# How to find angular velocity?

1. Apr 16, 2013

### Sneakatone

a) I converted 33 km/h ---> 33/3.6=9.16m/s

b)2*1/2(1200kg)(18.33)^2=403186.68 J

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2. Apr 16, 2013

### Dick

Don't delete the homework help problem format. Show the formulas you are using and show your complete working. Please?

3. Apr 16, 2013

### Sneakatone

formulas
a) angular velocity w=v/r

b)kenetic energy 2*1/2(I)(w)^2 since 2*I{initial}*w{initial}=I{final}w{final}

4. Apr 16, 2013

### Dick

Your formula for angular velocity is of limited use. It's only good for circular motion. More to the point is a formula for angular momentum. Have you got one of those? Explain the concept you would use to solve the problem before you start plugging numbers into something. Use words, not numbers.

5. Apr 16, 2013

### Sneakatone

the formular for angular momentum is L=mvr in a circular orbit.

6. Apr 16, 2013

### Dick

Circular is not going to do you much good here. The initial motion isn't circular. Have you got another one?

7. Apr 18, 2013

### Sneakatone

P=m*v

8. Apr 18, 2013

### Dick

That's linear momentum. Not angular.

9. Apr 18, 2013

### Sneakatone

I dont suppose L=Iw would work since were trying to find w.
L=r*P might work since theres an origin .

10. Apr 18, 2013

### Dick

You will use L=Iw eventually, after the collision occurs and the motion is circular. It's not circular before they collide. So you need something like L= r x p.

11. Apr 18, 2013

### cmcraes

Arent L=rp and L=lw the same thing?
L=Iw=I*v/r=(mr^2)(v/r)=mvr=rp
lw=rp

12. Apr 18, 2013

### Sneakatone

to find w, I first used the equation L=r x p
L=1*(1200kg*9.16m/s) ---> L=10992
applying the equation L=Iw

13. Apr 18, 2013

### Sneakatone

for kinetic energy should I do
[1/2*Iw^2]-[1/2mv^2] to get the difference of initial and final kinetic energy?

14. Apr 18, 2013

### Dick

Basically, yes. Compute the total initial kinetic energy and then subtract the rotational kinetic energy of the wreck.

15. Apr 18, 2013

### Sneakatone

when I used [1/2*Iw^2]-[1/2mv^2] it came out to
[1/2*(2500)(4.39)^2]-[1/2(1200)(9.16)^2]=-26253.24
can you get negative kinetic energy?

16. Apr 18, 2013

### Dick

Actually it should be even more negative. There are TWO cars of mass 1200kg moving at 9.16m/s. You are subtracting the final kinetic energy from the initial kinetic energy. Why do YOU think it's negative?

17. Apr 18, 2013

### Sneakatone

the amount of energy used for rotational was higher that when it was linear.

18. Apr 18, 2013

### Dick

Oh, come on. You subtracted linear from rotational and got a negative. That means rotational is LESS than linear.

19. Apr 18, 2013

### Sneakatone

so is this method wrong?

20. Apr 18, 2013

### Dick

No, the method of finding the difference in kinetic energies by subtracting final from initial is fine. You just aren't doing that. You are subtracting 1/2 of the initial kinetic energy from the final kinetic energy. That's wrong in two ways. Can you find them?