- #1

bcjochim07

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## Homework Statement

A dumbbell consists of two mass points of equal mass m at opposite ends of a rigid rod of length l and negligible mass. Calculate the angular momentum of this dumbbell as it rotates with constant angular velocity [tex]\omega[/tex] about an axis that makes a constant angle [tex]\theta[/tex] with the dumbbell and passes through its center of mass.

## Homework Equations

In my calculation, I'll use r=l/2

## The Attempt at a Solution

I'm not sure if this is correct, but here's what I did:**L**=

**r**X

**p**

Since there are two masses:

**L**=[tex]\Sigma[/tex]

**L**

_{i}=

**r**X

**p1**+

**r**X

**p2**= m[(

**r**X

**v1**)+ (

**r**X

**v2**)]

=m[(

**r**X(rsin[tex]\theta[/tex]

**[tex]\omega[/tex]**)+

**r**X(rsin[tex]\theta[/tex]

**[tex]\omega[/tex]**]

=mrsin[tex]\theta[/tex][

**r**X

**[tex]\omega[/tex]**+

**r**X

**[tex]\omega[/tex]**]

||

**L**||=mrsin[tex]\theta[/tex][r[tex]\omega[/tex]sin[tex]\theta[/tex]+r[tex]\omega[/tex]sin[tex]\theta[/tex]]

=2mr

^{2}sin

^{2}[tex]\theta[/tex][tex]\omega[/tex], r=l/2

**L**=(1/2)ml

^{2}sin

^{2}[tex]\theta[/tex] [tex]\widehat{n}[/tex] ;

where [tex]\widehat{n}[/tex] is a unit vector perpendicular to both

**r**and

**[tex]\omega[/tex]**

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