GYROSCOPE - Darshan Institute of Engineering & Technology · 2020. 7. 23. · 4. Gyroscope Dynamics...

Department of Mechanical Engineering Prepared By: Vimal Limbasiya Darshan Institute of Engineering & Technology, Rajkot Page 4.1

4 GYROSCOPE

Course Contents

4.1 Principle of Gyroscope

4.2 Angular Velocity

4.3 Angular Acceleration

4.4 Gyroscopic Couple (Torque)

4.5 Gyroscopic effect on

Aeroplane

4.6 Gyroscopic effect on Naval

ship

4.7 Stability of an Automobile

4.8 Stability of Two Wheel

Vehicle

4.9 Rigid disc at an angle fixed

to a rotating shaft

4. Gyroscope Dynamics of Machinery (3151911)

Prepared By: Vimal Limbasiya Department of Mechanical Engineering Page 4.2 Darshan Institute of Engineering & Technology, Rajkot

4.1 Principle of Gyroscope

If the axis of spinning or rotating body is given an angular motion about an axis

perpendicular to the axis of spin, an angular acceleration acts on the body about

the third perpendicular axis. The torque required to produce this acceleration is

known as active gyroscopic torque.

A reactive gyroscopic torque or couple also acts similar to the concept of

centripetal and centrifugal forces on a reacting body. The effect produced by the

reactive gyroscopic couple is known as gyroscopic effect. Thus aeroplanes, ships,

automobiles, etc., that have rotating parts in the form of wheels or rotors of

engines experiences this effect while taking turn, i.e., when the axes of spin is

subjected to some angular motion.

4.2 Angular Velocity

The angular velocity of a rotating body is specified by

the magnitude of velocity

the direction of the axis of rotor

the sense of rotation of the rotor, i.e., clockwise or counter-clockwise

Angular velocity is represented by a vector in the following manner:

(i) Magnitude of the velocity is represented by the length of the vector.

(ii) Direction of axis of the rotor is represented by drawing the vector parallel

to the axis of the rotor or normal to the plane of the angular velocity.

(iii) Sense of rotation of the rotor is denoted by taking the direction of the

vector in a set rule. The general rule is that of a right-handed screw, i.e., if a

screw is rotated in the clockwise direction, it goes away from the viewer

and vice-versa.

Fig. 4.1

For example, Fig. 4.1(a) shows a rotor which rotates in the clockwise direction

when viewed from the end A. Its angular motion has been shown vectorially in

Fig.4.1 (b). The vector has been taken to a scale parallel to the axis of the rotor.

The sense of direction of the vector is from a to b according to the screw rule.

However, if the direction of rotation of the rotor is reversed, it would be from b to

a [Fig.4.1(c)].

Dynamics of Machinery (3151911) 4. Gyroscope


4.3 Angular Acceleration:

Let a rotor spin (rotate) about the horizontal axis Ox at a speed of ω rad/s in the direction

as shown in Fig.(a). Let oa represent its angular velocity Fig. (b).

Fig. 4.2

Now, if the magnitude of the angular velocity changes to (ω+δω) and the direction of the

axis of spin to Ox’ (in time δt). The vector ob would represent its angular velocity in the

new position. Join ab which represents the change in the angular velocity of the rotor. The

vector ab can be resolved into two components:

(i) ac representing angular velocity change in a plane normal to ac or x-axis, and

(ii) cb representing angular velocity change in a plane normal to cb or y-axis.

Change of angular velocity,

( )cosac

Rate of change of angular velocity

( )cos

t

0

( )coslimt

Angular accelerationt

𝐴𝑠𝛿𝑡 → 0, 𝛿𝜃 → 0 𝑎𝑛𝑑 cos 𝛿𝜃 → 1

Angular acceleration

0limt

d

t dt

Change of angular velocity,

𝑐𝑑 = (𝜔 + 𝛿𝜔) sin 𝛿𝜃

Rate of change of angular velocity

( )sin

t




0

( )sinlimt t

𝐴𝑠𝛿𝑡 → 0, 𝛿𝜃 → 0 𝑎𝑛𝑑sin𝛿𝜃 → 𝛿𝜃


0

( )sinlimt

d

t dt

Total angular acceleration,

d d

dt dt

This shows that the total angular acceleration of the rotor is the sum of

1. dω/dt, representing change in the magnitude of the angular velocity of the rotor

2. ω.dθ/dt, representing change in the direction of the axis of spin, the direction of cb is

from c to b in the vector diagram (being a component of ab), the acceleration acts

clockwise in the vertical plane xy. (When viewed from front along the y-axis)

4.4 Gyroscopic Couple (Torque)

Let I be the moment of inertia of a rotor and ω its angular velocity about a

horizontal axis of spin Ox in the direction as shown in Fig.4.3 (a). Let this axis of

spin turn through a small angle δθ in the horizontal plane (xy) to the position Ox' in

time δt.

