How to Calculate Angular Velocity?

[Sneakatone]

a) I converted 33 km/h ---> 33/3.6=9.16m/s
9.16/0.5=18.33 rad/s

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

 

[Dick]

a) I converted 33 km/h ---> 33/3.6=9.16m/s
9.16/0.5=18.33 rad/s

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

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

 

[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}

 

[Dick]

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}

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.

 

[Sneakatone]

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

 

[Dick]

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

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

 

[Sneakatone]

P=m*v

 

[Dick]

P=m*v

That's linear momentum. Not angular.

 

[Sneakatone]

I don't suppose L=Iw would work since were trying to find w.
L=r*P might work since there's an origin .

 

[Dick]

I don't suppose L=Iw would work since were trying to find w.
L=r*P might work since there's an origin .

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.

 

[cmcraes]

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

 

[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
10992/2500=4.39 rad/s

 

[Sneakatone]

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

 

[Dick]

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

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

 

[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?

 

[Dick]

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?

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?

 

[Sneakatone]

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

 

[Dick]

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

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

 

[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

so is this method wrong?

 

[Dick]

so is this method wrong?

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?

 

[Sneakatone]

I don't really notice any errors except removing the 1/2 from the final.

 

[Dick]

I don't really notice any errors except removing the 1/2 from the final.

You are also subtracting final minus initial. Of course, that will be negative. Subtract initial minus final to get the kinetic energy lost. Final is rotational after the collision. Initial is linear, before the collision.

 

[Sneakatone]

so it should be
[1/2mv^2]-[Iw^2]?

 

[Dick]

so it should be
[1/2mv^2]-[Iw^2]?

You aren't really listening are you? It should be 2*[(1/2)mv^2]-[(1/2)Iw^2]. There are two cars and there's a (1/2) factor in the rotational energy as well.

 

[Sneakatone]

You aren't really listening are you? It should be 2*[(1/2)mv^2]-[(1/2)Iw^2]. There are two cars and there's a (1/2) factor in the rotational energy as well.

using that equation i ended up with 52506.48 J, which is two times what I originally had.

 

[Dick]

using that equation i ended up with 52506.48 J, which is two times what I originally had.

I've got to confess I'm not even tracking the numbers anymore. I've been trying to check the concepts more than the numbers.