A light string is wrapped around a solid cylinder

[robbyrandhawa]

Homework Statement

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

Homework Equations

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

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.

any help please!

 

[Kishlay]

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

 

[ehild]

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

No, it cancels

ehild

 

[Kishlay]

ok...

 

[Kishlay]

ya i got it...

 

[ehild]

Homework Statement

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

Homework Equations

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

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.

any help please!

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

 

[robbyrandhawa]

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

 

[ehild]

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

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

 

[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?

 

[haruspex]

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?

I would assume the tension asked for is after release.

 

[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?

 

[haruspex]

OK.
ΣF = FT - mg = ma

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

 

[Jam51]

Oh, right. That's torque...

The mass would be accelerating downward, and cylinder clockwise.

 

[haruspex]

The mass would be accelerating downward, and cylinder clockwise.

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 F_T-mg?

 

[Jam51]

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

 

[haruspex]

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

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?

 

[Jam51]

Ah,

So it should be mg - FT = ma?

 

[haruspex]

Ah,

So it should be mg - FT = ma?

Yes.

 

[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

 

[Jam51]

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

 

[Jam51]

FTr = I α
FT r = 1/2MR^2 (a/r)
FT = 1/2M a
M = 2FT / a = 2(2.9) / 0.12 = 48.3kg

?

 

[haruspex]

FTr = I α
FT r = 1/2MR^2 (a/r)
FT = 1/2M a
M = 2FT / a = 2(2.9) / 0.12 = 48.3kg

?

Looks right.

 

[Jam51]

Great. Thanks for your help!