Gyroscope

Question 1
The figure shows an arrangement of a heavy propeller shaft in a ship. The combined polar mass moment of inertia of the propeller and the shaft is 100 kg.m^2. The propeller rotates at \omega =12 rad/s. The waves acting on the ship hull induces a rolling motion as shown in the figure with an angular velocity of 5 rad/s. The gyroscopic moment generated on the shaft due to the motion described is _______N.m (round off to the nearest integer).

A
0
B
1
C
2
D
4
GATE ME 2021 SET-1   Theory of Machine
Question 1 Explanation: 
As the axes of the rolling of the ship and the axes of the rotor are parallel, there is no precession of the axis of spin.

Gyroscopic couple =I\cdot (\vec{\omega _s} \times \vec{\omega _p})=0
Question 2
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 \omega, when viewed from the right-hand side bearing, Q. The axle precesses about a vertical axis at \omega_p=\omega /10 in the clockwise direction when viewed from above. Let R_P \; and \; R_Q (positive upwards) be the resultant reaction forces due to the mass and the gyroscopic effect, at bearings P and Q, respectively. Assuming \omega^2 r=300m/s^2 and g=10m/s^2, the ratio of the larger to the smaller bearing reaction force (considering appropriate signs) is_______
A
-2
B
-3
C
-6
D
-9
GATE ME 2019 SET-2   Theory of Machine
Question 2 Explanation: 
Given:
\begin{aligned} &\omega_{\mathrm{p}}=\frac{\omega}{10} \\ &\omega^{2}=300 \mathrm{m} / \mathrm{s}^{2} \\ &\mathrm{g}=10 \mathrm{m} / \mathrm{s}^{2} \\ &\Sigma \mathrm{F}_{\mathrm{y}}=0 \Rightarrow \mathrm{R}_{\mathrm{p}}+\mathrm{R}_{\mathrm{Q}}=\mathrm{mg} \ldots(i)\\ &\Rightarrow \mathrm{R}_{\mathrm{Q}}(1.5 \mathrm{r})=(\mathrm{mg} \times 1.5) \frac{\mathrm{r}}{2}+\mathrm{I} \omega . \omega_{\mathrm{p}} \\ &\Rightarrow(1.5 \mathrm{r}) \mathrm{R}_{\mathrm{Q}}=\mathrm{mg} \times \frac{1.5 \mathrm{r}}{2}+\frac{\mathrm{mr}^{2}}{2} \frac{\omega^{2}}{10} \\ &\Rightarrow(1.5 \mathrm{R}) \mathrm{R}_{\mathrm{Q}}=\mathrm{mg} \times \frac{1.5 \mathrm{r}}{2}+\frac{\mathrm{m}}{2} \frac{\left(\mathrm{r} \omega^{2}\right) \mathrm{r}}{10} \\ &\Rightarrow 1.5 \mathrm{R}_{\mathrm{Q}}=\frac{1.5 \mathrm{mg}}{2}+\frac{\mathrm{m}}{2} \times \frac{300}{10} \\ &\Rightarrow R_{Q}=15 \mathrm{m}\ldots(ii) \end{aligned}
From equation (i) and (ii)
\mathrm{R}_{\mathrm{P}}+15 \mathrm{m}=10 \mathrm{m}
R_{p}=-5 \mathrm{m}
\Rightarrow \frac{R_{\text {larger }}}{R_{\text {smaller }}}=\frac{R_{Q}}{R_{p}}=\frac{15 \mathrm{m}}{-5 \mathrm{m}}=-3
Question 3
The rotor of a turbojet engine of an aircraft has a mass 180 kg and polar moment of inertia 10 kg\cdot m^2 about the rotor axis. The rotor rotates at a constant speed of 1100 rad/s in the clockwise direction when viewed from the front of the aircraft. The aircraft while flying at a speed of 800 km per hour takes a turn with a radius of 1.5 km to the left. The gyroscopic moment exerted by the rotor on the aircraft structure and the direction of motion of the nose when the aircraft turns, are
A
1629.6 N-m and the nose goes up
B
1629.6 N-m and the nose goes down
C
162.9 N-m and the nose goes up
D
162.9 N-m and the nose goes down
GATE ME 2019 SET-1   Theory of Machine
Question 3 Explanation: 
\begin{aligned} \text{mass of rotor }&=180 \mathrm{kg} \\ \mathrm{I}&=10 \mathrm{kg}-\mathrm{m}^{2} \\ \omega_{\mathrm{s}}&=1100 \mathrm{rad} / \mathrm{s} \\ \mathrm{v}&=800 \mathrm{kmph} \\ \mathrm{R}&=1.5 \mathrm{km} \\ \end{aligned}


\mathrm{I}=\mathrm{I} \omega_{\mathrm{S}} \omega_{\mathrm{p}}=1629.62 \mathrm{N}-\mathrm{m}
Nose goes down
Question 4
A car is moving on a curved horizontal road of radius 100 m with a speed of 20 m/s. The rotating masses of the engine have an angular speed of 100 rad/s in clockwise direction when viewed from the front of the car. The combined moment of inertia of the rotating masses is 10 kg-m^{2}. The magnitude of the gyroscopic moment (in N-m) is __________
A
100
B
200
C
300
D
400
GATE ME 2016 SET-1   Theory of Machine
Question 4 Explanation: 
\begin{aligned} R&=100\;m \\ v&=20\;m/sec \\ \omega _p &=\frac{v}{R} \\ &= 0.2\; rad/sec\\ \omega _s &=100\; rad/sec \\ I&= 10\; Kg-m^2\\ &\text{Gyroscopic moment,}\\ I\omega _s\omega _p &=10 \times 0.2 \times 100\; N-m\\ &=200\; N-m \end{aligned}
There are 4 questions to complete.

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