Fig. 4.3

Fig.4.3(b) shows the vector diagram. oa represents the angular velocity vector

when the axis is Ox and ob when the axis is changed to Ox'. Then ab represent the

change in the angular velocity due to change in direction of the axis of spin of the

rotor.

This change in the angular velocity is clockwise when viewed from a towards b and

is in the vertical plane xz. This change results in angular acceleration, the sense and

direction of which are the same as that of the change in the angular velocity.



Change in the angular velocity, 𝒂𝒃 = 𝜔 × 𝛿𝜃

Angular acceleration,

d

dt

In the limit, when𝛿𝑡 → 0,

d

dt

Usually, dθ/dt the angular velocity of the axis of spin is called the angular velocity

of precession and is denoted by ωp.

Angular acceleration 𝛼 = 𝜔.ωp

The torque required to produce this acceleration is known as the gyroscopic torque

and is a couple which must be applied to the axis of spin to cause it to rotate with

angular velocity ωp about the axis of precession Oz.

Acceleration torque, 𝑇 = 𝐼 ∙ 𝛼 = 𝐼 ∙ 𝜔.ωp

For the configuration of Fig.4.3(a)

Ox is known as the axis of spin

Oz is known as the axis of precession

Oy is known as the axis of gyroscopic couple

yz is plane of spin (parallel to plane of rotor)

xy is plane of precession

yz is plane of gyroscopic couple

The torque obtained above is that which is required to cause the axis of spin to

precess in the horizontal plane and is known as the active gyroscopic torque or the

applied torque.

A reactive gyroscopic torque or reaction torque is also applied to the axis which

tends to rotate the axis of spin in the opposite direction i.e., in the counter-

clockwise direction in the above case.

Just as the centrifugal force on a rotating body tends move the body tends to move

the body outwards, while a centripetal acceleration (and thus centripetal force)

acts on it inwards, in the same way, the effects of active and reactive gyroscopic

torques can be understood.

The effect of the gyroscopic couple on a rotating body is known as the gyroscope

effect on the body. A gyroscope is a spinning body which is free to move in other

directions under the action of external forces.



4.5 Gyroscopic effect on Aeroplanes

Fig. 4.4 (a) shows an aeroplane in space. Let the propeller be rotating in the

clockwise direction when viewed from the rear end. The angular momentum

vector oa due to the angular velocity is shown in Fig 4.4. (b).

If the plane takes a left turn, the angular momentum vector is shifted and may be

represented by the vector ob. The change is shown by the vector ab and is the

active gyroscopic couple. This vector is in the horizontal plane and is perpendicular

to the vector oa in the limit. The reactive vector is given by b'a' which is equal and

opposite to the vector ab. The interpretation of this vector shows that the couple

acts in the vertical plane and is counter-clockwise when viewed from the right-

hand side of the plane. This indicates that it tends to raise the nose and depress

the tail of the aeroplane.

Fig. 4.4

Fig.4.4 (c) shows the gyroscopic effect, when the aeroplane takes the right turn.

The change is shown by the vector cd and is the active gyroscopic couple. It is

perpendicular to the vector oc in the limit in the horizontal plane. The reactive

couple is given by d'c'. The couple acts in the vertical plane and is clockwise when

viewed from the right-hand side of the plane. Thus, it tends to dip the nose and

raise the tail of the aeroplane.

If the rotation of the engine is reversed, i.e., it rotates counter-clockwise when

viewing from the rear end, the angular momentum vector is oe as shown in Fig.4.4

(d). On taking a left turn, it changes to of. The active gyroscopic vector is ef and the

reactive fe'. Viewing from the right-hand side of the plane, it indicates that the

nose is dipped and the tail is raised. Similarly, when the plane takes a right turn,

the effect is indicated in Fig.4.4 (e). The nose is raised and the tail is depressed.



It can be concluded from the above cases that if the direction of either the spin of

the rotor or of the precession is changed, the gyroscopic effect is reversed, but if

both are changed, the effect remains same.

4.6 Gyroscopic Effect on Naval Ships

Some of the terms used in connection with the motion of naval ships or sea vessels

are given below.

Bow is the fore or the front end.

Stern or aft is the rear end.

Starboard is the right hand side when looking from stern.

Port is the left-hand side when looking form the stern.

