Question

Been working on this one for a while - and can only get through part A...

for part B, I am trying 44.1+(28.836)(4.8x10^-2x2pi)=52.8 and 44.1-(28.836)(4.8x10^-2x2pi)=28.18 --> it keeps marking this wrong.

A demonstration gyroscope wheel is constructed by removing the tire from a bicycle wheel 0.655 m in diameter, wrapping lead wire around the rim, and taping it in place. The shaft projects 0.200 m at each side of the wheel, and a woman holds the ends of the shaft in her hands. The mass of the system is 9.00 kg ; its entire mass may be assumed to be located at its rim. The shaft is horizontal, and the wheel is spinning about the shaft at 5.10 rev/s .

A)Find the force each hand exerts on the shaft when the shaft is at rest.

B)Find the force each hand exerts on the shaft when the shaft is rotating in a horizontal plane about its center at 4.80×10^-2 rev/s .

C)Find the force each hand exerts on the shaft when the shaft is rotating in a horizontal plane about its center at 0.320 rev/s .

D)At what rate must the shaft rotate in order that it may be supported at one end only?

Answer 1

a) If the shaft is at rest, there is no net torque on the gyroscope, and each hand exerts an upward force with magnitude equal to half the weight mg

F = mg/2 = (9.00kg) (9.8m/s²)/2 = 44.1N

(b)

The net torque about the center will be the product of the precession frequency ? and the horizontal component of the wheel’s angular momentum ,

Thus the difference between the magnitudes of the applied forces, multiplied by the distance d = 0.200m, is the product

Denoting the two forces F_L and F_R (for left and right), we have equations

Dividing the first by d and adding to the second, and then subtracting as well, yields

For

(c) for

(d)

From relation ,

When ,  the force that one hand exerts goes to zero

the shaft must rotate at