Option 4 : No gyroscopic effect

__Explanation:__

Gyroscopic moment is given by,

M = Iωωp

Here it is a case of rolling.

In the case of rolling of a ship the axis of precession (ie longitudinal axis) is always parallel to the axis of spin of the stabilising rotor for all position.

Hence there is no gyroscopic moment acting on the body of a ship.

Option 1 : Rolling

**Explanation:**

For the effect of the gyroscopic couple to occur, the axis of precession should always be perpendicular to the axis of spin.

If, however, the axis of precession becomes parallel to the axis of spin, there will be **no effect of the gyroscopic couple** acting on the body of the ship.

In the case of rolling 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.

Option 2 : 5 rad/s

__Concept:__

When a spinning disc undergoes precession then the torque applied is gyroscopic couple.

Gyroscopic couple = I × ω × ω_{p}

Where, ω = spinning angular velocity, ω_{p} = angular velocity of precession, I = mass moment of Inertia

__Calculation:__

**Given:**

T = 100 N.m, ω = 20 rad/s, I = 1 kg-m^{2}

T = I × ω × ω_{p}

⇒ 100 = 1 × 20 × ω_{p}

Option 3 : 2000 Nm

__Concept:__

Gyroscopic Couple \( = I \times \omega \times {\omega _p}\)

Where I is the moment of inertia of the ship, ω_{p} is the angular velocity of precision and ω is the rotational speed of the ship

__Calculation:__

**Given:**

I = 200 kgm^{2}, ω = 200 rad/s

\({\omega _p} = \frac{v}{r} = \frac{5}{{100}} = 0.05\)

Gyroscopic Couple = 200 × 200 × 0.05 = **2000 Nm**

When a ship travels in a sea, which of the following effects is more dangerous?

Option 3 : Rolling

**Explanation:**

**Types of motion in Naval ships:**

**Pitching:**

- It is the limited angular motion of the ship
**about the transverse axis.**

Steering:

- Steering is turning on the side.

**Rolling:**

- It is the limited angular motion of the ship
**about the longitudinal axis**.

- Rolling motion usually occurs because of the difference in buoyancy on the two sides of a ship due to a wave. This is periodic in nature and associated with all the moving ship and is the most dangerous motion for ships.

**The stability of the ship depends on metacentric height. Greater is metacentric height greater is the stability.**

Metacentric height:

It is defined as the point about which a body starts oscillating when the body is tilted by a small angle. The meta-centre may also be defined as the point at which the line of action of the force of buoyancy will meet the normal axis of the body when the body is given a small angular displacement.

The distance MG i.e. the distance between the meta-centre of a floating body and the centre of gravity of the body is called metacentric height. It is measured along with the line BG.

Mathematically,

\(GM= \frac{I}{V} - BG\)

As **I _{PITCHING }> I _{ROLLING} **

**GM _{PITCHING } > GM_{ ROLLING }**

TPITCHING < T ROLLING

Therefore** ship in pitching is more stable as compared to rolling it means in rolling it is dangerous**.

__Additional Information__

Gyroscopic Effect:

- If the axis of a spinning body is given an angular motion about an axis perpendicular to the axis of spin then angular action acts on the body about the third perpendicular axis.
- The torque which produces this acceleration is the active gyroscopic torque, a reactive gyroscopic couple acts on the rotating body, the effect produced by the reactive gyroscopic couple is known as the gyroscopic effect.
- The gyroscopic effect arises when there is the precession of the rotating rotor. The gyroscopic couple is generated which changes the bearing reactions. Hence gyroscopic effects to be considered while designing bearing.

Net reaction of ground on wheels due to gyroscopic couple, due to wheels and the dead weight and centrifugal force of a vehicle negotiating a curve is

Option 2 : decreased on inner wheels and increased on outer wheels

__Explanation:__

When a four-wheeler takes a turn, the axis of rotating parts (i.e. wheels and engine) undergoes precession and the gyroscopic couple will be acting on these rotating parts.