Steering is turning on the side when viewing from the top.

Pitching is limited angular motion of the ship about the transverse axis.

Rolling is limited angular motion of the ship about the longitudinal axis.

Let the plane of spin the rotor and other rotating masses be horizontal and across

the breath of the ship. Assume 𝜔 to the angular velocity of the rotor in the clockwise

direction when viewed from stern(rear end).

Fig. 4.5

1. Gyroscopic effect during Pitching

When the ship turns left the angular momentum vector changes from oa to ob. The

reaction couple is found to be b ̍ a ̍which tends to raise the bow and lower the stern.

On the turning right the reaction couple is revered. So that bow is lower and stern is

raised.

2. Gyroscopic Effect on Pitching

Pitching of the ship is usually considered to take place with simple harmonic motion.

A simple harmonic motion is represented by x = X sin ω0t



Such a motion is obtain by the projection of a rotating vector X on a diameter while

rotating around a circle with a constant angular velocity ω0 and where x is a

displacement from the time mean position in time t.

In such the way angular displacement θ of the axis of the spin from its mean position

is given by 𝜃 = φ sin 𝜔0𝑡.

Where φ = amplitude (angular) of swing or the maximum angle turned from the

mean position in radius

02

Angular Velocity of SHMTime Period

0 0Angular Velocity of Precession, . cosd

tdt

This is maximum when 𝑐𝑜𝑠𝜔0𝑡 = 1

Therefore, maximum angular velocity of precession 𝜔𝑝 = φ ∙ 𝜔0

2Gyroscopic Couple

Time PeriodpI I

When the bow is rising, the reaction couple is clockwise on viewing from top and

thus the ship would move towards right or starboard side. Similarly, when the bow is

lowered, the ship turns toward left or port side (Fig. 4.4(c)].

20 0Angular . sinacceleration t

20Maximum angular .acceleration

3. Gyroscopic Effect on Rolling

As the axes of the rolling of the ship of the rotor are parallel, there is no precession

of the axis of spin and thus there is no gyroscopic effect.

In the same way, the effect on steering, pitching or rolling can be observed when the

plane of the spin of the rotating masses is horizontal but along the longitudinal axis

of the vessel or the axis is vertical.

4.7 Stability of an Automobile

In case of a four wheeled vehicle, it is essential that no wheel is lifted off the ground

while the vehicle takes a turn. The condition is fulfilled as long as the vertical

reaction of the ground on any the wheels is positive (or upwards). Fig.4.6 shows a four wheeled vehicle having a mass m. assuming that the weight is

equally divided among the four wheels,

Weight on each wheel ( )4 4

W mgdownwards

Reaction of ground on each wheel, R ( )4 4

w

W mgupwards



Fig. 4.6

1. Effect of Gyroscopic Couple

Gyroscopic couple due to four wheels,

2

C 4 4w w w p wv

I IrR

Where 𝐼𝑤 = mass moment of inertia of each wheel

𝜔𝑤 = angular velocity of wheel = 𝑣/𝑟

𝜔𝑝 = angular velocity of precession = 𝑣/𝑅

𝑣 = linear velocity of the vehicle

𝑅 = radius of curvature

Gyroscopic couple due to engine rotating parts, Ce e e p e w pI I G

where G is the gear ratio e

w

Total gyroscopic couple, CG w eC C

Positive sign is used when the engine parts rotate in same direction as the wheels

and the negative sign when they rotate in the opposite.

Assuming that CG is positive and the vehicle takes a left turn, the reaction gyroscopic

couple on it is clockwise when viewed from the rear of the vehicle. The reaction

couple is provided by equal and opposite forces on the outer and the inner wheels

of the vehicle.



Forces on the two outer wheels (downwards)GC

w

Forces on the two inner wheels (upwards)GC

w

Forces on each of the outer wheels (downwards)2

GC

w

Forces on each of the inner wheels (upwards)2

GC

w

Thus the forces on each of the outer wheels is similar to the weight. On the inner

wheels it is in the opposite direction. Thus,

Reaction of ground on each outer wheel, R (upwards)2

GG

C

w

Reaction of ground on each inner wheel, R (downwards)2

GG

C

w

2. Effect of Centrifugal Couple

As the vehicle moves on a curved path, a centrifugal force also acts on the vehicle in

the outward direction at the centre of mass of the vehicle.

2 22Centrifugal Force, p

v vmR mR m

R R

This force would tend to overturn the vehicle outwards and the overturning couple

will be

2

2C p

vC mR h m h

R

This is equivalent to a couple due to equal and opposite forces on outer and inner

wheels.

