Question 1 |

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

rotate slowly (compared to \omega ) about the negative z-axis direction | |

rotate slowly (compared to \omega ) about the positive z-axis direction | |

rotate slowly (compared to \omega ) about the negative y-axis direction | |

rotate slowly (compared to \omega ) about the positive y-axis direction |

Question 1 Explanation:

The spin vector will chase the couple on torque vector and produce precision in system.

Hence precision will be -y direction. Rotate slowly (compared to \omega ) about negative z-axis direction.

Question 2 |

The figure shows a schematic of a simple Watt
governor mechanism with the spindle O_1O_2 rotating
at an angular velocity \omega about a vertical axis. The
balls at P and S have equal mass. Assume that there
is no friction anywhere and all other components are
massless and rigid. The vertical distance between
the horizontal plane of rotation of the balls and the
pivot O_1
is denoted by h . The value of h=400 mm
at a certain \omega . If \omega is doubled, the value of h will be
_________ mm.

50 | |

100 | |

150 | |

200 |

Question 2 Explanation:

h_1
= 400 mm, h_2
= ?

\omega _1=\omega \;\;\;\omega _2=2\omega

For Watt governor,

\begin{aligned} h&=\frac{g}{\omega ^2}\\ h\propto \frac{1}{\omega ^2}\\ \Rightarrow h_1\omega _1^2&=h_2\omega _2^2\\ 400 \times \omega ^2&=h_2 \times (2\omega )^2\\ h_2&=100mm \end{aligned}

\omega _1=\omega \;\;\;\omega _2=2\omega

For Watt governor,

\begin{aligned} h&=\frac{g}{\omega ^2}\\ h\propto \frac{1}{\omega ^2}\\ \Rightarrow h_1\omega _1^2&=h_2\omega _2^2\\ 400 \times \omega ^2&=h_2 \times (2\omega )^2\\ h_2&=100mm \end{aligned}

Question 3 |

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).

0 | |

1 | |

2 | |

4 |

Question 3 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

Gyroscopic couple =I\cdot (\vec{\omega _s} \times \vec{\omega _p})=0

Question 4 |

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_______

-2 | |

-3 | |

-6 | |

-9 |

Question 4 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

\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 5 |

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

1629.6 N-m and the nose goes up | |

1629.6 N-m and the nose goes down | |

162.9 N-m and the nose goes up | |

162.9 N-m and the nose goes down |

Question 5 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

\mathrm{I}=\mathrm{I} \omega_{\mathrm{S}} \omega_{\mathrm{p}}=1629.62 \mathrm{N}-\mathrm{m}

Nose goes down

There are 5 questions to complete.