok.........

5. Dec 15, 2013

### Kishlay

ya i got it.....

6. Dec 15, 2013

### ehild

You did not specify, but the cylinder should rotate about a fixed horizontal axis. There is some relation between the torque acting on it and its angular acceleration. There is also some relation about the angular acceleration of the cylinder and the linear acceleration of the mass.

ehild

7. Dec 15, 2013

### robbyrandhawa

so i have to use t=rF and I=1/2mr^2 and then solve for m??

8. Dec 16, 2013

### ehild

No. You do not need and can not solve for the mass of the cylinder. I meant that there is some relation between the angular acceleration of the cylinder and the linear acceleration of the hanging body. They are connected with the string. The hanging mass can descend only when the string unwinds from the coil. See the figure. The hanging mass moves downward and the cylinder rotates clockwise. There is some tension T in the string. What are the forces acting on the hanging mass? What torque acts on the cylinder?

ehild

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9. Jan 19, 2016

### Jam51

I am working on this problem, and when solving for the tension in the string, am I assuming it is prior to release? where: ΣF = FT - mg = 0, so T = 2.94 Nm? Or should I be solving for ΣF = FT - mg = ma and use Δy of 0.54m in 3s to solve for a?

10. Jan 19, 2016

### haruspex

I would assume the tension asked for is after release.

11. Jan 19, 2016

### Jam51

OK.

So, am I correct in saying:
y = 1/2at^2
0.54m = 1/2a(3)^2
a = 0.12m/s^2

Then,
ΣF = FT - mg = ma
FT = m(g+a)
= .3kg (9.8 + 0.12m/s^2)
= 2.976 Nm

I just realized with a being so small it didn't make much of a difference. Did I get a wrong?

12. Jan 19, 2016

### haruspex

Which way is the mass accelerating?
Also, units for tension would be N, not Nm.

13. Jan 19, 2016

### Jam51

Oh, right. That's torque...

The mass would be accelerating downward, and cylinder clockwise.

14. Jan 19, 2016

### haruspex

Right.
It's always a bit hard commenting on such equations without knowing how you are defining directions.
From your first calculation, it looks like you are taking down as positive for a, and you later use a positive value for g. So which way is the net force FT-mg?

15. Jan 20, 2016

### Jam51

I thought if acceleration is positive downward than I can use a positive g? So the net force would be down as well...

16. Jan 20, 2016

### haruspex

Right, so in your equation $F_T-mg=ma$, which way is the force on the left and which way is the acceleration on the right?

17. Jan 20, 2016

### Jam51

Ah,

So it should be mg - FT = ma?

18. Jan 20, 2016

### haruspex

Yes.

19. Jan 20, 2016

### Jam51

So I get:
ΣF = mg - FT = ma
FT = m(g - a)
= .3kg (9.8 - 0.12m/s^2)
= 2.9 N

Then for part c (mass of cylinder), I tried using α = a/r, τ = FTr, I = 1/2MR^2, and τR = Iα to solve for mass, but I'm still left with r from α = a/r

FTr x R = I α
FT R^2 = 1/2MR^2 α
FT = 1/2M a/r

20. Jan 20, 2016

### Jam51

Nevermind, I don't know where I came up with τR = Iα