Q1: A rigid body in the X-Y plane consists of two point masses (1 kg each) attached to the ends of two massless rods, each of 1 cm length, as shown in the figure. It rotates at 30 RPM counter-clockwise about the Z-axis passing through point O. A point mass of √2 kg, attached to one end of a third massless rod, is used for balancing the body by attaching the free end of the rod to point O. The length of the third rod is ________ cm. [GATE ME 2022 SET-2]
(a) 1
(b) √2
(c) 1/√2
(d) 1/2√2
Ans: (a)
m1 = 1kg, m2 = 1kg, r1 = 1cm, r2 = 1cm
Balancing mass, mb = √2kg
Making force polygon for complete balance
From the right angle triangle,
rb = 1
Q2: 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 ω 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 [GATE ME 2022 SET-2]
(a) rotate slowly (compared to ω) about the negative z-axis direction
(b) rotate slowly (compared to ω) about the positive z-axis direction
(c) rotate slowly (compared to ω) about the negative y-axis direction
(d) rotate slowly (compared to ω) about the positive y-axis direction
Ans: (a)
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 ω) about negative z−axis direction.
Q3: The figure shows a schematic of a simple Watt governor mechanism with the spindle O1O2 rotating at an angular velocity ω 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 O1 is denoted by h. The value of h = 400 mm at a certain ω. If ω is doubled, the value of hh will be _________ mm. [GATE ME 2022 SET-1]
(a) 50
(b) 100
(c) 150
(d) 200
Ans: (b)
h1 = 400mm, h2 = ?
ω1 = ω ω2 = 2ω
For Watt governor,
h = g/ω2
h ∝ 1/ω2
400 x ω2 = h2 x (2ω)2
h2 = 100 mm
Q1: 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.m2. The propeller rotates at ω = 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). [GATE ME 2021 SET-1]
Ans: 0 to 0
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 = 
Q2: Consider a reciprocating engine with crank radius R and connecting rod of length L. The secondary unbalance force for this case is equivalent to primary unbalance force due to a virtual crank of _____ [GATE ME 2021 SET-1]
(a) radius L2/4R rotating at half the engine speed
(b) radius R/4 rotating at half the engine speed
(c) radius R2/4L rotating at twice the engine speed
(d) radius L/2 rotating at twice the engine speed
Ans: (c)
Unbalanced secondary force,
Balancing radius 
Balancing crank speed = 2ω
Question for GATE Past Year Questions: Balancing & Gyroscope
Try yourself:The rotor of turbojet engine of an air craft has a mass 180 kg and polar moment of inertia 10 kg.m2 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 k m 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[2019]
Explanation
Given ,
m = 180 k g,
I = 10 k g.m 2,
V = 800 k m /h r.,
W = 1100 rad/s,
R = 1.5 km = 1500 m
We know that , Gyroscopic couple G = Iωωp
Effect ⇒ Nose goes down
= 1629.63 Nm
G = 1629.63 N .m
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Question for GATE Past Year Questions: Balancing & Gyroscope
Try yourself:A cantilever type gate hinged at Q is shown in the figure. P and R are the centers of gravity of the cantilever part and the counterweight respectively. The mass of the cantilever part is 75 kg. The mass of the counterweight, for static balance, is
[2008]
Explanation
For balancing the moments about the hinge Q,
75 × 2 = m R × 0.5
⇒ mR = 300 k g
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