Gyroscope MCQ Quiz in தமிழ் - Objective Question with Answer for Gyroscope - இலவச PDF ஐப் பதிவிறக்கவும்

Explanation:

• When the engine or propeller rotates in an anticlockwise (clockwise from front) direction when viewed from the rear or tail end and the airplane takes a left turn, then the effect of the reactive gyroscopic couple will be to dip the nose and raise the tail of the airplane.

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 (i.e. longitudinal axis) is always parallel to the axis of spin of the stabilizing rotor for all positions. 

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

Concept:

Gyroscopic Couple = I × ω × ω_p

Where I is the moment of inertia of the flywheel, ωp is the angular velocity of spin and ω is the rotational speed of the flywheel

Calculation:

Given, W = \(\frac{{981}}{\pi }\) kgf ⇒ m =  \(\frac{{981}}{\pi }\)kg, radius of gyration k = 100 cm = 1m, ω = 6rad/s, N = 100 r.p.m,

\(I = m{k^2} = \frac{{981}}{{\pi }} × {\left( 1 \right)^2}\) kg-m²

\(I = \frac{{981}}{\pi }\;kg-\;{m^2}\)

\(ω_p = \frac{{2\pi N}}{{60}} = \frac{{2\pi × 100}}{{60}} = \frac{{20\pi }}{6}\;rad/sec\)

\(T = Iω \;{ω _p} = \frac{{981}}{\pi } × \frac{{20\pi }}{6} × 6= 19620 \ kg \ m^2/s^2 = 19620\;Nm\)

1 kgf = 9.81 N

⇒ T = 2000 kgf-m

Concept:

The Reactive Gyroscopic couple,

\( ⇒ \overrightarrow {{G}} = I\left( {\overrightarrow {{\omega _s}} \times \overrightarrow {{\omega _p}} } \right)\;\;\;\;\; \ldots \left( 1 \right)\)

Calculation:

Given:

m = 4 kg, k = 0.060 m, ωs = 209 rad/s, Np = 50 rpm, L = 6 m

Solution :

Given :- m = 4 kg, N = 800 rpm, K = 0.06 m,  N_P = 50 rpm, l = 80 mm = 0.08 m

⇒ I = mk² = 4 × (0.06)² = 0.0144 kg - m²

• The angular speed of the axle,

\(\omega = \frac{{2\pi × 800}}{{60}} = 83.78 \ rad/s; \ \omega _p = \frac{{2\pi \times 50}}{{60}} = 5.24 \ rad/s\)

• The Reactive Gyroscopic couple,

⇒ G = 0.0144 × 83.78 × 5.24

⇒ G = 6.32 N.m

• This reactive couple (G) will be acting in a clockwise sense as shown below so because of this couple a reaction force will be generated in an upward direction to the axle at bearing B and in the downward direction on the left side bearing A. This reaction force will have the magnitude of G/L.

The FBD for the bearings A and B, disc.

• Also, because the rotor is located at the center, the reaction force on the axle on bearing A side will be in the upward direction.
• So, Reaction at force at the axle on Bearing A side,

\(R_A =( \frac{G}{l} - \frac{{mg}}{2})\downarrow\)

Now, this will be the same as the net force acting on bearing A.

\({R_A} = \frac{{6.32}}{{0.08}} - \frac{{4 \times 9.81}}{2} = 59.4 \ N \downarrow\)

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-m², 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   

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 (ωp) and axis about which precession takes place is known as the axis of precession.

Applied gyroscopic couple:

• The couple which appears due to a 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 = Iωωp
• The angular momentum of this disk during precession = Iω

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

Concept:

Gyroscopic couple = Iωω_p

Calculation:

Given data;

I = 10 kg m²

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:

Concept:

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 moment of inertia of the rotor is given by

I = mk2;

During steering, the angular velocity of precession is given by

ωp = v/r;

The total gyroscopic couple will be

G = I⋅ω⋅ωp;

Calculation:

Given:

m = 2.2 tonnes = 2200 kg, N = 1800 rpm, R = 250 m, k = 0.32 m

\(V = 25\ km/h = \frac{{25000}}{{36000}} = 6.94 \ m/s\)

The moment of inertia of the rotor,

⇒ I = mk² = 2200 × (0.32²) = 225.3 kg.m²

Now, the angular velocity of the precision,

\({w_p} = \frac{v}{r} = \frac{{6.94}}{{250}} = 0.0278 \ rad/s\)

Now, the  gyroscopic Couple is given by,

\(G = Iw{w_p} = 225.3 \times \frac{{2\pi \times 1800}}{{60}} \times 0.0278\)

⇒ G = 1180 N.m

Concepts:

Breakwaters:

Breakwaters means a barrier built out into the sea to protect a coast or harbour from the force of waves. Breakwaters are all about erosion control. These structures will not provide any emergency protection to a beach or a home that is caught in the line of the storm. It might reduce the severity of storm waves but it will not save structure from high waves.

There are two basic types of breakwater: floating and fixed. Installing fixed breakwaters may create long-term ecological hazards. Fixed breakwaters are often an eyesore, it's an aesthetically displeasing sight on the shoreline.

Breakwaters are intended to protect homes and beaches, but as man-made structures they have some aesthetic and environmental disadvantages.

 A well-designed floating breakwater can stop waves that are up to 6.5 feet high, while fixed breakwater can stop waves up to 10 feet high depending on their design and the local environment.

Additional Information

Dock: It is artificially enclosed basin into which vessels are brought for inspection and repair.

Quay: A quay is a long structure, usually built of stone, where boats can be tied up to take on and off their goods.

Explanation:

Given:

I_p = 20 kg – m²

\({\omega _s} = \frac{{2\pi \times 1000}}{{60}} = 104.67\;rad/sec\;\)

\({\omega _p} = \frac{V}{r} = \frac{{200 \times 1000}}{{3600 \times 150}} = 0.37\;rad/sec\)

Now,

Gyroscopic acceleration = ω_s × ω_p

∴ Gyroscopic acceleration = 104.67 × 0.37

∴ Gyroscopic acceleration = 38.7279 rad²/sec²