Angular Velocity Homework: Cylinder, 10kg Mass, 50m High

[mollybethe]

Homework Statement

a cylinder, r= 0.5 m with a mass 100 kg which is hanging 50 m above the ground. A rope of negligible mass which is 25 m long is wrapped around the cylinder. At the end of the rope a 10 kg mass is hanging. The hanging mass will fall and the rope will spin off the cylinder. What is ω at that point?

Homework Equations

I=.5MR^2
W=Fd
F=ma
Energy: W=.5Iw^2

The Attempt at a Solution

Initial angular velocity is 0 rad/s
r=.5m
F=ma=10*9.8=98N
W=98*25=2450 J
I=.5*100*.5^2=12.5
2450=.5*12.5w^2
w=19.79rad/s?

 

[tiny-tim]

welcome to pf!

hi mollybethe! welcome to pf!

(have an omega: ω and try using the X² icon just above the Reply box )

Initial angular velocity is 0 rad/s
r=.5m
F=ma=10*9.8=98N
W=98*25=2450 J
I=.5*100*.5^2=12.5
2450=.5*12.5w^2
w=19.79rad/s?

too difficult to read

use conservation of energy

(and you'll need an equation relating v and ω)​

 

[mollybethe]

Homework Equations

I=.5Mr²
Work=Fd
Force=ma
Energy: mgh=.5Iω²

The Attempt at a Solution

Initial angular velocity is 0 rad/s
r=.5m
F=ma=(10)(9.8)=98N
W=(98)(25)=2450 J
I=(.5)(100)(.5²)=12.5
2450=(.5)12.5ω²
ω=19.79rad/s---is that even close to right?

 

[gneill]

The gravitational potential energy that comes from the mass descending is going to end up as kinetic energy in both the mass (as its velocity) and the cylinder (as its rotation). Your solution is putting all of the energy into the cylinder, so the rotation rate you're finding is a bit too high.

 

[mollybethe]

Does that mean I need to use the energy equation:

mgh=Iω²+.5mv²

To find v=at; have to find X_f=.5at²
t=2.26s; v=9.8(2.26)=22.15m/s

(10)(9.8)(25)=(12.5)ω²+.5(10)22.15²
but that gives me -.16 for ω²??

Is my mgh correct?

 

[gneill]

Rotational energy is (1/2)Iω².

Yes, you want to equate the potential energy to the total energy as you surmised:

mgh = (1/2)Iω² + (1/2)mv²

Now, if you can find a relationship between ω and v, you can replace one of them and solve for for the other.

 

[gneill]

v=rω.

But isn't .5mv² the energy in the falling mass, I wasn't given the radius...

Yes, that's the energy of the falling mass.

The problem statement said, "Imagine a cylinder of radius 0.5 m with a mass of 100 kg". So you have the radius of the cylinder.

 

[mollybethe]

Can I do that? Can I replace ω with (v/r)? Is it the same velocity? So then my answer would be:

mgh=(1/2)Iω2 + (1/2)mv2

2450=v²(30)

v=9.04 m/s

ω=(9.04/.5)=18.07 rev/s?

 

[gneill]

Yup. That looks fine.

The radius of the cylinder is r, and the rope will be unspooling with velocity v. This is the same as having the circumference of the cylinder rotating with tangential velocity v. So that v = ωr.

 

[mollybethe]

Thank you for your help, it shouldn't be radians/sec, should I convert it?

 

[gneill]

Thank you for your help, it shouldn't be radians/sec, should I convert it?

Ah, I missed checking your units. Sorry about that. The 18.07 is in radians per second, not revolutions per second. If you want some other units you'll have to convert.