# Help with a simple conversation of energy problem?

1. Oct 31, 2006

### the_quack

energy is conserved, no slipping or air resistance or anything.

There is a pulley, on one side is a mass resting on the ground, and on the other side is a larger mass above the ground. If the larger mass is released, what is it's speed when it hits.

okay, here is what I have:

PgM = Pgm + KM + KR + Km
Mgh = mgh + v[ srqt(.5M) + sqrt(.5I)/r + sqrt(.5m) ]

I=
.5mr^2 = .5(7.50kg)(.260m)^2 = 0.2535kg*m^2
h=3
m=18
M=26.5
so
PgM = 779.1 J
Pgm = 529.2 J

My answer is v(M)=49.84m/s, completely illogical, and the book answer is 3.22m/s. What did I do wrong? All the numbers look right, and it appears to make sense, but the answer is way off...

Thanks for any help!

Last edited: Oct 31, 2006
2. Oct 31, 2006

### Noein

The smaller mass also has kinetic energy just before the larger mass hits the ground.

3. Oct 31, 2006

### geoffjb

No conversation about energy is ever simple.

4. Oct 31, 2006

### drpizza

So, you're changing the potential energy of the large mass into:

potential energy for the small mass
energy in the pulley
kinetic energy of the large mass

There's one more thing with energy... if the large mass is moving, and it's attached to the small mass via a string and pulley...

5. Oct 31, 2006

### the_quack

hmm, I added the kinetic energy of the small mass and that only changed it a little...

Do you see any problems with my algebra? Is the formula correct?
PgM = Pgm + KM + KR + Km
Mgh = mgh + v[ srqt(.5M) + sqrt(.5I)/r + sqrt(.5m) ]

6. Oct 31, 2006

### the_quack

Any last-minute tips?

I have to quit pretty soon...

7. Nov 1, 2006

### matthew baird

I dont think so, you seem to think that you can just pull out that V and sqrt the other terms. Thats a problem. Also, the V in the energy for the pulley is radial velocity or w(omega) that is not the same velocity as the boxes.

8. Nov 1, 2006

### the_quack

I know, the V in the equations was squared, wasn't it?

And w = v/r

9. Nov 1, 2006

### BerryBoy

Something to think about. The potential energy of the large mass is going to:

1 - Give the smaller mass potential energy to raise to height (h).
2 - Give the system of masses (M+m) kinetic energy for a speed (v).
3 - Give the pulley rotational energy at angular speed (w = v/r)

You've written that (I assume) as:

PgM = Pgm + KM + Km + KR

If I you write this out without the square-root bits you get:

$$Mgh - mgh = \frac{1}{2} (M + m) v^2 + \frac{1}{2}I \omega ^2$$

Regards,
Sam

10. Nov 1, 2006

### BerryBoy

Now its over to you to re-arrange and solve for v.

Hope that helps :-)