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An aeroplane moving with a velocity of 300 kmph negotiates a turn along a circular path of radius 600 m. If the angular velocity of the rotating parts of the plane is 120 rad/s and the moment of inertia of the rotating parts is 60 kg.m2, the gyroscopic couple is?
- a)500 N.m
- b)1000 N.m
- c)3000N.m
- d)5000 N.m
Correct answer is option 'B'. Can you explain this answer?
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An aeroplane moving with a velocity of 300 kmph negotiates a turn alo...
A gyroscope is a device for measuring or maintaining orientation, based on the principles of angular momentum. Mechanically, a gyroscope is a spinning wheel or disk in which the axle is free to assume any orientation. Although this orientation does not remain fixed, it changes in response to an external torque much less and in a different direction than it would without the large angular momentum associated with the disk's high rate of spin and moment of inertia. After a short interval of time t, let the disc be spinning about the new axis of spin OX’ (at an angle δθ ) with an angular velocity (ω +δω). Using the right hand screw rule, initial angular velocity of the disc ω is represented by vector ox; and the final angular velocity of the disc (ω +δω ) Gyroscopic Couple 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 the plane of spinning.
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Gyroscopic couple = Iωωc
ωc = 120 rad/sec
ω = u/r = 300 × 5/18 ÷ 600 = 0.1388
I = 60 kgm2
Gyroscopic couple = 60 x 120 x 0.1388 = 1000 N.m
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An aeroplane moving with a velocity of 300 kmph negotiates a turn alo...
Given,
Velocity of the aeroplane, v = 300 kmph = 83.33 m/s
Radius of the circular path, r = 600 m
Angular velocity of the rotating parts, ω = 120 rad/s
Moment of inertia, I = 60 kg.m^2
The gyroscopic couple acting on the plane is given by the formula,
Gyroscopic couple = Iω(dω/dt)
Here, dω/dt is the rate of change of angular velocity. In this case, since the plane is moving along a circular path, the rate of change of angular velocity is given by,
dω/dt = v/r
Substituting the given values, we get,
Gyroscopic couple = Iω(v/r)
= 60 x 120 x (83.33/600)
= 1000 N.m
Therefore, the gyroscopic couple acting on the plane is 1000 N.m, which is option B.
Velocity of the aeroplane, v = 300 kmph = 83.33 m/s
Radius of the circular path, r = 600 m
Angular velocity of the rotating parts, ω = 120 rad/s
Moment of inertia, I = 60 kg.m^2
The gyroscopic couple acting on the plane is given by the formula,
Gyroscopic couple = Iω(dω/dt)
Here, dω/dt is the rate of change of angular velocity. In this case, since the plane is moving along a circular path, the rate of change of angular velocity is given by,
dω/dt = v/r
Substituting the given values, we get,
Gyroscopic couple = Iω(v/r)
= 60 x 120 x (83.33/600)
= 1000 N.m
Therefore, the gyroscopic couple acting on the plane is 1000 N.m, which is option B.
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An aeroplane moving with a velocity of 300 kmph negotiates a turn along a circular path of radius 600 m. If the angular velocity of the rotating parts of the plane is 120 rad/s and the moment of inertia of the rotating parts is 60 kg.m2, the gyroscopic couple is?a)500 N.mb)1000 N.mc)3000N.md)5000 N.mCorrect answer is option 'B'. Can you explain this answer?
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An aeroplane moving with a velocity of 300 kmph negotiates a turn along a circular path of radius 600 m. If the angular velocity of the rotating parts of the plane is 120 rad/s and the moment of inertia of the rotating parts is 60 kg.m2, the gyroscopic couple is?a)500 N.mb)1000 N.mc)3000N.md)5000 N.mCorrect answer is option 'B'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about An aeroplane moving with a velocity of 300 kmph negotiates a turn along a circular path of radius 600 m. If the angular velocity of the rotating parts of the plane is 120 rad/s and the moment of inertia of the rotating parts is 60 kg.m2, the gyroscopic couple is?a)500 N.mb)1000 N.mc)3000N.md)5000 N.mCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An aeroplane moving with a velocity of 300 kmph negotiates a turn along a circular path of radius 600 m. If the angular velocity of the rotating parts of the plane is 120 rad/s and the moment of inertia of the rotating parts is 60 kg.m2, the gyroscopic couple is?a)500 N.mb)1000 N.mc)3000N.md)5000 N.mCorrect answer is option 'B'. Can you explain this answer?.
An aeroplane moving with a velocity of 300 kmph negotiates a turn along a circular path of radius 600 m. If the angular velocity of the rotating parts of the plane is 120 rad/s and the moment of inertia of the rotating parts is 60 kg.m2, the gyroscopic couple is?a)500 N.mb)1000 N.mc)3000N.md)5000 N.mCorrect answer is option 'B'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about An aeroplane moving with a velocity of 300 kmph negotiates a turn along a circular path of radius 600 m. If the angular velocity of the rotating parts of the plane is 120 rad/s and the moment of inertia of the rotating parts is 60 kg.m2, the gyroscopic couple is?a)500 N.mb)1000 N.mc)3000N.md)5000 N.mCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An aeroplane moving with a velocity of 300 kmph negotiates a turn along a circular path of radius 600 m. If the angular velocity of the rotating parts of the plane is 120 rad/s and the moment of inertia of the rotating parts is 60 kg.m2, the gyroscopic couple is?a)500 N.mb)1000 N.mc)3000N.md)5000 N.mCorrect answer is option 'B'. Can you explain this answer?.
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