GATE 2019 ME (ME2) – Question 51
A uniform disk of radius r and mass m is mounted centrally on a negligible-mass horizontal axle of length 1.5r. It spins counterclockwise at ω viewed from right bearing Q. The axle precesses clockwise viewed from above at $\omega_p=\omega/10$. Reactions $R_P,R_Q$ are positive upward and include weight and gyroscopic effects. Given $\omega^2r=300$ m/s² and g = 10 m/s², find the larger/smaller bearing-reaction ratio, with sign.
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Correct answer: -3 to -3
Explanation
Gyroscopic couple is $I\omega\Omega=(mr^2/2)\omega^2/10$. The corresponding equal-and-opposite bearing force is this couple divided by 1.5r, namely $mr\omega^2/30=10m$. Weight adds 5m upward at each bearing. Reactions are 15m and −5m, so the signed larger/smaller ratio is **−3**.