Forces on each outer wheels (downwards)2

CC

w

Forces on each inner wheels (upwards)2

CC

w

Again, the force on each of the outer wheels is similar to the weight on each of the

inner wheels, it is opposite.

Reaction of ground on each outer wheel, R (upwards)2

CC

C

w

Reaction of ground on each inner wheel, R (downwards)2

CC

C

w



Vertical reaction on each outer wheel (upwards)4 2 2

G CC CW

w w

Vertical reaction on each inner wheel (upwards)4 2 2

G CC CW

w w

It can be observed that there are chances that the reaction of the ground on the

inner wheels may not be upwards and thus the wheels are lifted from the ground.

For positive reaction, the Conditions will be

04 2 2 4 2

G C G CW G C

C C C CW Wor or R R R

w w w

4.8 Stability of Two Wheel Vehicle

The case of two-wheel vehicle can be taken in the same way as that of an auto

mobile. However, it easier to tilt such a vehicle inwards to neutralise the overturning

effect and the vehicle can stay in equilibrium while taking a turn. Let a vehicle take a left turn as shown in fig.4.7 (a). The vehicle is inclined to the

vertical (inwards) for equilibrium. The angle of inclination of the vehicle to the

vertical is known as the angle of heel.

Fig.4.7

Let, v = linear velocity of vehicle on the track

r = radius of the wheel

R = radius of the track

Iw = moment of inertia of each wheel

Ie = moment of inertia of rotating parts of the engine

m = total mass of the vehicle and the rider

𝜔𝑤 = angular velocity of the wheel

𝜔𝑒 = angular velocity of rotating parts of engine

G = Gear ratio



h = height of center of mass of the vehicle and the rider

θ = inclination of vehicle to the vertical (angle of heel)

As the axis of spin is not horizontal but inclined to the vertical at an angle θ and

the axis of precession is vertical, it is necessary to take the horizontal component

of the spin vector.

Spin vector (horizontal) cos (2 )cosw w w e eI I I

2

gyroscopic couple (2 )cos

(2 ) cos

(2 ) cos

w w e w p

w e

w e

I I G

v vI GI

r R

vI GI

rR

The reaction couple b’ a’ is clockwise when viewed for the rear (back) of the vehicle

and tends to overturn it in the outward direction.

2

Overturning couple due to Centrifugal Force cosv

m hR

2 2

2

Total Overturning couple (2 ) cos cos

2 cos

w e

w e

v vI GI m h

rR R

I GIvmh

R r

𝑅𝑖𝑔ℎ𝑡𝑒𝑛𝑖𝑛𝑔 (𝑏𝑎𝑙𝑎𝑛𝑐𝑖𝑛𝑔)𝑜𝑢𝑝𝑙𝑒𝑑𝑢𝑒𝑡𝑜𝑡ℎ𝑒𝑤𝑒𝑖𝑔ℎ𝑡 = 𝑚𝑔ℎ𝑠𝑖𝑛θ.

2 2For equalibrium, cosw e

I GIvmh mg h sin

R r

From this Relation, the angle of heel θ can be determined to avoid skidding of the

vehicle.

4.9 Rigid disc at an angle fixed to a rotating shaft

Consider a circular disc fixed rigidly to a rotating shaft in such a way that the polar

axis of the shaft makes angel θ with the axis of the shaft (Fig.4.8). Assume that the

shaft rotates clockwise with angular velocity ω when viewed along the left end of

the shaft.

Fig. 4.8



Let OX be the axis of the shaft

OP be the polar axis of the disc and

OD the horizontal diametral axis of the disc

Also, let m, r and t be the mass, radius and the thickness of the disc. Then

2

moment of inertia of disc about polar axis OP2

p

mrI

2 2 2

moment of inertia of disc about diametral axis

12 4 4

dI

t r mrm

(i) First consider the spinning about the polar axis

Angular velocity of spin = Angular velocity of Disc about the polar axis OP

= ω cos θ

Angular velocity of precession = Angular velocity of disc about the diametral axis

OD

= ω sin θ

2

1gyroscopic couple cos sin sin2

2p pI I

Its effect is to rotate the disc counter-clockwise when viewing from the top.

(ii) Now consider the spinning about the diametral axis

Angular velocity of spin = Angular velocity of disc about the diametral axis OD =

ω sin θ Angular velocity of precession = Angular velocity of disc about the polar

axis OP

= ω cos θ

21

gyroscopic couple sin cos sin22

d dI I

Its effect is to rotate the disc clockwise when viewing from the top. (Angular

velocity of precession is counter-clockwise when viewing from the right end along

OP.)

