Question

A toy gyroscope has a ring of mass M and radius R attached to the axle by lightweight spokes. The end of the axle is distance R from the center of the ring. The gyroscope is spun at angular velocity v, then the end of the axle is placed on a support that allows the gyroscope to precess. a. Find an expression for the precession frequency Ω in terms of M, R, v, and g. b. A 120 g, 8.0-cm-diameter gyroscope is spun at 1000 rpm and allowed to precess. What is the precession period?

Short answer

Part a

The expression for precession frequency is $\Omega =\frac{g}{R\omega }$.

Part b

The precession period is$5.4s$.

Step-by-step solution

Given information

The mass of the gyroscope is $M=120g$

role="math" localid="1650463286647" $R=8cm=0.08m$

The spin angular frequency of the gyroscope is role="math" localid="1650463469116" $\omega =1000rpm=104.7rad/s$

Part a

Consider a gyroscope as a hoop with a massless spoke.

The moment of inertia of a hoop is $I=M{R}^2\dots \left(1\right)$

The precession frequency is given by role="math" localid="1650462099722" $\Omega =\frac{Mdg}{I\omega }\dots \left(2\right)$

From (1) and (2),

role="math" localid="1650462308740" $$\begin{gathered} \Omega =\frac{Mdg}{M{R}^2\omega } \\ =\frac{Rg}{R^2\omega } \\ =\frac{g}{R\omega } \end{gathered}$$

Therefore, the expression for the precession frequency is$\Omega =\frac{g}{R\omega }$.

Part b

Substitute the given values into $\Omega =\frac{g}{R\omega }$.

role="math" localid="1650463338918" $$\begin{gathered} \Omega =\frac{9.80}{0.08\times 104.7} \\ =1.17rad/s \end{gathered}$$

The period is given by $T=\frac{2\mathrm{\pi }}{\Omega }$.

role="math" localid="1650463361689" $$\begin{gathered} T=\frac{2\mathrm{\pi }}{1.17} \\ =5.4s \end{gathered}$$

Therefore, the precession period is$5.4s$.