A gyroscope is a device that uses angular momentum to detect orientation and maintain stability. It consists of a spinning wheel or disk whose axis is free to orient in any direction. Gyroscopes are used for navigation and stabilization in ships, airplanes, drones, and other vehicles. They work by producing a gyroscopic effect - as the spinning axis rotates about another axis, conservation of angular momentum causes a reactive torque perpendicular to the plane of rotation. This effect counters external forces and helps maintain the orientation of the device.

1. A gyroscope is a device for measuring or maintaining orientation, based
on the principles of conservation of angular momentum.
In essence, a mechanical gyroscope is a spinning wheel or disk whose
axle is free to take any orientation.
 Gyroscopes are installed in ships in order to minimize the rolling and
pitching effects of waves. They are also used in aeroplanes, monorail
cars, gyrocompasses etc.

2. Gyroscope is a mechanism comprising a rotor journaled to spin
about one axis, the journals of the rotor being mounted in an inner
gimbal or ring; the inner gimbal is journaled for oscillation in an outer
gimbal for a total of two gimbals.
The outer gimbal or ring, which is the gyroscope frame, is mounted
so as to pivot about an axis in its own plane determined by the
support.
Outer gimbal possesses one degree of rotational freedom and its
axis possesses none.
The inner gimbal is mounted in the gyroscope frame so as to pivot
about an axis in its own plane that is always perpendicular to the
pivotal axis of the gyroscope frame (outer gimbal). This inner gimbal
has two degrees of rotational freedom.

3. Applications of gyroscopes include navigation for the stabilization of
flying vehicles like radio-controlled helicopters.
Due to their high precision, gyroscopes are also used to maintain
direction in tunnel mining.
Gyroscopes are also used in Air & Land Vehicles, Ships, Hovercrafts etc.
Vastly used in UAV’s (Unmanned Aerial Vehicle) commonly named as
DRONES.

4. The angular velocity of the axis of spin (i.e.dθ/dt) is known as angular
velocity of precession and is denoted by ωP.
The axis, about which the axis of spin is to turn, is known as axis of
precession.
The angular motion of the axis of spin about the axis of precession is
known as precessional angular motion.

5. The Cause of Precession: - Newton’s 1st Law of Motion.
- Law of Conservation of Angular Momentum.

6. The Cause of Precession: - Newton’s 1st Law of Motion.
- Law of Conservation of Angular Momentum.

7. Consider a disc spinning with an angular velocity ω rad/s about the axis
of spin OX, in anticlockwise direction when seen from the front.
Since the plane in which the disc is rotating is parallel to the plane YOZ,
therefore it is called plane of spinning.
The horizontal plane XOZ is called plane of precession and OY is the axis
of precession.

8. Since the angular momentum is a vector quantity, therefore it may be
represented by the vector ox. The axis of spin OX is also rotating
anticlockwise when seen from the top about the axis OY.
Let the axis OX is turned in the plane XOZ through a small angle δθ
radians to the position OX ′ , in time δt seconds. Assuming the angular
velocity ω to be constant, the angular momentum will now be
represented by vector ox′.

9. The couple I.ω.ωp, in the direction of the vector xx′ is the active
gyroscopic couple, which has to be applied over the disc when the axis of
spin is made to rotate with angular velocity ωP about the axis of
precession.
When the axis of spin itself moves with angular velocity ωP, the disc is
subjected to reactive couple whose magnitude is same but opposite in
direction to that of active couple. This reactive couple to which the disc is
subjected when the axis of spin rotates about the axis of precession is
known as reactive gyroscopic couple.
The gyroscopic couple is usually applied through the bearings which
support the shaft.

10. The fore end of the ship is called bow and the rear end is known as stern
or aft.
The left hand and right hand sides of the ship, when viewed from the
stern are called port and star-board respectively.

11. Steering is the turning of a complete ship in a curve towards left or
right, while it moves forward.
Consider the ship taking a left turn, and rotor rotates in the
clockwise direction when viewed from the stern.

12. When the rotor of the ship rotates in the clockwise direction when
viewed from the stern, it will have its angular momentum vector in the
direction ox as shown below.
As the ship steers to the left, the active gyroscopic couple will change the
angular momentum vector from ox to ox′. The vector xx′ now represents
the active gyroscopic couple and is perpendicular to ox.

13. Thus the plane of active gyroscopic couple is perpendicular to xx′ and its
direction in the axis OZ for left hand turn is clockwise.
The reactive gyroscopic couple of the same magnitude will act in the
opposite direction (i.e. in anticlockwise direction).
The effect of this reactive gyroscopic couple is to raise the bow and lower
the stern.
When the ship steers to the right under similar conditions as discussed,
the effect of the reactive gyroscopic couple will be to raise the stern and
lower the bow.

14. Pitching is the movement of a complete ship up and down in a vertical
plane about transverse axis. In this case, the transverse axis is the axis of
precession.
The pitching of the ship is assumed to take place with simple harmonic
motion i.e. the motion of the axis of spin about transverse axis is simple
harmonic.
When the pitching is upward, the effect of the reactive gyroscopic
couple, as shown in Fig.(b) will try to move the ship toward star-board.
 On the other hand, if the pitching is downward, the effect of the reactive
gyroscopic couple, as shown in Fig.(c), is to turn the ship towards port side.

