Chapter 11: Q68P (page 325)

A top spins at $30rev/s$ about an axis that makes an angle of $30°$ with the vertical. The mass of the top is $50kg$ , its rotational inertia about its central axis is ${5.0×10-4kg.m}^{2}$ , and its centre of mass is $4 .0 c m$ from the pivot point. If the spin is clockwise from an overhead view,
(a) what are the precession rate?
(b) what are the direction of the precession as viewed from overhead?

Short Answer

Expert verified

Precession rate is $2.08 {rad/s}^{2}$ .
Direction of precession is clockwise.

Step by step solution

Step 1: Given

$\begin{matrix} f = 3 0rev/sec \\ m = 0.5 kg \\ I = 5 \times 10^{- 4} kg . m^{2} \\ r = 0.04 m \end{matrix}$

Determining the concept

Use formula for precession rate in terms of mass, gravitational acceleration, spin angular frequency, and moment of inertia to find out the precession rate.

Formula are as follow:

$\begin{array}{l} ф = \frac{m \times g \times r}{Iω} \\ ω = 2 \times π \times f \end{array}$

where, $f$ is frequency, $m$ is mass, $g$ is acceleration due to gravity, $r$ is radius, $I$ is moment of inertia, $ω$ is angular frequency and $ф$ is precession rate.

Determining the precession rate

(a)

$\begin{array}{l} ω = 2 \times π \times f \\ ω = 2 π \times 30 \\ ω = 188.4 9rad/sec \end{array}$

Now,

$\begin{array}{l} ф = \frac{m \times g \times r}{Iω} \\ ф = \frac{0.5 \times 9.81 \times 0.04}{5 \times 10^{- 4} \times 188.49} \\ ф = 2.0 8rad/sec \end{array}$