Question

A demonstration gyroscope wheel is constructed by removing the tire from a bicycle wheel 0.650 m in diameter, wrapping leas wire around the nm, and taping it in place. The shaft projects 0.210 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 6.00 kg; its entire mass may be assumed to be located at its nm. The shaft is horizontal and the wheel is spinning about the shaft at 5.30 rev/s. Find the force each hand exerts on the shaft when the shaft is rotating in a horizontal plane about its center at 0.300 rev/s. Enter your answers numerically separated by a comma. F_L, F_R = At what rate must the shaft rotate in order that it may be supported at one end only? Ohm =

Answer 1

The vertical forces must sum to zero.

Denote the upward forces that the hands exert as F_L and F_R. $\dpi{100} \small \tau$ = (F_L − F_R )r, where
r = 0.210 m.

The conditions that F_L and F_R must satisfy are $\dpi{100} \small F_{L}+F_{R}=w$ and $\dpi{100} \small F_{L}-F_{R}=\Omega \frac{I\omega }{r}$ , where the
second equation is $\dpi{100} \small \tau =\Omega L$ , divided by r. These two equations can be solved for the forces by first adding
and then subtracting, yielding $\dpi{100} \small F_{L}=\frac{1}{2}\left ( w+\Omega \frac{I\omega }{r} \right )$ and $\dpi{100} \small F_{R}=\frac{1}{2}\left ( w-\Omega \frac{I\omega }{r} \right )$ . Using the values

w=mg = (6 kg)(9.8 m/s²) = 58.8 N and

Radius of the bicycle wheel, a = 0.325 m

$\dpi{100} \small \frac{I\omega }{r}=\frac{6 \times 0.325^{2} \times 5.30 \times 2\pi}{0.210}=100.5 kg.m/s$

Gives

F_L = 29.4 N + $\dpi{100} \small \Omega$ (50.25N.s), F_R = 29.4 N - $\dpi{100} \small \Omega$ (50.25N.s).

Part C

To find the force each hand exerts on the shaft when the shaft is rotating in a horizontal plane about its center at 0.300 rev/s

$\dpi{100} \small \Omega$ = 0.3 rev/s =1.89 rad/s, F_L = 124.37 N, F_R = − 65.57 N, with the minus sign indicating a downward
force.

Part D

To find the rate of the shaft rotates in order that it may be supported at one end only

F_R = 0, gives  $\dpi{100} \small \Omega =\frac{29.4}{50.25}=0.585 rad/s$, which is 0.0931 rev/s.

$\dpi{100} \small \Omega =0.0931 rev/s$