# Homework Help: Pulley with 2 hanging masses

1. Apr 28, 2014

### amondellio

1. The problem statement, all variables and given/known data
Use conservation of energy to find the final velocity of m1.
m1=0.05kg m2=0.02kg m3=0.5kg r=16cm h=2m

2. Relevant equations
Kinetic Energy=1/2m*v2
Rotational Kinetic Energy=1/2I*ω2
Gravitational Potential Energy=mgh

3. The attempt at a solution
0=(1/2m1v2) + (1/2m2v2) + (1/4m3)

I simplify that to:m1gh / m1+m2+1/4m3=v2

Using that equation I get an answer that seems highly unlikely and its different from the answer I got when I used rotational kinematics, which was v=-1.92 m/s

Can somebody please tell me what I'm doing wrong? Thanks for any help. I attached a picture with problem drawn out and my actual work so far which probably makes more sense than what I wrote here.

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2. Apr 28, 2014

### paisiello2

What happened to the 1/2 term?

3. Apr 28, 2014

### Staff: Mentor

Both masses are changing height (one goes down but the other goes up). So rethink the amount of energy supplied due to the change in gravitational potential energy.

4. Apr 28, 2014

### amondellio

Oh I just forgot to type it in. I've actually reworked the problem considerably. Starting with m1*gh + m2*gh = (1/2m1*w^2*r^2)+(1/2m2*w^2*r^2) + (1/2(1/2m3*w^2*r^2) and then factored and simplified that to: m1*gh + m2*gh = r^2(m1 + m2 + m3) +m2*gh(final height here)

I figure that the total initial energy is the sum of each block's gravitational potential, and that is equal to kinetic energy of both blocks and the pulley plus the final gravitational potential of m2. Using that equation I'm getting WAY closer to the target answer, coming up with 1.86.

5. Apr 28, 2014

### Staff: Mentor

Remember: One block goes up, the other goes down...

6. Apr 28, 2014

### amondellio

I've tried putting a negative sign like everywhere and its only changing the answer by a hundredth at most :/

7. Apr 28, 2014

### Staff: Mentor

Let's examine your work in more detail. First, what value do you calculate for the total energy obtained from the gravitational potential energy?

8. Apr 28, 2014

### amondellio

Initial gravitational potential I get .981 for m1 and .3924 for m2, total of 1.3734. Final gravitational potential is 0 for m1 and -0.7848 for m2, since I've set downward y motion as positive and the counterclockwise rotation of the pulley is positive. Still don't think that's right though. Should the kinetic energy be negative for one of the terms?

9. Apr 28, 2014

### Staff: Mentor

Let me rephrase. What is the overall net change in gravitational potential energy? One number.

No, kinetic energy is always positive.

10. Apr 28, 2014

### amondellio

ΔU=-0.7848-1.3734=-2.1582 and looking at that it seems like that's way too big of a difference. Should I switch the positive direction of motion? So that I have ΔU=0.7848-1.3734=-0.5886 that seems much more likely to me? Oh man I'm so confused by this

11. Apr 28, 2014

### Staff: Mentor

A simple way to look at it is to assume a zero reference for potential energy at the initial position for each mass, then calculate the change in potential energy for each. The one that falls will result in a gain in KE for the system. The one that rises will result in a loss of KE for the system. Knowing that one represents a gain and the other a loss means you only need to know the magnitudes of the changes, and can assign signs to the values according to whether they are a gain (+) or loss (-).

Your -0.5886 number has the right magnitude for the net change in PE. Note that the change in KE must have the opposite sign, since what's lost in PE results in a gain in KE.

12. Apr 29, 2014

### amondellio

That just made me even more unsure of what I'm doing. I appreciate your help so much, is there anyway to "dumb it down" further? I have been working on this problem for hours I'm getting so frustrated.

13. Apr 29, 2014

### amondellio

I understand that the blocks are moving different directions but mgh are each positive for both blocks so I don't understand what causes the negative.

14. Apr 29, 2014

### Staff: Mentor

What part is confusing? Can you be specific?

15. Apr 29, 2014

### Staff: Mentor

No, if the blocks are moving in opposite directions then one is moving an amount +h and the other an amount -h in whatever coordinate system you've chosen.

But if you do wish to consider the magnitudes only, you only need to assign a sign to each in order to calculate the net change according to whether they are adding to or subtracting from the change in KE:
$$\Delta KE = m_1 g |\Delta h_1| - m_2 g |\Delta h_2|$$

16. Apr 29, 2014

### amondellio

Uim1 = .05*9.81*2 = .981
Uim2 = .02*9.81*2 = .3924

Ufm1 = .02*9.81*0 = 0
Ufm2 = .02*9.81*4 = .7848

ΔU = Uf-Ui = ?

For the final PE calculations should I be using Δh (m1= -2/m2= +2) instead of h (m1= 0/m2= 4)?

17. Apr 29, 2014

### Staff: Mentor

If you use the initial positions of the masses as the zero reference for their PE's then both will have initial PE of zero (so you don't need to calculate and deal with the initial values of PE for them), and you only need to know by how much they move, which is -2m for m1 and +2m for m2.

18. Apr 29, 2014

### amondellio

So (.05*9.81*-2) - (.02*9.81*2) = -.5886

But you said: Your -0.5886 number has the right magnitude for the net change in PE. Note that the change in KE must have the opposite sign, since what's lost in PE results in a gain in KE.

So that sounds like it shouldn't be -.5886, but instead .5886?

19. Apr 29, 2014

### Staff: Mentor

PE + KE = constant

ΔKE = - ΔPE

If your change in PE is negative then the change in KE will be positive.

20. Apr 30, 2014

### amondellio

I figured out the solution yesterday, thanks for the help. I've posted a picture in case anybody is interested.

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21. Apr 30, 2014

### BiGyElLoWhAt

ok, assuming they start from rest, you have:
$E_{total} = KE + PE = 0 + U_{m_{1}} + U_{m_{2}} (+U_{pulley})$ ...although the pulley drops out.

Now take what you know from your initial conditions and apply CoE.

$E_{initial} \pm \Delta W = E_{final}$
So if you take the pulley, the 2 masses, and the mass of the earth as your system, and you set you GPE at the initial height of the masses (2m) what's your total energy initially?

Now what's your work? Keep in mind that $W = \int_{a}^{b}\vec{F} \cdot d\vec{r}$ and for a constant force parallel to the displacement this is equal to $F\Delta y$
How much work is done on the system? *Hint $M_{earth}$ is part of your system

So now that you've figured out your initial total energy and your work, you know what your final total energy should be. You also know what your final potential energy is. Now look at the first equation again: $E_{total} = KE + PE$
rearranging gives:
$E_{total} - PE = KE$
You know the left side, and hence know the right. This, however, is the total kinetic energy of your system.