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## Homework Statement

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

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

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 = (r

^{3}/3)(sinφ)(ωM/l) |

^{1/2}

_{-1/2}

so after plugging everything in I get:

**L = (sinφ ω M l**

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.

^{2})/12MY 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 ∫

(diagram are included)

Thanks