A simple gyroscope consists of a wheel (R = 0.75 m, I = 0.9 MR²)rotating on a horizontal shaft.If the shaft is balanced on a pivot 0.50 m from the wheel, and the wheel rotates at 500 rev/min, what is the precession frequency of the wheel? A)1.9 x 10^-2 rad/s B)2.9 x 10^-2 rad/s C)0.19 rad/s D)0.28 rad/s E)cannot be calculated without knowing the mass of the wheel

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Quizplus · Accepted Answer

The precession frequency of a gyroscope can be calculated using the formula $\Omega = \frac{r\omega^2}{g}$, where $\Omega$ is the precession frequency, $r$ is the distance from the pivot to the center of mass, $\omega$ is the angular velocity of the wheel, and $g$ is the acceleration due to gravity. Given the rotation rate of 500 rev/min, we first convert this to rad/s: $500 \text{ rev/min} \times \frac{2\pi \text{ rad}}{1 \text{ rev}} \times \frac{1 \text{ min}}{60 \text{ s}} = \frac{500 \times 2\pi}{60} \text{ rad/s}$. Plugging the values into the formula, we can solve for the precession frequency without needing the mass of the wheel, as it cancels out in the calculation.

A Simple Gyroscope Consists of a Wheel (R = 0