Question
A uniform disc with radius r and a mass of m kg is mounted centrally on a horizontal axle of negligible mass and length of 1.5r. The disc spins counter-clockwise about the axle with angular speed ω, when viewed from the right-hand side bearing, Q. The axle precesses about a vertical axis at ωp = ω/10 in the clockwise direction when viewed from above. Let RP and RQ (positive upwards) be the resultant reaction forces due to the mass and the gyroscopic effect, at bearings P and Q, respectively. Assuming ω2r = 300 m/s2 and g = 10 m/s2, the ratio of the larger to the smaller bearing reaction force (considering appropriate signs) is ______ .
Answer :
-3
ωp = ω/10 , ω²r = 300m/s², g = 10m/s², l = 1.5r
Gyroscopic couple, C = Iωωp
C = (mr²/2)ω.ωp = (mr /2).(rω²/10)= (mr/2) (300/10) = 15mr
Reaction due to weight, RP = RQ = mg/2 = 10m/2 = 5m (due to symmetry)
Reaction due to gyroscopic couple, RQ = 15MR/1.5R = 10m, RP = -10m
Now net reaction, RQ = 10m + 5m = 15m
RP = 5m - 10m = -5m
RQ /RP = 15m/5-m = -3
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