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Gyroscopic Effects Questions and Answers

Answer: C

Precession is the slow change in the direction of the axis of rotation of a spinning object. In a gyroscope a torque applied perpendicular to the spin axis makes the axis turn about a third axis, which is the precession. Wobbling about the axis is nutation, the spinning motion is the spin, and the applied force is the torque and not the precession, so the other options describe other things.

The rate of precession equals the applied torque divided by the angular momentum of the spinning rotor, so a faster spin gives a slower precession.

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2.

A massive uniform rigid circular disc is mounted on a frictionless bearing at the end E of a massive uniform rigid shaft AE which is suspended horizontally in a uniform gravitational field by two identical light inextensible strings AB and CD as shown, where G is the center of mass of the shaft-disc assembly and \(g\) is the acceleration due to gravity. The disc is then given a rapid spin \(\omega\) about its axis in the positive x-axis direction as shown, while the shaft remains at rest. The direction of rotation is defined by using the right-hand thumb rule. If the string AB is suddenly cut, assuming negligible energy dissipation, the shaft AE will

Horizontal shaft AE along the x axis, hung from fixed points B and D by vertical strings AB and CD of length L. G lies between A and C, close to C. The disc at end E spins at omega about x. z points up, g acts downward

Exam Year(s)

GATE ME - 2022,

Answer: A

The shaft will rotate slowly about the negative z-axis direction: once AB is cut, the spinning disc makes the assembly precess about the vertical through C instead of falling.

With AB gone, string CD alone holds the assembly. G lies between A and C, so the weight \(W\) acts a distance \(a\) on the negative-x side of C. Taking x towards E, z up and y completing a right-handed set, the gravity torque about C is

\[ \begin{aligned} \vec{\tau} &= (-a\hat{i}) \times (-W\hat{k}) \\ &= aW(\hat{i} \times \hat{k}) = -aW\hat{j} \end{aligned} \]

The spin angular momentum is \(\vec{H} = I\omega\hat{i}\). For steady precession at rate \(\Omega\hat{k}\), \(\vec{\Omega} \times \vec{H} = \vec{\tau}\):

\[ \begin{aligned} \Omega I\omega\,(\hat{k} \times \hat{i}) &= \Omega I\omega\,\hat{j} \\ &= -aW\hat{j} \end{aligned} \]

so \(\Omega = -aW/(I\omega)\): rotation about the negative z axis, and slow because the rate falls as \(\omega\) rises.

Rotation about the y axis would be the shaft simply swinging down under gravity, which is what happens without the spin. The positive z direction would need the spin, or G's side of C, to be reversed.

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3.

The rotor of an aeroplane engine has a mass moment of inertia 1.0 kg \(\text{m}^{2}\). The engine rotates at a speed of 500 RPM in the clockwise direction if viewed from the front of the aeroplane. If the aeroplane while flying at 1200 km/hr turns with a radius of 2 km at same elevation, then the magnitude of the gyroscopic moment exerted by the rotor on the aeroplane structure in N m is

Exam Year(s)

GATE ME - 2026,

Answer: A

The gyroscopic moment is about 8.73 N m.

The spin speed is \(\omega = 2\pi(500)/60 = 52.36\) rad/s. The precession rate is the aeroplane's angular velocity in the turn, \(\omega_{p} = v/R\), with \(v = 1200/3.6 = 333.3\) m/s:

\[ \begin{aligned} \omega_{p} &= 333.3/2000 = 0.1667\ \text{rad/s} \\ C &= I\omega\omega_{p} = 1.0 \times 52.36 \times 0.1667 = 8.73\ \text{N m} \end{aligned} \]

17.46 and 26.19 are two and three times the value, and 4.37 half of it. The clockwise sense of rotation decides the direction of the moment (nose up or down), not its size.

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