Resultant gyroscopic couple on the disc,

2 22 2

22

1 1sin2 sin2

2 2 2 4

sin28

p d

mr mrC I I

mr



Example 4.1 : A uniform disc of diameter 300 mm and of mass 5 kg is mounted on one end

of an arm of length 600 mm. The other end of the arm is free to rotate in a universal

bearing. If the disc rotates about the arm with a speed of 300 r.p.m. clockwise, looking from

the front, with what speed will it precess about the vertical axis?

d = 300 mm

m = 5 kg

l = 600 mm

N = 300 rpm

2 2 300

60 60

N=31.42 rad/s

the mass moment of inertia of the disc, about an axis through its centre of gravity and

perpendicular to the plane of disc,

2 25 (0.15)

2 2

mrI = 0.056 kg-m2

Couple due to mass of disc,

C = m.g.l = 5 × 9.81 × 0.6 = 29.43 N-m

Couple (C) = I.ω.ωP

29.43 = 0.056 × 31.42 × ωP = 1.76 ωP

Speed of precession, ωP = 29.43/1.76 = 16.7 rad/s

Example 4.2 : A uniform disc of 150 mm diameter has a mass of 5 kg. It is mounted centrally

in bearings which maintain its axle in a horizontal plane. The disc spins about it axle with a

constant speed of 1000 r.p.m. while the axle precesses uniformly about the vertical at 60

r.p.m. The directions of rotation are as shown in Fig. 4.9. If the distance between the

bearings is 100 mm, find the resultant reaction at each bearing due to the mass and

gyroscopic effects.

Fig 4.9

d = 150 mm

m = 5 kg

x = 100 mm

N = 1000 rpm

Np = 60 rpm



2 2 1000

60 60

N=104.7 rad/s (anticlockwise)

2 2 60

60 60

p

p

N=6.284 rad/s (anticlockwise)

The mass moment of inertia of the disc, about an axis through its centre of gravity and

perpendicular to the plane of disc,

2 25 (0.075)

2 2

mrI = 0.014 kg-m2

Gyroscopic Couple acting on the disc,

(C) = I.ω.ωP = 0.014 × 104.7 × 6.284 = 9.2 N-m

The direction of the reactive gyroscopic couple is shown in Fig. 4.10 (b). Let F be the force at

each bearing due to the gyroscopic couple.

∴ F = C/x = 9.2/0.1 = 92 N

The force F will act in opposite directions at the bearings as shown in Fig. 4.10 (a). Now let

RA and RB be the reaction at the bearing A and B respectively due to the weight of the disc.

Since the

disc is mounted centrally in bearings, therefore,

RA = RB = 5/2 = 2.5 kg = 2.5 × 9.81 = 24.5 N

Fig. 4.10

Let RA1 and RB1 = Resultant reaction at the bearings A and B respectively.

Since the reactive gyroscopic couple acts in clockwise direction when seen from the front,

therefore its effect is to increase the reaction on the left hand side bearing (i.e. A) and to

decrease the reaction on the right hand side bearing (i.e. B).

∴ RA1 = F + RA = 92 + 24.5 = 116.5 N (upwards)

and RB1 = F – RB = 92 – 24.5 = 67.5 N (downwards)



Example 4.3 : An aeroplane makes a complete half circle of 50 metres radius, towards left,

when flying at 200 km per hr. The rotary engine and the propeller of the plane has a mass of

400 kg and a radius of gyration of 0.3 m. The engine rotates at 2400 r.p.m. clockwise when

viewed from the rear. Find the gyroscopic couple on the aircraft and state its effect on it.

R = 50 m v = 200 km/hr

m = 400 kg k = 0.3 m

N = 1000 rpm

2 2 2400

60 60

N= 251 rad/s

The mass moment of inertia of the engine and the propeller

I = m.k2 = 400(0.3)2 = 36 kg-m2

Angular velocity of precession,

55.6

50p

v

R= 1.11 rad/s

Gyroscopic Couple acting on the aircraft,

C = I. ω. ωP = 36 × 251.4 × 1.11 = 100 46 N-m

= 10.046 kN-m Ans.

When the aeroplane turns towards left, the effect of the gyroscopic couple is to lift the

nose upwards and tail downwards.

Example 4.4: An aeroplane flying at 240 km/hr turns towards the left and completes a

quarter circle of radius 60 m. The mass of the rotary engine and its propeller is 500 kg and

radius of gyration is 0.35m. The engine speed is 2200 rpm clockwise looking from tail end.

Find the gyroscopic couple on the aeroplane and state its effect.