Also as the vehicle is moving along a curved path, the centrifugal force will also be acting on the vehicle. This force will also give a couple.

The total reaction at the wheels will be due to the weight of the vehicle, due to centrifugal force and due to the gyroscopic couple.

1) The weight of the vehicle is equally divided among four wheels, then weight on each wheel is W/4 in a downward direction and the reaction of the ground at each wheel due to the weight W = W/4 in an upward direction.

2) Couple due to centrifugal force.

The vehicle is moving along a curved path, hence a centrifugal force will also be acting on the vehicle in the outward direction. This force acts through the center of mass of the vehicle (i.e. through CG).

Here the magnitude of centrifugal force is P.

Therefore, couple due to centrifugal force, Cc = P × h

where h = distance from ground level to CG

The above is the overturning couple acting on the vehicle. The resisting couple will be acting on the wheels which is equal to the couple due to equal and opposite forces acting on outer or inner wheels.

Therefore, the force on each inner wheel is P/4 in an upward direction.

3) Gyroscopic effects on four-wheel vehicles taking a left turn

W = weight of the vehicle, \(\frac{W}{4}\) = reaction from the road on each wheel, Iw = Moment of inertia of each wheel, IE = Moment of inertia of engine flywheel, ωw = Angular velocity of each wheel, ωE = Angular velocity of each engine, ωP = Angular velocity of precision

The gyroscopic couple on a vehicle is,

C = Cw + CE

C = 4 × IW × ωW × ωp ± IE × ωE × ωP

**Due to gyroscopic couples, there will be vertical reactions on wheels, on outer wheels, it would be positive, and on inner wheels, it would be negative.**

Option 4 : raising of the nose and lowering of the tail

__Explanation:__

Point the right-hand thumb in the direction of the arrow of the reactive gyroscopic couple, then the curling finger will give the effect of the reactive gyroscopic couple.

Following points are to be noted when the aeroplane takes left turn

- As the propeller is rotating in the ox axis, the axis ox is called the axis of spin. The plane is taking the left turn that means axis ox is rotating towards the axis oz.
- If we cover axis of spin (ox in this case) with the right hand as the curved fingers indicate the direction of rotating propeller and rotate that axis in the direction of the aeroplane is taking a turn, then the curved fingers show the direction of the active gyroscopic couple and the direction of the reactive gyroscopic couple is exactly opposite but equal in magnitude as shown in the figure.
**Thus it results in rising of the nose and falling of the tail of the aeroplane.**- The magnitude of gyroscopic is given as,

G = Iωsωp

I = Moment of inertia of rotor, kg/m^{2}, ω_{s} = angular velocity of rotor, rad/s ω_{p} = angular velocity of precession, rad/s

**Alternate solution:**

**Explanation:**

The reactive gyroscopic couple (opposite to the active gyroscopic couple),

\( ⇒ {{\vec G}} = {{I}}\left( {\overrightarrow {{{{\omega }}_{{s}}}} \times \overrightarrow {{{{\omega }}_{{p}}}} \;} \right)\;\;\;\;\; \ldots \left( 1 \right)\)

∵ The Aircraft turns to the left so the precession angular velocity of the turn,

\(\overrightarrow {{{\rm{\omega }}_{\rm{p}}}} = \omega_{p}{\rm{\;}}\left( { + {\rm{\hat j}}} \right){\rm{\;rad}}/{\rm{s}}\)

Where,

\({\rm{\vec G}} = Reactive\ {{Gyroscopic\;couple\;vector}}\)

I = M × k2 =The moment of inertia of the gyroscope's spinning **shaft**

\(\overrightarrow {{{\rm{\omega }}_{\rm{s}}}} = {\rm{Spin\;anugular\;velocity\;vector\;of\;the\;shaft}}\)

\(\overrightarrow {{{{\omega }}_{\rm{p}}}} = {{The\;precession\;}}{{angular}}\;{{velocity}}\)

∵ The engine shaft of an aircraft rotates in counter-clockwise direction, when viewed from nose..

