The rotor (flywheel) of a toy gyroscope has a mass of 0.160 k g . Its moment of

Question

The rotor (flywheel) of a toy gyroscope has a mass of 0.160kg . Its moment of inertia about its axis is 1.15

The rotor (flywheel) of a toy gyroscope has a mass of 0.160kg . Its moment of inertia about its axis is 1.15 times 10?4kg?m2 . The mass of the frame is 2.90 times 10?2kg . The gyroscope is supported on a single pivot (see figure) with its center of mass a horizontal distance of 4.35cm from the pivot. The gyroscope is precessing in a horizontal plane at the rate of 1.20 revolution in 2.50s. Find the angular speed with which the rotor is spinning about its axis, expressed in rev/min

Explanation / Answer

know that I=(1/2)*M*R^2 so I found the radius of the rotor to be .0464 meters.

I have torque=I*alpha and L=I*omega
Therefore I can say that torque/alpha = L/omega
Solving for omega, i get omega=(L*alpha)/torque
and since L=r*p and p=m*v, I get L=r*m*v
Also, I know that torque=r*F
Plugging the L and torque into the omega equation, I get omega = r^2*m*v*a/(r*F) which simplifies into omega=r*m*v*a/F

ec au = rac{dec L} {dt}

Notice that the torque is at right angles to the angular momentum vector. When the time derivative of some vector is normal to the vector, the vector's length doesn't change but its direction does. It rotates or precesses. The precession is related to the torque by

ec au = ec omega_p imes ec L

You computed the torque, and you already know the precession rate. Since the vectors are at nice ninety degree angles, you can divide the torque by the precession rate to determine the angular momentum. From that point you should be able to compute the angular velocity.

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