# A solid disk, spinning counter-clockwise, has a mass of 15 kg and a radius of 6 m. If a point on the edge of the disk is moving at 2 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

Dec 30, 2016

The angular momentum of the disk is $90 \frac{k g {m}^{2}}{s}$, and the angular momentum is $\frac{1}{3} \frac{r a d}{s}$.

#### Explanation:

Angular momentum is given by $\vec{L} = I \omega$, where $I$ is the moment of inertia of the object, and $\omega$ is the angular velocity of the object.

The moment of inertia of a solid disk is given by $I = \frac{1}{2} m {r}^{2}$, and angular velocity is given by $\frac{v}{r}$, where $v$ is the tangential velocity and $r$ is the radius.

We are given that $m = 15 k g$, $r = 6 m$, and $v = 2 \frac{m}{s}$. We can use these values to calculate the moment of inertia and angular velocity, and ultimately the angular momentum.

$\omega = \frac{v}{r} = \frac{2 \frac{m}{s}}{6 m}$

$\implies \omega = \frac{1}{3} \frac{r a d}{s}$

This is the angular velocity. Note this is a positive quantity as the disk spins counter-clockwise, which by convention is the positive direction.

Note: The radian is a defined unit. It's definition of a ratio of two lengths makes it a pure number without dimensions. The unit of angle, be it radians, degrees, or revolutions, is really just a name to remind us that we're dealing with angle. Consequently, the radian unit seems to appear or disappear without warning. Above we have velocity divided by distance, which we would expect to have units of ${s}^{-} 1$, but because this division gives an angular quantity, we've inserted a dimensionless unit (radians) to give $\omega$ the appropriate units(1). Do not mistake this for a frequency and multiply by $2 \pi$.

$I = \frac{1}{2} m {r}^{2} = \frac{1}{2} \left(15 k g\right) {\left(6 m\right)}^{2}$

$\implies I = 270 k g {m}^{2}$

This is the moment of inertia.

We can now calculate the angular momentum:

$\vec{L} = I \omega = \left(270 k g {m}^{2}\right) \cdot \left(\frac{1}{3} \frac{r a d}{s}\right)$

$\implies \vec{L} = 90 \frac{k g {m}^{2}}{s}$

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1. Knight, R. D. In Physics for Scientists and Engineers ; Houston, A., Ed.; Pearson: Glenview, IL, 2013; pp 99, 102.