The rotor (flywheel) of a toy gyroscope hasmass 0.140 kg. Its moment of inertia about its axis is . The mass of the frame is 0.0250 kg. The gyroscopeis supported on a single pivot javascript:void(0); with its center of mass a horizontaldistance of 4.00 cm from the pivot. The gyroscope is precessing ina horizontal plane at the rate of one revolution in 2.20 s. Find the upward force exerted by thepivot. Find the angular speed with which the rotoris spinning about its axis, expressed in rev/min. Comments

Question

Alara Mccarthy · Accepted Answer

force exerted by the pivot is F=MΩ2r M is the mass of the fly wheel , ? is the precessionangular speed r is the horizontal distance for angular speed we can apply the formula ω=WrIΩ W is the weight = m g=0.0250kg×9.8ms ris the horizontal distance = 0.04 m I is the moment of inertia = 1.20×10−4kg×m2 ? is the precession angular speed = 1 rev/ 2.20s= 2? rad /2.20 s = 2.85 rad/s you plug the values andismplify you will get the answer

Accepted Answer

Omega=2πrad/T=2πrad/2.20s=2.856rad/s     A. FP=Wtoot=(0.165kg)(9.80m/s²)=1.62N  B. W=wr/|Omega=(0165kg)(9.8m/s²)(0.400m)/(1.20X10^-4kg•mg²)(2.856rad/s)  =189rad✓