A thin rod of length l and mass M rotates about a vertical axis through its center with angular velocity ω. The rod makes an angle φ with the rotation axis. Determine the magnitude and direction of L (angular momentum).
So we're given: mass - M, length - l, angular velocity - ω, angle - φ
2. Homework Equations
I know that I have to use L = r × p = r × vm
In this case L = I⋅ω is not valid because L and ω vectors are not perpendicular
The Attempt at a Solution
So it tried this...
I used the L = r × vm
but because my object is a rod with uniform mass I take the derivative of L = r × vm to get dL = r × vdm
because it's a derivative I have an infinitely small mass dm, so I said that dm=(M/l)dr
r is now the radius of a point mass on the rod from the center of mass
v is the velocity of that point mass, and since we know ω we can use v = ωr
and so I get that v = (r sinφ)ω [rsinφ is the perpendicular distance to dm from the axis]
1. I start here
dL = r × vdm
2. I now integrate both sides to get L
∫ dL = ∫ r(rsinf)wdm =
3. I integrate from -l/2 to l/2 because that is the length of the rod
L = ∫1/2-1/2 (r sinφ) w(M/l)dr
4. I now plug in the two values
L = (r3/3)(sinφ)(ωM/l) |1/2-1/2
so after plugging everything in I get: L = (sinφ ω M l2)/12
MY QUESTION: IS THIS CORRECT? I did a lot of guessing because I could not find anything similar to this in my physics book or online.
1. Is my thinking correct?
2. Am I allowed to integrate ∫ r(rsinf)wdm because r × vm is a cross product.
(diagram are included)