# A gyroscope consists of a uniform disk with a 49 cm radius at the center of an axle that is 9 cm long and of negligible mass. It is horizontal and supported at one end. The disk is spinning around the axle at 990 rev/min, what is the precession rate?

Feb 8, 2016

$\setminus {\omega}_{p} = \setminus \frac{2 g r}{\setminus \omega {R}^{2}} = 0.03148$ rad/s $= 0.3007$ rev/min.

#### Explanation:

If $\setminus \vec{{F}_{}}$ is the force generating the torque $\setminus \vec{\setminus {\tau}_{}}$ on the gyroscope, the precession frequency $\setminus {\omega}_{p}$ is related to the angular momentum $\setminus \vec{{L}_{}}$ as follows. The angular momentum itself is related to the moment-of-inertia $I$ and spin frequency $\setminus \omega$.

Torque : $\setminus \vec{\setminus {\tau}_{}} \setminus \equiv \setminus \vec{{r}_{}} \setminus \times \setminus \vec{{F}_{}} = \setminus \frac{d \setminus \vec{{L}_{}}}{\mathrm{dt}}$,
Angular Momentum: $\setminus \vec{{L}_{}} \setminus \equiv I \setminus \omega$

Precission Frequency: $\setminus {\omega}_{p} = \frac{1}{L} \setminus \frac{\mathrm{dL}}{\mathrm{dt}} = \setminus \frac{\tau}{L}$

The torque generating force is the weight ( $F = M g$) of the spinning disk which acts vertically down. Since the gyroscope's axle is horizontal to the ground, $\setminus \vec{{r}_{}}$ and $\setminus \vec{{F}_{}}$ are perpendicular to each other.
$\setminus \tau = r . F . \sin {90}^{o} = r . M g . \sin {90}^{o} = M g r$

Moment of inertia of a disk of mass $M$ and radius $R$ is $I = \setminus \frac{M {R}^{2}}{2}$

Angular momentum $L = I . \setminus \omega = \frac{1}{2} M {R}^{2} \setminus \omega$

Precessional frequency $\setminus {\omega}_{p} = \setminus \frac{\tau}{L} = \setminus \frac{M g r}{\frac{1}{2} M {R}^{2} \setminus \omega} = \setminus \frac{2 g r}{\setminus \omega {R}^{2}}$

Given : $R = 49 c m = 0.49 m$;$\setminus \quad$$r = 4 c m = 0.04 m$$\setminus \quad$$g = 9.8 m {s}^{- 2}$
$\setminus \omega = 990$ rev/min $= 33 \setminus \pi$ rad/s,

$\setminus {\omega}_{p} = \setminus \frac{2 g r}{\setminus \omega {R}^{2}} = 0.03148$ rad/s $= 0.3007$ rev/min.