# What is the angular momentum of a gyroscope that is a solid cylinder with a radius of .24 m, a mass of 15 kg and a angular velocity of 140 rad/sec?

Sep 11, 2015

$\vec{L} = I \times \vec{\omega}$
or
L=Ixxomega =60.48 kg m^2 s^-1

#### Explanation:

The moment of inertia $I$ for a solid cylinder is $\frac{1}{2} m {r}^{2}$.

Substitute into the equation $L = I \times \omega$ to get;

$L = \frac{1}{2} m {r}^{2} \omega$.

Crunch in the numbers and you should get $60.48 k g {m}^{2} {s}^{-} 1$.

The formula for moment of inertia can be derived by exploiting the definition of moment of inertia, $I = \oint {r}^{2} \mathrm{dm}$.

Hope this helps. I would be happy to explain more on the derivation if you want. Cheers.