# What is the angular momentum of a rod with a mass of 2 kg and length of 9 m that is spinning around its center at 12 Hz?

Jun 26, 2018

The angular momentum is given by

$\vec{L} = I \vec{\omega}$

where

$I = \text{moment of inertia } \left(k g \cdot {m}^{2}\right)$

omega="angular velocity "("rad/s")

For a thin rod with uniform density rotating about the centre, the moment of inertia is given by

$I = \frac{1}{12} \cdot M {L}^{2}$

$M = \text{the total mass of the rod } \left(k g\right)$

$L = \text{length of the rod } \left(m\right)$

You can find moment of inertia formulas for different shaped objects tabulated. In this case

$I = \frac{1}{12} \cdot 2 \cdot {9}^{2} = 13.5 \text{ kg} \cdot {m}^{2}$

We want to work in SI units, so convert the frequency to angular velocity using

$\omega = 2 \pi f = 2 \pi \cdot 12 = 24 \pi \text{ rad/s}$

This arises because each rotation is $2 \pi$ radians and there are 12 rotations per second, so you are rotating through $24 \pi$ radians per second.

Finally, substitute these values into the equation to calculate the angular momentum

vec L=Ivec omega=13.5*24pi=1017.876=1.02*10^3" "(" kg"*m^2)/s

Notice that we ditch the radians in the units. This is a a thing that I don't really understand but is due to the definition of the radian being a ratio of two lengths (so units cancel out). It gets used and dropped sporadically.

#equalrightsforallunits

Careful with significant figures too.