16. For the effect of gyroscopic couple to occur, the axis of precession
should always be perpendicular to the axis of spin.
In case of rolling of a ship, the axis of precession (i.e. longitudinal axis) is
always parallel to the axis of spin for all positions. Hence, there is no
effect of the gyroscopic couple acting on the body of a ship.

17. Consider the four wheels A, B, C and D of an
automobile locomotive taking a turn towards left as
shown in the figure. The wheels A and C are the inner
wheels whereas B and D are the outer wheels. The
C.O.G. of the vehicle lies vertically above the road
surface.
Let,
M = mass of the vehicle in kg
W = weight of the vehicle in N = mg
R = radius of curvature in meters (R > r w )
X = width of track in meters

18. h = distance of C.O.G vertically above the road surface in meters
I W = mass of moment of inertia of one wheel in kgm2
ω W = angular velocity of the wheels or velocity of the spin in rad/s
I E = mass moment of inertia of the rotating parts of the engine in kgm^2
ω E = angular velocity of the rotating parts of the engine in rad/s
G = Gear ratio = ωE / ωw
v = linear velocity of the vehicle in m/s = ω W . r w
r w = radius of the wheel in meters
The weight of the vehicle (W) will be equally distributed over the four
wheels which will act downwards.
The reaction between each wheel and the road surface of the same
magnitude will act upwards.

19. Effect of the gyroscopic couple:
Velocity of precession, ωp= v/R
∴ Gyroscopic couple due to 4 wheels, C w = 4 I w ωw.ωp
and gyroscopic couple due to the rotating parts of the engine,
CE = IE.ωE.ωP = IE.G ωw.ωp.... ( G = ωE/ωW)
∴ Net gyroscopic couple,
C = C w ± CE = 4 I w ωw.ωp ± IE.G ωw.ωp = ωw.ωp (4 I w ± G.IE)
Due to the gyroscopic couple, vertical reaction on the road surface will be
produced.
Let the magnitude of this reaction at the two outer or inner wheels be ‘P’ N. Then
P × x = C or P = C/x
∴ Vertical reaction at each of the outer or inner wheels,
P /2 = C/ 2x

20. Effect of the centrifugal couple:
Centrifugal force,
FC =mv2 /R
∴ The couple tending to overturn the vehicle,
CO = FC × h= mv2 /R × h
This overturning couple is balanced by vertical reactions, which are vertically
upwards on the outer wheels and vertically downwards on the inner wheels.
Let the magnitude of this reaction at the two outer or inner wheels be Q.
Then
Q × x = CO or Q= CO /x= mv2 h/(Rx)
∴ Vertical reaction at each of the outer or inner wheels, Q/2= mv2 h/(2Rx)
∴ Total vertical reaction at each of the outer wheel, PO =W/4 +P/2 +Q/2
Total vertical reaction at each of the inner wheel, P1 =W/4 -P/2 -Q/2

21. Consider a two wheel vehicle (say a scooter or motor cycle) taking a right
turn.
Let
m = Mass of the vehicle and its rider in kg,
W = Weight of the vehicle and its rider in newtons = m.g

22. h = Height of the centre of gravity of the vehicle and rider,
r W = Radius of the wheels,
R = Radius of track or curvature,
I W =Mass moment of inertia of one of the wheels
ω W =Angular velocity of the wheels or velocity of spin in rad/s,
I E =Mass moment of inertia of the rotating parts of the engine
ω E= Angular velocity of the rotating parts of the engine in rad/s,
G = Gear ratio = ω E / ω W

23. Effect of gyroscopic couple:
v = ω W × r W or ω W = v / r W
ω E =G.ω W = G × v/ r W
∴ Total (I × ω) = 2 IW × ωW ± IE× ωE
=2 IW × v / r W ± IE × G × v/ r W = v/ r W (2 IW ±G. IE )
velocity of precession, ω P = v /R
When the wheels move over the curved path, the vehicle is always inclined
at an angle θ with the vertical plane as shown in Fig.(b). This angle is known
as angle of heel.
Gyroscopic couple,
C1 =I. ω cosθ × ω P = v/ r W (2 IW ±G. IE ) cosθ × v /R
=v2 /R. r W (2 IW ±G. IE ) cosθ

24. Effect of centrifugal couple:
Centrifugal force,
FC =mv2 /R
Centrifugal couple, C2 = FC × h cos θ = (mv2 /R)× h cos θ
centrifugal couple has a tendency to overturn the vehicle,
Total overturning couple,
C0 = Gyroscopic couple + Centrifugal couple
= v2 /R.r W (2 IW + G. IE ) cosθ + mv2 /R × h cosθ
= v2 /R [(2 IW + G. IE )/ r W + m.h] cosθ
Balancing couple = m.g.h sin θ
The balancing couple acts in clockwise direction when seen from the front of the
vehicle. Therefore for stability, the overturning couple must be equal to the
balancing couple, i.e.
v2 /R [(2 IW + G. IE )/ r W + m.h] cos θ = m.g.h sin θ