In what way effect changed when the

(i) aeroplane turns towards right

(ii) engine rotates clockwise when viewed from the front (nose end) and the aeroplane turns

(a) left and (b) right

v = 240 km/hr

R = 60 m

m = 450 kg

k = 320 mm

N = 2000 rpm

2 2 2000

60 60

N= 209.4 rad/s

240 1000

3600v = 66.67 m/s



I = m.k2 = 450(0.32)2 = 46.08 kg-m2

66.67

60p

v

R= 1.11 rad/s

C = I. ω. ωP = 46.08 × 209.4 × 1.11 = 10713 N-m

This couple acts in the vertical plane and tends to raise the nose and depress the tail of the

aeroplane.

(i) If the aeroplane takes a right turn, the nose is depressed and the tail is raised.

(ii) (a) when the engine rotates or spins clockwise on viewing from the nose end and

aeroplane turns left, the nose is depressed and the tail is raised.

(b) When the aeroplane takes a right turn, the tail is depressed and the nose is raised.

Example 4.5 : The mass of the turbine rotor of a ship is 20 tonnes and has a radius of

gyration of 0.60 m. Its speed is 2000 r.p.m. The ship pitches 6° above and 6° below the

horizontal position. A complete oscillation takes 30 seconds and the motion is simple

harmonic. Determine the following: 1. Maximum gyroscopic couple, 2. Maximum angular

acceleration of the ship during pitching, and 3. The direction in which the bow will tend to

turn when rising, if the rotation of the rotor is clockwise when looking from the left.

m = 20000 kg

k = 0.6 mm

N = 2000 rpm

φ = 60 = 6 x π/180 = 0.105 rad

tp = 30 s

2 2 2000

60 60

N= 209.4 rad/s

I = m.k2 = 20000(0.6)2 = 7200 kg-m2

Angular velocity of simple harmonic motion

1

2 2

30pt= 0.21 rad/s

Maximum angular velocity of precession,

ωPmax = φ.ω1 = 0.105 × 0.21 = 0.022 rad/s

Maximum gyroscopic couple,

Cmax = I.ω.ωPmax = 7200 × 209.5 × 0.022 = 33 185 N-m

= 33.185 kN-m

maximum angular acceleration during pitching

= φ(ω1)2 = 0.105 (0.21)2 = 0.0046 rad/s2

when the rotation of the rotor is clockwise when looking from the left (i.e. rear end or stern)

and when the bow is rising (i.e. pitching is upward), then the reactive gyroscopic couple acts

in the clockwise direction which tends to turn the bow towards right (i.e. towards star-

board).



Example 4.6 : A ship propelled by a turbine rotor which has a mass of 5 tonnes and a speed

of 2100 r.p.m. The rotor has a radius of gyration of 0.5 m and rotates in a clockwise

direction when viewed from the stern. Find the gyroscopic effects in the following

conditions:

1. The ship sails at a speed of 30 km/h and steers to the left in a curve having 60 m radius.

2. The ship pitches 6 degree above and 6 degree below the horizontal position. The bow is

descending with its maximum velocity. The motion due to pitching is simple harmonic and

the periodic time is 20 seconds.

3. The ship rolls and at a certain instant it has an angular velocity of 0.03 rad/s clockwise

when viewed from stern.

Determine also the maximum angular acceleration during pitching. Explain how the

direction of motion due to gyroscopic effect is determined in each case.

m = 5000 kg k = 0.5 mm

N = 2100 rpm v = 30 km/hr = 8.33 m/s

R = 60 m φ = 60 = 6 x π/180 = 0.105 rad

tp = 20 s

2 2 2100

60 60

N= 220 rad/s

8.33

60p

v

R= 0.14 rad/s

I = m.k2 = 5000(0.5)2

= 1250 kg-m2

Gyroscopic Couple, C = I. ω. ωP = 1250 × 220 × 0.14

= 38500 N-m

When the rotor in a clockwise direction when viewed from the stern and the ship steers to

the left, the effect of reactive gyroscopic couple is to raise the bow and lower the stern.


1

2 2

20pt= 0.3142 rad/s

maximum angular velocity of precession,

ωPmax = φ.ω1 = 0.105 × 0.3142

= 0.033 rad/s

∴ Maximum gyroscopic couple,

Cmax = I.ω.ωPmax = 1250 × 220 × 0.033

= 9075 N-m

Since the ship is pitching with the bow descending, therefore the effect of this maximum

gyroscopic couple is to turn the ship towards port side.



