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Four particles each of mass m are placed at the corners of a square of side of length l. The radius of gyration of the system about an axis perpendicular to the square and passing through its centre is
- a)l/ √ 2
- b)l / 2
- c)l
- d)√2 l
Correct answer is option 'D'. Can you explain this answer?
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Four particles each of mass m are placed at the corners of a square of...
The moment of inertia of a particle of mass m about an axis passing through its centre of mass and perpendicular to its motion is given by:
I = (1/2) m (r^2)
where r is the distance of the particle from the axis of rotation. For a square of side length l, the distance of each particle from the axis passing through the centre of the square is given by:
r = l/2
Therefore, the moment of inertia of each particle is:
I = (1/2) m (l/2)^2 = (1/8) ml^2
The total moment of inertia of the system is the sum of the moments of inertia of the four particles:
I_total = 4I = (4/8) ml^2 = (1/2) ml^2
The radius of gyration of the system about an axis passing through its centre of mass and perpendicular to the square is given by:
k^2 = I_total/M
where M is the total mass of the system:
M = 4m
Substituting the values, we get:
k^2 = (1/2) ml^2 / (4m) = (1/8) l^2
Taking the square root, we get:
k = (1/2) (l/√2)
Therefore, the radius of gyration of the system about an axis perpendicular to the square and passing through its centre is (1/2) (l/√2) or approximately 0.3535 l.
I = (1/2) m (r^2)
where r is the distance of the particle from the axis of rotation. For a square of side length l, the distance of each particle from the axis passing through the centre of the square is given by:
r = l/2
Therefore, the moment of inertia of each particle is:
I = (1/2) m (l/2)^2 = (1/8) ml^2
The total moment of inertia of the system is the sum of the moments of inertia of the four particles:
I_total = 4I = (4/8) ml^2 = (1/2) ml^2
The radius of gyration of the system about an axis passing through its centre of mass and perpendicular to the square is given by:
k^2 = I_total/M
where M is the total mass of the system:
M = 4m
Substituting the values, we get:
k^2 = (1/2) ml^2 / (4m) = (1/8) l^2
Taking the square root, we get:
k = (1/2) (l/√2)
Therefore, the radius of gyration of the system about an axis perpendicular to the square and passing through its centre is (1/2) (l/√2) or approximately 0.3535 l.
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Four particles each of mass m are placed at the corners of a square of side of length l. The radius of gyration of the system about an axis perpendicular to the square and passing through its centre isa)l/√ 2b)l /2c)ld)√2 lCorrect answer is option 'D'. Can you explain this answer?
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Four particles each of mass m are placed at the corners of a square of side of length l. The radius of gyration of the system about an axis perpendicular to the square and passing through its centre isa)l/√ 2b)l /2c)ld)√2 lCorrect answer is option 'D'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Four particles each of mass m are placed at the corners of a square of side of length l. The radius of gyration of the system about an axis perpendicular to the square and passing through its centre isa)l/√ 2b)l /2c)ld)√2 lCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Four particles each of mass m are placed at the corners of a square of side of length l. The radius of gyration of the system about an axis perpendicular to the square and passing through its centre isa)l/√ 2b)l /2c)ld)√2 lCorrect answer is option 'D'. Can you explain this answer?.
Four particles each of mass m are placed at the corners of a square of side of length l. The radius of gyration of the system about an axis perpendicular to the square and passing through its centre isa)l/√ 2b)l /2c)ld)√2 lCorrect answer is option 'D'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Four particles each of mass m are placed at the corners of a square of side of length l. The radius of gyration of the system about an axis perpendicular to the square and passing through its centre isa)l/√ 2b)l /2c)ld)√2 lCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Four particles each of mass m are placed at the corners of a square of side of length l. The radius of gyration of the system about an axis perpendicular to the square and passing through its centre isa)l/√ 2b)l /2c)ld)√2 lCorrect answer is option 'D'. Can you explain this answer?.
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Four particles each of mass m are placed at the corners of a square of side of length l. The radius of gyration of the system about an axis perpendicular to the square and passing through its centre isa)l/√ 2b)l /2c)ld)√2 lCorrect answer is option 'D'. Can you explain this answer?, a detailed solution for Four particles each of mass m are placed at the corners of a square of side of length l. The radius of gyration of the system about an axis perpendicular to the square and passing through its centre isa)l/√ 2b)l /2c)ld)√2 lCorrect answer is option 'D'. Can you explain this answer? has been provided alongside types of Four particles each of mass m are placed at the corners of a square of side of length l. The radius of gyration of the system about an axis perpendicular to the square and passing through its centre isa)l/√ 2b)l /2c)ld)√2 lCorrect answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
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