Angular velocity of the engine shaft of the Aircraft (using right-hand thumb rule for angular velocity direction about x-axis)

\(\overrightarrow {{{\rm{\omega }}_{\rm{s}}}} = \omega_{s}{\rm{\;}}\left( { + {\rm{\hat i}}} \right){\rm{\;rad}}/{\rm{s}}\)

Using equation (1),

\( ⇒ {\rm{\vec G}} =I\left( {\omega_{s}{\rm{\;}}\left( { + {\rm{\hat i}}} \right) \times \omega_{p}\left( { {\rm{+\hat {j}}}} \right){\rm{\;}}} \right)\)

\( ⇒ {{\vec G}} = I\omega_{s}\omega_{p}\ (+\ \hat{k})\ N.m\)

⇒ Using right-hand thumb rule for Reactive Gyroscopic torque will be + z direction as shown in the figure because of which, the nose of the aircraft will rise and tail of the falling of the aircraft will get dip.

Option 3 : Three planes perpendicular to one another

**Concept:**

**Axis of spin:**

The rotor of engine of the vehicle is rotated about a particular axis, that **axis of rotation** referred to as the **axis of spin.**

In general, the axis of rotation of any engine considered as the axis of spin as a reference axis for the further calculation.

**Axis of precession:**

The **motion of direction of initial angular momentum** is known as** precession** and axis about which precession takes place is known as the **axis of precession.**

**Applied gyroscopic couple:**

The couple which appears due to change in the direction of angular momentum is known as an applied gyroscopic couple or gyroscopic torque.

And **to achieve the effective gyroscopic effect**, above **three planes must be perpendicular to each other.**

Option 1 : tend to raise the tail and depress the nose

__Explanation:__

Let the propeller rotate clockwise as seen from the rear (it is the tail end of an aeroplane).

The **gyroscopic couple C** acting on plane = **I**ω**ω _{p}**

Where I is the moment of inertia of engine and propeller, ω_{p} is the angular speed of precision and ω is the angular speed of the engine.

**For the direction of gyroscopic couple**

**oa** is the angular momentum vector before turning, **ob** is the angular momentum vector after turning right, **ab** is the active gyroscopic couple and **a’b’** is the reactive gyroscopic couple which is equal but opposite to the active gyroscopic couple.

The vector **ab** is perpendicular to the plane of the applied couple. Its sense is anticlockwise when seen from the left side of the plane. Using the **right-hand screw rule**, the reactive couple will be **clockwise**, which will **lower the nose and raise the tail of the aeroplane.**

Option 2 : 100 m kgf

__Concept:__

The gyroscopic couple is given as,

C = Iωω_{p}

where I = moment of inertia, ω = Angular velocity or spin velocity, ωp = Angular velocity of precession.

**Calculation:**

**Given:**

I = 30 kgm^{2}, N = 800 rpm, R = 170 m, V = 240 km/hr

ω = \(\frac{2\pi N}{60}=\frac{2\pi × 800}{60}=83.77~rad/sec\) and

\(\omega_p=\frac{V}{R}= \frac{66.66}{170}=0.392~rad/sec\)

Therefore, C = 30 × 83.77 × 0.392 = 985.1352 Nm

⇒ \(C=\frac{985.1352}{9.81}\approx100~m~kgf\)

The disc is spinning with angular velocity ω rad/sec about the axis of spin. The couple applied to the disc causing precision will be :

where ω_{p} = angular velocity of precision of axis of spin and I = mass momentum inertia of disc.

Option 4 : Iωω_{p}

**Explanation:**

Axis of spin:

- The disc is rotated about a particular axis, that axis of rotation (ω) referred to as the axis of spin.
- In general, the axis of rotation of the disc is considered as the axis of spin as a reference axis for the further calculation.

