A demonstration gyroscope wheel is constructed by removing the tire from a bicycle wheel 0.650 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 8.00 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 5.00 rev/s. Find the magnitude and direction of the force each hand exerts on the shaft (a) when the shaft is at rest; (b) when the shaft is rotating in a horizontal plane about its center at 0.050 rev/s; (c) when the shaft is rotating in a horizontal plane about its center at 0.300 rev/s. (d) At what rate must the shaft rotate in order that it may be supported at one end only?

Solution 102CP Step 1 of 5: The conditions that F L and F R must satisfy are F L F =Rw and F F = L , R Ir Where the second equation is = L, divided by r. These two equations can be solved for the force 1 I 1 I by the first adding and the subtracting, F =L(w +2 ) r and F =L(w 2 ) r w = mg = 8 kg × 9.8 m/s 2 = 78.4 N 1 I F L (w2+ ) r 1 8 kg×(0.325) (5 rev/s)(2×3.14 rad/rev) F L [(28.4 N) + ( 0.2 m )] F L [(28.4 N) + (132.7 kg.m/s)] F = 39.2 N + (66.4 N.s), L similarly, F R 39.2 N (66.4 N.s) Step 2 of 5: a) when shaft at rest, = 0 F L 39.2 N + 0(66.4 N.s) F L 39.2 N and F R 39.2 N + 0(66.4 N.s) F R 39.2 N Step 3 of 5: b) = 0.05 rev/ s = 0.314 rad/s, F L 39.2 N + 0.314(66.4 N.s) F L 60 N and F R 39.2 N 0314(66.4 N.s) F R 18.4 N