Since the ship rolls at an angular velocity of 0.03 rad / s, therefore angular velocity of

precession

when the ship rolls,

ωP = 0.03 rad /s

∴ Gyroscopic couple,

C = I.ω.ωP = 1250 × 220 × 0.03

= 8250 N-m

In case of rolling of a ship, the axis of precession is always parallel to the axis of spin for all

positions, therefore there is no effect of gyroscopic couple.

Maximum angular acceleration during pitching

αmax = φ (ω1)2 = 0.105 (0.3142)2

= 0.01 rad/s2

Example 4.7 : The turbine rotor of a ship has a mass of 2000 kg and rotates at a speed of

3000 r.p.m. clockwise when looking from a stern. The radius of gyration of the rotor is 0.5

m. Determine the gyroscopic couple and its effects upon the ship when the ship is steering

to the right in a curve of 100 m radius at a speed of 16.1 knots (1 knot = 1855 m/hr).

Calculate also the torque and its effects when the ship is pitching in simple harmonic

motion, the bow falling with its maximum velocity. The period of pitching is 50 seconds and

the total angular displacement between the two extreme positions of pitching is 12°. Find

the maximum acceleration during pitching motion.

m = 2000 kg k = 0.5 mm

N = 3000 rpm v = 16.1 knots = 8.3 m/s

R = 100 m tp = 20 s

2φ = 120 or φ = 6 x π/180 = 0.105 rad

2 2 3000

60 60

N= 314.2 rad/s

8.3

100p

v

R= 0.083 rad/s

I = m.k2 = 2000(0.5)2

= 500 kg-m2

Gyroscopic Couple, C = I. ω. ωP = 500 × 314.2 × 0.083

= 13040 N-m

when the rotor rotates clockwise when looking from a stern and the ship steers to the right,

the effect of the reactive gyroscopic couple is to raise the stern and lower the bow.


12 2

50pt= 0.1257 rad/s

Maximum angular velocity of precession,



ωPmax = φ.ω1 = 0.105 × 0.1257

= 0.0132 rad/s

∴ Torque or maximum gyroscopic couple during pitching,

Cmax = I.ω.ωPmax = 500 × 314.2 × 0.0132

= 2074 N-m

When the pitching is downwards, the effect of the reactive gyroscopic couple is to turn the

ship towards port side.

Maximum acceleration during pitching

αmax = φ (ω1)2 = 0.105 (0.1257)2

= 0.00166 rad/s2

Example 4.8: Each wheel of a four-wheeled rear engine automobile has a moment of inertia

of 2.4 kg.m2 and an effective diameter of 660 mm. The rotating parts of the engine have a

moment of inertia of 1.2 kg.m2. The gear ratio of engine to the back wheel is 3 to 1. The

engine axis is parallel to the rear axle and the crankshaft rotates in the same sense as the

road wheels. The mass of the vehicle is 2200 kg and the center of mass is 550 mm above the

road level. The track width of the vehicle is 1.5 m. Determine the limiting speed of the

vehicle around a curve with 80 m radius so that all the four wheels maintain contact with

the road surface.

Iw = 2.4 kg.m2 r = 0.33 m

Ie = 1.2 kg.m2 G = ωe/ωw = 3

m = 2200 kg h = 0.55 m

w = 1.5 m R = 80 m

(i)

2200 9.81

Reaction due to weight, 4 4

w

mgR

= 5395.5 N (upwards)

(ii) 2

Reaction due to gyroscopic couple, 4w wv

C IrR

22 4 2.4 0.364

0.33 80w

vC v

22 1.2 3 0.136

0.33 80e e w p

vC I G v

2 2 0.364 0.136G w eC C C v v

= 0.5 v2

20.5Reaction on each outer wheel, R

2 2 1.5G

GO

C v

w = 0.167 v2 (upwards)

Reaction on each inner wheel, RGi = 0.167 v2 (downwards)



(iii)

2 22200

Reaction due centrifugal couple, C 0.5580

C

mv vh

R = 15.125 v2


2 2 1.5e

CO

C v


Reaction on each inner wheel, Rci = 5.042 v2 (downwards)

For maximum safe speed, the condition is

Rw = RGi + Rci

5395.5 = (0.167 +5.042) v2

v2 = 1035.8

v = 32.18 m/s or 115.9 km/hr

Example 4.9: A four wheeled trolley car has a total mass of 3000 kg. Each axle with its two

wheels and gears has a total moment of inertia of 32 kg.m2. Each wheel is of 450 mm radius.

The centre distance between two wheels on an axle is 1.4 m. Each axle is driven by a motor

with a speed ratio of 1:3. Each motor along with its gear has a moment of inertia of 16 kg.m2

and rotates in the opposite direction to that of the axle. The centre of mass of the car is 1 m

above the rails. Calculate the limiting speed of the car when it has to travel around a curve

of 250 m radius without the wheels leaving the rails.

