The rotor (flywheel) of a toy gyroscope has mass 0.140 kg. Its moment of inertia about its axis is The mass of the frame is 0.0250 kg. The gyroscope is supported on a single pivot (?Fig. E10.51?) 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. (a) Find the upward force exerted by the pivot. (b) Find the angular speed with which the rotor is spinning about its axis, expressed in rev/min. (c) Copy the diagram and draw vectors to show the angular momentum of the rotor and the torque acting on it.

Solution 53E Step 1: a) The upward force is nothing but torque on the axis, = Force × Lever distance = mg r Mass of the system, m = mass of the gyroscope + mass of the frame = 0.140 kg + 0.0250 kg m = 0.165 kg Acceleration due to gravity, g = 9.8 m/s 2 Lever distance, r = 4 cm = 0.04 m 2 Therefore, torque, = 0.165 kg × 9.8 m/s × 0.04 m = 0.065 Nm The upward force exerted on the axis, = 0.065 Nm Step 2: b) The precession rate, = Torque / Angular momentum = mgr / I Where, I = moment of inertia of the gyroscope = Angular velocity of the rotor about its own axis. Provided, I = 1.20 × 10 kg m 2 Rate of precession, = 1 rev / 2.20 s = 0.454 rev/ s 1 rev = 2 rad Therefore, = 0.454 × 2 rad /s = 2.851 rad/s Substituting all these values in the above equation, -4 2 2.851 rad/s = 0.065 Nm / ( 1.20 × 10 kg m × ) -4 2 Rearranging the equation, we get, = 0.065 Nm / ( 1.20 × 10 kg m × 2.851 rad/s ) The angular velocity of the rotor, = 190 rad/s 1 rad = 1/2 revolutions 1 s = 1/60 min Therefore, 1 rad/s = (1/2) / ( 1/60) rev/min = 9.554 rev/min Therefore, angular velocity of the rotor, = 190× 9.554 rev/min = 1815.26 rev/ min