# Homework Help: Moment of inertia help.

1. Apr 29, 2010

### pat666

1. The problem statement, all variables and given/known data

The pulley in the diagram has a radius of 0.160 m and a moment of inertia 0.480 kg.m². The rope does not slip on the pulley rim. Use energy methods to calculate the speed of the 4.00 kg block just before it strikes the floor

2. Relevant equations

3. The attempt at a solution

not sure how to start this could someone give me a hand please.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

#### Attached Files:

• ###### pic.doc
File size:
44 KB
Views:
100
2. Apr 29, 2010

### vela

Staff Emeritus
Start by identifying the different energies involved in the problem.

3. Apr 29, 2010

### pat666

taking a stab gravitational potential and kinetic??

4. Apr 29, 2010

### pat666

any more detailed help?? what is the energy in the pulley?

5. Apr 29, 2010

### vela

Staff Emeritus
Yes, those are the two types involved. Now write down an expression for the total kinetic energy in the system and the total potential energy of the system.

6. Apr 29, 2010

### pat666

i can do most of that but i dont understand the energy for the pulley.

7. Apr 29, 2010

### pat666

normally if there was no pulley it would be gph of 4kg = ke of 2kg

8. Apr 29, 2010

### pat666

ok so i can figure out the mass of the pulley from i=1/2mr^2 but that wouldnt help would it \??

9. Apr 29, 2010

### Jolsa

What is the kinetic energy of a rotating body?

10. Apr 29, 2010

### pat666

ok can you please look at this and get back to me soonish.....thanks
m_1* g*h_1+m_2 *g*h_2+1/2 I*ω^2=1/2 m_1 *v_1^2+1/2 m_2* v_2^2+1/2 I*ω^2 final
the left side is initial and the right final

11. Apr 29, 2010

### pat666

i was just trying to solve that equation i wrote when i realized there are two unknown velocitys. its not ok to say that the velocity of block to will be zero at the top is it?

12. Apr 29, 2010

### pat666

Just thought of something, if attached the two masses would have equal but opposite velocity's???????????

13. Apr 29, 2010

### vela

Staff Emeritus
Please put your equation between [ tex] and [ /tex] tags so it's easier to read.

Your equation isn't quite right. It's better if you approach this systematically. The total energy of the system is given by

$$E = m_1gh_1+m_2gh_2+\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2 v_2^2+\frac{1}{2}I\omega^2$$

That's just the potential energies of the two masses plus the kinetic energies of the two masses and the pulley. Now conservation of energy says that $E_i = E_f$. Each side of the equation will initially have the five terms. On the LHS, the heights and velocities will be their initial values, and on the RHS, you will have their final values. Some terms you will be able to erase because they're zero. At this point, you'll have the equation you were trying to write down earlier.

Now, as you already noted, you still have too many unknowns. Think about how the velocities of mass 1 and mass 2 are related. And how do they relate to the angular speed of the pulley?

14. Apr 29, 2010

### vela

Staff Emeritus
Yes! Now how is their speed v related to the angular speed ω of the pulley?

15. Apr 29, 2010

### pat666

Thankyou so much!!!!!!!!!!!!!!!! I can do it from here i think.

16. Apr 29, 2010

### pat666

sorry one more thing would you take the height of m2 to be 5m at the top?? (since there different physical sizes) and just to be sure the relation of speed and angular velocity is omega =speed(v)/radius since it is not sliping??

Last edited: Apr 29, 2010
17. Apr 29, 2010

### vela

Staff Emeritus
Yes and yes.

18. Apr 30, 2010

### pat666

Hey did you get 3.182m/s??

19. Apr 30, 2010

### vela

Staff Emeritus
No, I got 2.81 m/s.

20. Apr 30, 2010

### vela

Staff Emeritus
What's the original equation you started from? You either started with the wrong expression or you made a bunch of algebra errors.

21. Apr 30, 2010

### pat666

ok ill put it up now just wondering. if i went wron anywhere i think it was with the omega stuff i just subbed v^2/R^2 where omega^2 was

22. Apr 30, 2010

### pat666

[ tex]196.2=98.1+2v-v+0.24ω^2[ /tex]

23. Apr 30, 2010

### pat666

ok i am not sure how to display an equation properly sorry.

24. Apr 30, 2010

### pat666

ok i figured it out myself.... i cant do simple math was the problem, i hust solved my original equation with the 89 and got 2.76, thats close enough to yours to be accepted isnt it?

25. Apr 30, 2010

### vela

Staff Emeritus
The last term should have (v/r)^2, not v^2/r. The velocity is squared in every term, so the sign doesn't matter.