Axis of precession:

- The motion of direction of initial angular momentum is known as precession (ω
_{p}) and axis about which precession takes place is known as the axis of precession.

Applied gyroscopic couple:

- The couple which appears due to change in the direction of angular momentum is known as an active gyroscopic couple or precession couple.
- The active gyroscopic couple or precession couple is given by,
**T = Iωω**_{p}

where ω= angular velocity of the disc, ωp = angular velocity of precision of axis of spin and I = mass momentum inertia of disc.

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 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 of to the nearest integer).

__Concept:__

Gyroscopic moment is given by,

M = Iωω_{p}

Here it is a** case of rolling.**

In the case of rolling of a** ship the axis of precession (ie longitudinal axis) is always parallel to the axis of spin for all position**.

Hence there is **no gyroscopic moment acting on the body of a ship**.

Option 2 : the torque required to cause the axis of spin to precess in the horizontal plane.

The active gyroscopic torque in the gyroscope about a horizontal axis represents the torque required to cause the axis of spin to precess in the horizontal plane.

While running through a bend a bicycle remains stable due to _____

Option 3 : centrifugal action

__Explanation:__

- Centripetal force is that force that is required to move a body in a circular path with uniform speed. The force acts on the body along the radius and towards the centre.
- Centrifugal force is the apparent force that is felt by an object moving in a curved path. This force acts outwardly away from the center of rotation.

**Turning a Bicycle:**

- A bicycle held straight up will tend to go straight. It is
**tempting to say**that it**stabilized by the gyroscopic action**of the bicycle wheels, but the**gyroscopic action is quite small.** - If the rider leans left, a torque will be produced which causes a counterclockwise precession of the bicycle wheel, tending to turn the bicycle to the left.

- This is a good visual example of the
**directions of the angular momenta and torques,**but the**gyroscopic torques of bicycle wheels are apparently quite small.**The gyroscopically motivated descriptions like**"leaning left turn it left"**are**more appropriate to motorcycles.**

**If you lean left, you turn left**

- A rider leaning left will produce a
**torque**that will cause the bicycle wheel to**precess counterclockwise**as seen from above,**turning the bicycle left.**The angular momentum of the bicycle wheels is to the left. The torque produced by leaning is to the rear of the bicycle, as may be seen from the**right-hand rule.**This gives a rearward change in the**angular momentum vector,**turning the bicycle on the left.

- In terms of the stability of the bicycle when riding, the association with leaning and turning does hold true.
- The construction of a bicycle is such that a left lean does cause the front wheel to turn left, contributing a kind of self-stability to the bicycle.
- If you feel unbalanced and leaning left, then turning left does help you correct the imbalance because the
**centrifugal force**associated with the turn does tend to push the top of the bicycle back toward the vertical. - Part of the process of learning to ride a bicycle would then seem to be the learning of how to turn the front wheel to produce the needed centrifugal balancing force to bring you back to an upright and balanced orientation.
- More drastic turns are needed at low speeds to get the necessary centrifugal force which depends upon the inverse of the radius of curvature.
- Much more gentle turns are sufficient at higher speeds since the centrifugal force depends upon the square of the velocity.

The figure given below shown a crusher having several cylindrical rollers of weight W. the crushing force due to each roller will be:

Option 3 : More than W

**Concept:**

Crushing Force = Weight of the roller + Weight of the gyroscopic action

\(F = W + \frac{{I{\omega ^2}}}{r}\)

__Concept:__

**Gyroscopic couple:**

If the axis of a spinning or rotating body is gives 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 couple required to produce this acceleration is known as gyroscopic couple or torque.