Iw = 32/2 = 16 kg.m2 r = 0.45 m

Im = 16 kg.m2 G = 3

m = 3000 kg h = 1 m

w = 1.4 m R = 250 m

(i)

3000 9.81

Reaction due to weight, 4 4

w

mgR

= 7357.5 N (upwards)

(ii) 2

Reaction due to gyroscopic couple, 4w wv

C IrR

22 4 16 0.569

0.45 250w

vC v

22 2 2 16 3 0.853

0.45 250m m w p

vC I G v

(as there are two motors)

2 2 0.569 0.853G w mC C C v v

= - 0.284 v2


2 2 1.4G

GO

C v

w = 0.1014 v2 (downwards)

Reaction on each inner wheel, RGi = 0.1014 v2 (upwards)



(iii)

2 23000

Reaction due centrifugal couple, C 180

C

mv vh

R = 12 v2

212Reaction on each outer wheel, R

2 2 1.4e

CO

C v


Reaction on each inner wheel, Rci = 4.286 v2 (downwards)

Total reaction on outer wheel = 7357.5 - 0.1014 v2 + 4.286 v2

= 7357.5 + 4.1846 v2

Total reaction on inner wheel = 7357.5 - 0.1014 v2 - 4.286 v2

= 7357.5 - 4.1846 v2

Thus, the reaction on the outer wheel is always positive (upwards). There are chances that

the inner wheels leave the rails.

For maximum speed,

7357.5 - 4.1846 v2 = 0 v2 = 1758.2

v = 41.93 m/s or 151 km/hr

Example 4.10: Each wheel of a motorcycle is of 600 mm diameter has a moment of inertia

of 1.2 kg.m2.The total mass of the motorcycle and the rider is 180 kg and the combined

center of mass is 580 mm above the ground level when the motorcycle is upright. The

moment of inertia of the rotating parts of the engine is 0.2 kg.m2. The engine speed is 5

times the speed of the wheels and is in the same sense. Determine the angle of heel

necessary when the motorcycle takes a turn of 35 m radius at a speed of 54 km/hr.

Iw = 1.2 kg.m2 r = 0.3 m

Ie = 0.2 kg.m2 G = ωe/ωw = 5

m = 180 kg h = 0.58 m

v = 54 km/hr R = 35 m

2

Gyroscopic couple, (2 ) cosG w ev

C I GIrR

2(15) (2 1.2 5 0.2) cos 72.86cos

0.3 35GC

2 2180 (15)

Centrifugal couple, C cos 0.58 cos35

C

mvh

R = 671.14 cosθ

Total overturning couple = (72.86 + 671.14) cosθ

= 744 cosθ

Rightening couple = mgh sinθ = 180 x 9.81 x 0.58 sinθ

= 1024 sinθ



1024 sinθ = 744 cosθ

744

tan 0.7271024

θ = 360.

Example 4.11: Find the angle of inclination with respect to the vertical of a two wheeler

negotiating a turn. Given : combined mass of the vehicle with its rider 250 kg ; moment of

inertia of the engine flywheel 0.3 kg-m2 ; moment of inertia of each road wheel 1 kg-m2 ;

speed of engine flywheel 5 times that of road wheels and in the same direction ; height of

centre of gravity of rider with vehicle 0.6 m ; two wheeler speed 90 km/h ; wheel radius 300

mm ; radius of turn 50 m.

Iw = 1 kg.m2 r = 0.3 m

Ie = 0.3 kg.m2 G = ωe/ωw = 5

m = 250 kg h = 0.6 m

v = 90 km/hr R = 50 m

2

Gyroscopic couple, (2 ) cosG w ev

C I GIrR

2(25) (2 1 5 0.3) cos 146cos

0.3 50GC

2 2250 (25)Centrifugal couple, C cos 0.6 cos

50C

mvh

R = 1875 cosθ

Total overturning couple = (146 +1875) cosθ

= 2021 cosθ

Rightening couple = mgh sinθ

= 250 x 9.81 x 0.6 sinθ

= 1471.5 sinθ

2021 sinθ = 1471.5 cosθ

1471.5

tan 1.37342021

θ = 53.940.

Date post:	25-Jan-2021
Category:	Documents
Upload:	others
View:	2 times
Download:	1 times

GYROSCOPE - Darshan Institute of Engineering & Technology · 2020. 7. 23. · 4. Gyroscope Dynamics...

Documents