Pulley problem with definite mass rope

[Hopelessmin]

Hi, I've been working on this problem for more than 4 hours now. and I haven't got a clue on how to finish the problem. ANY help would be tremendously appreciated!
Thanks!

Homework Statement

a pulley has a circumference of 1.20m and mass of 2.25kg. pulley is a solid uniform disk. a heavy rope, 8.00m in length with a mass of 4.80kg hangs over the pulley with one end of the rope 0.600m lower than the other end of the rope. the pulley is initially at rest. when the pulley is released, the pulley turns as the lower end of the rope accelerates downward. assume no slippage. what is the angualr velocity of the pulley at the moment the end of the rope leaves pulley?

Homework Equations

v=wr
I=mR^2
L1+L2+1/2(circumference)=8.00m
PEi=KE(translation)+KE(Rotation)

The Attempt at a Solution

ok so we know L1+L2+1/2(circumference)=8.00m; therefore, L1=3.4m, and L2=4.0m.
the initial potential energy is mg(4.00m) because the rope will leave pulley after it moves 4.00m more. The problem is that I don't know which mass to use. Do i use the mass of the whole rope? the mass changes as the rope falls over the pulley, as well as its acceleration.

ignoring the ambiguity of mass, our KE(trans)+KE(rotat)=1/2(mass1)(v^2)+1/2(mass of pulley)R^2)(w^2)=1/2(m)(rw)^2+1/2(mR^2)(w^2). In this equation, I don't know what is mass1. is it the mass of entire rope?

I am not even sure if my PE is right. So any help would be truly appreciated! Thanks in advance, Physics Forum members :)

 

[tiny-tim]

Welcome to PF!

Hi Hopelessmin! Welcome to PF!

(have an omega: ω and a squared: ² )

a pulley has a circumference of 1.20m and mass of 2.25kg. pulley is a solid uniform disk. a heavy rope, 8.00m in length with a mass of 4.80kg hangs over the pulley with one end of the rope 0.600m lower than the other end of the rope. the pulley is initially at rest. when the pulley is released, the pulley turns as the lower end of the rope accelerates downward. assume no slippage. what is the angualr velocity of the pulley at the moment the end of the rope leaves pulley?
…
ok so we know L1+L2+1/2(circumference)=8.00m; therefore, L1=3.4m, and L2=4.0m.
the initial potential energy is mg(4.00m) because the rope will leave pulley after it moves 4.00m more. The problem is that I don't know which mass to use. Do i use the mass of the whole rope? the mass changes as the rope falls over the pulley, as well as its acceleration.

All PE is relative!

So starting with "the initial potential energy is …" is wrong … just state what your "base" of PE is (in this case, choose "base" PE as being when the two ends are of equal length), and just use the difference in PE from that.

(and once you do that, it is obvious how much of the rope you use! )

ignoring the ambiguity of mass, our KE(trans)+KE(rotat)=1/2(mass1)(v^2)+1/2(mass of pulley)R^2)(w^2)=1/2(m)(rw)^2+1/2(mR^2)(w^2). In this equation, I don't know what is mass1. is it the mass of entire rope?

How much of the rope is moving? That's what has KE.

Then solve the equation KE + PE = constant.

 

[thomate1]

Dear student,

First of all, it's great that you have been working on this problem for 4 hours. It shows dedication and persistence, which are important qualities for a scientist.

Now, let's break down the problem. We have a pulley with a mass and circumference, and a heavy rope with a mass and length. The pulley is initially at rest, and when released, the lower end of the rope accelerates downward. We want to find the angular velocity of the pulley when the end of the rope leaves it.

To solve this problem, we need to use the equations you have listed. However, there are a few things we need to clarify first.

1. The mass of the rope: You are correct in questioning which mass to use in your calculation. In this case, we can assume that the mass of the rope is distributed evenly along its length, and therefore, we can use the total mass of the rope in our calculations. This will simplify our problem and give us a more accurate result.

2. The initial potential energy: You have correctly identified that the initial potential energy is mg(4.00m). However, we need to be more specific about the value of g we are using. In this problem, we can assume that g is equal to 9.8 m/s^2, the standard gravitational acceleration on Earth.

3. The moment of inertia of the pulley: You have correctly identified the moment of inertia of the pulley as I=mR^2. However, we need to be more specific about the value of R we are using. In this case, we can use the radius of the pulley, which is equal to its circumference divided by 2π (since circumference = 2πr).

Now, let's plug in the values we know into our equations:

KE(trans)+KE(rotat)=1/2(mass1)(v^2)+1/2(mass of pulley)(R^2)(w^2)

KE(trans) = 0 (since the pulley is initially at rest)
KE(rotat) = 1/2(I)(w^2) = 1/2(mR^2)(w^2)
KE(rotat) = 1/2(2.25)(1.20/2π)^2(w^2) = 0.00156w^2

PEi = mg(4.00m) = (