A simple gyroscope consists of a wheel (M = 2.0 kg, R = 0.75 m, I = 1.1 kg∙m²)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)1.8 x 10^-2 rad/s B)2.7 x 10^-2 rad/s C)0.17 rad/s D)0.26 rad/s E)0.50 rad/s

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Quizplus · Accepted Answer

The precession frequency ($\Omega$) of a gyroscope can be calculated using the formula $\Omega = \frac{mgr}{I\omega}$, where $m$ is the mass of the wheel, $g$ is the acceleration due to gravity (approximately $9.8 m/s^2$), $r$ is the distance from the pivot to the center of mass of the wheel, $I$ is the moment of inertia of the wheel, and $\omega$ is the angular velocity of the wheel in rad/s. First, convert the rotation rate from rev/min to rad/s: $500 \text{ rev/min} = \frac{500 \times 2\pi}{60} \text{ rad/s}$. Then, substitute the given values into the formula to find $\Omega$.

A Simple Gyroscope Consists of a Wheel (M = 2