# What is the magnitude of the angular momentum of the bar?

1. Nov 10, 2012

### Toranc3

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

A rigid, uniform bar with mass m and length b rotates about the axis passing through the midpoint of the bar perpendicular to the bar. The linear speed of the end points of the bar is v . What is the magnitude of the angular momentum of the bar?

2. Relevant equations

L=Iw
I=1/12(ML^(2) This L is the length.
v=wr

3. The attempt at a solution

L=1/12*m*B^(2)*w

v=wr
it says that the linear speed of the bar at the end points is v
When I first read that I automatically thought that I would have to double v=wr because it has two endpoints. However, that is wrong. Why is it that I am only supposed to use a single v=wr?

2. Nov 10, 2012

### tiny-tim

Hi Toranc3!
v = ωr applies to the speed v of any point on the object at distance r from the centre of rotation

so v = ω(L/2) for one end, and v = ω(L/2) for the other end

3. Nov 10, 2012

### Toranc3

Hi tiny tim, thanks for replying.
So going back to angular momentum equation L=Iw, the omega applies only to a certain point of the rotation. Therefore when I substitute v=wr I am only replacing v/r=w at that certain point. So that is why I do not double it. Would it be correct to say that? The final answer is L=1/6mbv, forgot to add that in.

4. Nov 10, 2012

### tiny-tim

(try using the Quick Symbols box next to the Reply box )
no, ω is the same for every point of a rigid body

v is what varies

(ie the speed, v, is proportional to the distance, r, from the fixed point)

5. Nov 10, 2012

### Toranc3

Well I am still not so sure why but I am going to put this error that I did in my head so I will not make the same error again. Thanks though