he rotor (flywheel) of a toy gyroscope has mass 0.150 kg . Its moment of inertia

Question

he rotor (flywheel) of a toy gyroscope has mass 0.150 kg . Its moment of inertia about its axis is 1.20×10-4kgm2. The mass of the frame is 2.70×102 kg . The gyroscope is supported on a single pivot with its center of mass a horizontal distance of 4.00 cm from the pivot. The gyroscope is precessing in a horizontal plane at the rate of one revolution in 2.20 s .

Find the upward force exerted by the pivot.

Find the angular speed with which the rotor is spinning about its axis, expressed in rev/min.

Explanation / Answer

Given that :

mass of the toy gyroscope, m1 = 0.15 kg

mass of the frame, m2 = 0.027 kg

moment of inertia about its axis, I = 1.2 x 10-4 kg.m2

distance between pivot and centre of mass, r = 4 cm = 0.04 m

(a) The upward force exerted by the pivot which will be given as :

using an equation,    Fup = (m1 + m2) g                                                             { eq.1 }

inserting the values in eq.1,

Fup = [(0.15 kg) + (0.027 kg)] (9.8 m/s2)

Fup = (0.177 kg) (9.8 m/s2)

Fup = 1.73 N

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