τ = Iωω_{p} = I α

I = polar moment of inertia, ω = angular velocity of spin axis, ω_{p} = angular velocity of precision axis

∝ = Angular acceleration

**Calculation:**

I = 10 kg-m^{2}, ω = 100 rad/s

\({\omega _p} = \frac{V}{r} = \frac{{20}}{{100}} = 0.2\;rad/sec\)

τ = Iωωp

τ = 10 × 100 × 0.2 = 200 N-m

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 ω, when viewed from the right-hand side bearing, Q. The axle precesses about a vertical axis at ω_{p} = ω/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 ω^{2}r = 300 m/s^{2} and g = 10 m/s^{2}, the ratio of the larger to the smaller bearing reaction force (considering appropriate signs) is _______

__Concept:__

Gyroscopic couple is given by Iωω_{p}

Given a disc of mass m and radius r mounted on a shaft of length 1.5r as shown

\({\omega _p} = \frac{\omega }{{10}}\)

First let us understand the direction of Gyroscopic torque and its resulting reaction.

The ω and ω_{p} directions are as shown:

Apply right hand thumb rule pointing figure towards ω and closing towards w_{p}, we get torque in direction of thumb, which is as shown.

Now the torque applied on shaft due to bearings to restrict its motion must be opposite to that of obtained.

⇒ To produce this torque R_{Q} must be downward and R_{P} upward.

Also. Torque (Gyroscopic) = \(I\;w{w_p} = \frac{{m{r^2}}}{2}\omega \times \frac{\omega }{{10}} = \frac{{m{r^2}{\omega ^2}}}{{20}}\)

∵ No other forces are acting on shaft.

⇒ R_{P} + R_{Q} = 0 ⇒ R_{P} = - R_{Q}.

Also couple due to R_{P}, R_{Q} = (R_{P}) (1.5r)

\(\Rightarrow \left( {{R_P}} \right)\left( {1.5r} \right) = \frac{{m{r^2}{\omega ^2}}}{{20}} \Rightarrow \left( {{R_P}} \right)\left( {1.5} \right) = \frac{{m{\omega ^2}r}}{{20}} = \frac{{m\left( {300} \right)}}{{20}}\)

\({R_P} = \frac{{m\left( {300} \right)}}{3} = 10\;m.\; \Rightarrow {R_Q} = - 10m.\;\)

Also, the mass of disc will be supported by both the bearing equally (as load is in mid, leading to symmetric loading)

\({R_{{P_2}}} = \frac{{mg}}{2} = 5m\)

& \({R_{{Q_2}}} = 5m\)

⇒ Net reaction at Q and P are:

⇒ R_{Q,net} = - 10 m + 5 m = - 5m

R_{P,net} = 10m + 5m = 15 m

\(\Rightarrow \frac{{Larger\;Reaction}}{{Smaller\;Reaction}} = \frac{{15m}}{{ - 5m}} = - 3\)

The rotor of a turbojet engine of an aircraft has a mass 180 kg and polar moment of inertia 10 kg·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

Option 2 : 1629.6 N·m and the nose goes down

__Concept:__

Gyroscopic couple = Iωω_{p}

__Calculation:__

Given data;

I = 10 kg m^{2}

ω = 1100 rad/s

ν = 800 km/hr = 222.22 m/s

r = 1.5 km = 1500 m

\(\nu =r{{\omega }_{p}}\Rightarrow {{\omega }_{p}}=\frac{222.22}{1500}=0.148148~rad/s\)

Gyroscopic couple (CG) = (10) (1100) (0.148148) = 1629.63 Nm

Reactive gyroscopic couple will dip the nose.

__Key points:__

In an automobile, if the vehicle makes a left turn, the gyroscopic torque

Option 2 : increase the forces on the outer wheels

__Explanation:__

If the vehicle makes a turn towards the left **the gyroscopic torque increases the force on the outer wheel**.

