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GATE Mechanical Engineering (ME) 2019 Shift 2 Solved Paper

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Question : 37 of 65

Marks: +1, -0

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 processes about a vertical axis at

\omega_{p}=\frac{\omega}{10}

in the clockwise direction when viewed from above. Let

R_{e}

R_{Q}

(positive upwards) be the resultant reaction forces due to the mass and the gyroscopic effect, at bearings

P

Q

, respectively. Assuming

\omega^{2} r = 300 \frac{\text{m}}{\text{s}^{2}}

g = 10 \frac{\text{m}}{\text{s}^{2}}

, the ratio of the larger to the smaller bearing reaction force (considering appropriate signs) is

Your Answer:

Solution:

Gyroscope couple

(G C) = I \omega \omega_{p}

Using:

\omega_{p} = \frac{\omega}{10}

and

\omega^{2} r = 300

G C = \frac{m r^{2}}{2} \frac{\omega^{2}}{10} = 15 m r

Reaction at

P

and

Q

due to gravity

R_{p 1} + R_{q 1} = m g = 10 m

R_{p 1} = 5 m

\therefore R_{q 1} = 5 m

Reaction at P and

Q

due to GC

Taking momentum about point

p

.

R_{q 2} \times 1.5 r - 15 m r = 0

R_{q 2} = \frac{15 m r}{1.5 r} = 10 m

\therefore R_{p 2} = -10 m

Now,

\frac{R_{Q}}{R_{P}} = \frac{R_{q 1}+R_{q 2}}{R_{P 1}+R_{P 2}} = \frac{5+10}{5-10} = -3

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