**Effect of Gyroscopic couple in an automobile while turning to left -**

Where,

m = Mass of the vehicle in kg, W = Weight of the vehicle in Newton = mg, r_{W} = Radius of the wheels in meters, R = Radius of curvature in meters (R > r_{W})

h = Distance of center of gravity, vertically above the road surface, x = Width of track in meters

I_{W} = Mass moment of inertia of one of the wheels in kg-m^{2}

I_{E }= Mass moment of inertial of the rotating parts of the engine in kg-m^{2}

ω_{w }= Angular velocity of the wheels or velocity of spin in rad/s

ω_{e }= Angular velocity of the rotating parts of the engine in rad/s

G = Gear ratio = ω_{e} / ω_{w} , v = Linear velocity of the vehicle in m/s = ω_{w }.r_{w}

P = Magnitude of reaction at the two outer or inner wheels in Newton

Road reaction over each wheel = \(\frac{W}{4} = \frac{{mg}}{4}\;N\)

Since the vehicle takes a turn towards the left due to the precession and other rotating parts a gyroscopic couple will act -

**velocity of precession ω _{p }**

Gyroscopic couple due to 4 wheels, **C _{W} = 4**

Gyroscopic couple for rotating part of the engine,

Net Gyroscopic couple, C =

****Due to the gyroscopic couple, a vertical reaction on the road surface will be produced.

On equating the couple-

\(P\; \times \;x\; = \;C\)

\(P\; = \;C/x\)

**Hence, Vertical reaction at each of the outer or inner wheels = P/2 = C/2x. **

**The reaction will be vertically upwards on the outer wheels** **which increases the force on outer wheels** and vertically downwards on the inner wheels which decreases the force on the inner wheel.

Option 4 : \(\frac{W}{4} - \frac{P}{4} - \frac{Q}{4}\)

__Explanation:__

When a four-wheeler takes a turn, the axis of rotating parts (i.e. wheels and engine) undergoes precession and the gyroscopic couple will be acting on these rotating parts.

Also as the vehicle is moving along a curved path, the centrifugal force will also be acting on the vehicle. This force will also give a couple.

**The total reaction at the wheels will be due to the weight of the vehicle, due to centrifugal force and due to the gyroscopic couple. **

**1) The weight of the vehicle** is equally divided among four wheels, then weight on each wheel is W/4 in a downward direction and the reaction of the ground at **each wheel** due to the weight W = W/4 in an upward direction.

**2) Couple due to centrifugal force.**

The vehicle is moving along a curved path, hence a centrifugal force will also be acting on the vehicle in the outward direction. This force acts through the center of mass of the vehicle (i.e. through CG).

Here the magnitude of centrifugal force is P.

Therefore, couple due to centrifugal force, C_{c} = P × h

where h = distance from ground level to CG

The above is the overturning couple acting on the vehicle. The resisting couple will be acting on the wheels which is equal to the couple due to equal and opposite forces acting on outer or inner wheels.

Therefore, the force on **each inner wheel** is P/4 in an upward direction.

**3)** **Gyroscopic effects on four-wheel vehicles taking a left turn**

W = weight of the vehicle, \(\frac{W}{4}\) = reaction from the road on each wheel, I_{w} = Moment of inertia of each wheel, I_{E} = Moment of inertia of engine flywheel, ω_{w} = Angular velocity of each wheel, ω_{E} = Angular velocity of each engine, ω_{P} = Angular velocity of precision

The gyroscopic couple on a vehicle is,

C = C_{w} + C_{E}

C = 4 × I_{W } × ω_{W } × ω_{p }± I_{E }× ω_{E } × ω_{P}

Due to gyroscopic couples, there will be vertical reactions on wheels, on outer wheels, it would be positive, and on inner wheels, it would be negative.

Q = vertical reaction due to gyroscopic couple

Vertical reaction due to gyroscopic couple on **each inner wheel ** = \( - \frac{Q}{4}\)

Total vertical reaction on inner wheels = \( \frac{W}{4} - \frac{P}{4} - \frac{Q}{4}\)