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A rigid body can be hinged about any point on the x-axis. When it is hinged such that the hinge is at x, the moment of inertia is given by
I = 2x2 - 12x + 27 The x-coordinate of centre of mass is
- a)x = 2
- b)x = 0
- c)x = 1
- d)x = 3
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
A rigid body can be hinged about any point on the x-axis. When it is h...
Moment of inertia is minimum at centre of mass of the body.
I=minimum for centre of mass
Therefore x co-ordinate of centre of mass is 3 unit.
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A rigid body can be hinged about any point on the x-axis. When it is h...
Given:
Moment of Inertia, I = 2x^2 - 12x + 27
We need to find the x-coordinate of the Centre of Mass.
Formula:
Centre of Mass, Xcm = (Σmixi) / Σmi
Approach:
We need to find the mass distribution function for the given moment of inertia function.
Derivation for mass distribution function:
I = ∫r^2 dm
where 'r' is the distance from the axis of rotation and 'dm' is the mass element
let's consider a small element of mass 'dm' at a distance 'x' from the origin
I = ∫ r^2 dm = ∫ x^2 dm
Consider a small element of length 'dx' containing mass 'dm'
dm = λdx
where 'λ' is the linear mass density of the element
λ = m/L
where 'm' is the mass of the object and 'L' is the total length of the object
dm = (m/L)dx
r = x
I = ∫ x^2 dm = ∫ x^2 (m/L)dx = (m/L) ∫ x^2 dx
I = (m/L) [x^3/3]
m/L = 3I / x^3
Substitute this value in the dm equation
dm = (3I / x^3) dx
Now, we can find the total mass of the object by integrating the mass distribution function
M = ∫ dm = ∫ (3I / x^3) dx
M = 3I ∫ x^-3 dx
M = -3I [x^-2/2]
M = (3/2) I (1/x^2)
We can find the x-coordinate of the Centre of Mass using the formula mentioned above.
Calculation:
Σmi = (3/2) I (1/x^2) * x
Σmi = (3/2) I (1/x)
Σmi = (3/2) (2x^2 - 12x + 27) / x
Σmi = (3/2) (2x - 12 + 27/x)
Σmi = (3/2) [2(x-3) + 33/x]
Σmi = 3(x-3) + (99/x)
Σmi = (3x - 6 + 99) / x
Σmi = (3x + 93) / x
Xcm = (Σmixi) / Σmi
Xcm = [(3/2) I (1/x^2) * x^2] / [(3x + 93) / x]
Xcm = (3/2) I / (3x + 93)
Substitute the given value of I in the above equation
Xcm = (3/2) (2x^2 - 12x + 27) / (3x + 93)
Xcm = (3/2) [(2x-3) / (x+31)]
Xcm = 3(x-3) / (2x+62)
Hence, the x-coordinate of the Centre of Mass is x = 3.
Moment of Inertia, I = 2x^2 - 12x + 27
We need to find the x-coordinate of the Centre of Mass.
Formula:
Centre of Mass, Xcm = (Σmixi) / Σmi
Approach:
We need to find the mass distribution function for the given moment of inertia function.
Derivation for mass distribution function:
I = ∫r^2 dm
where 'r' is the distance from the axis of rotation and 'dm' is the mass element
let's consider a small element of mass 'dm' at a distance 'x' from the origin
I = ∫ r^2 dm = ∫ x^2 dm
Consider a small element of length 'dx' containing mass 'dm'
dm = λdx
where 'λ' is the linear mass density of the element
λ = m/L
where 'm' is the mass of the object and 'L' is the total length of the object
dm = (m/L)dx
r = x
I = ∫ x^2 dm = ∫ x^2 (m/L)dx = (m/L) ∫ x^2 dx
I = (m/L) [x^3/3]
m/L = 3I / x^3
Substitute this value in the dm equation
dm = (3I / x^3) dx
Now, we can find the total mass of the object by integrating the mass distribution function
M = ∫ dm = ∫ (3I / x^3) dx
M = 3I ∫ x^-3 dx
M = -3I [x^-2/2]
M = (3/2) I (1/x^2)
We can find the x-coordinate of the Centre of Mass using the formula mentioned above.
Calculation:
Σmi = (3/2) I (1/x^2) * x
Σmi = (3/2) I (1/x)
Σmi = (3/2) (2x^2 - 12x + 27) / x
Σmi = (3/2) (2x - 12 + 27/x)
Σmi = (3/2) [2(x-3) + 33/x]
Σmi = 3(x-3) + (99/x)
Σmi = (3x - 6 + 99) / x
Σmi = (3x + 93) / x
Xcm = (Σmixi) / Σmi
Xcm = [(3/2) I (1/x^2) * x^2] / [(3x + 93) / x]
Xcm = (3/2) I / (3x + 93)
Substitute the given value of I in the above equation
Xcm = (3/2) (2x^2 - 12x + 27) / (3x + 93)
Xcm = (3/2) [(2x-3) / (x+31)]
Xcm = 3(x-3) / (2x+62)
Hence, the x-coordinate of the Centre of Mass is x = 3.
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A rigid body can be hinged about any point on the x-axis. When it is hinged such that the hinge is at x, the moment of inertia is given byI = 2x2-12x + 27 The x-coordinate of centre of mass isa)x = 2b)x = 0c)x = 1d)x = 3Correct answer is option 'D'. Can you explain this answer? for Class 11 2025 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A rigid body can be hinged about any point on the x-axis. When it is hinged such that the hinge is at x, the moment of inertia is given byI = 2x2-12x + 27 The x-coordinate of centre of mass isa)x = 2b)x = 0c)x = 1d)x = 3Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A rigid body can be hinged about any point on the x-axis. When it is hinged such that the hinge is at x, the moment of inertia is given byI = 2x2-12x + 27 The x-coordinate of centre of mass isa)x = 2b)x = 0c)x = 1d)x = 3Correct answer is option 'D'. Can you explain this answer?.
A rigid body can be hinged about any point on the x-axis. When it is hinged such that the hinge is at x, the moment of inertia is given byI = 2x2-12x + 27 The x-coordinate of centre of mass isa)x = 2b)x = 0c)x = 1d)x = 3Correct answer is option 'D'. Can you explain this answer? for Class 11 2025 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A rigid body can be hinged about any point on the x-axis. When it is hinged such that the hinge is at x, the moment of inertia is given byI = 2x2-12x + 27 The x-coordinate of centre of mass isa)x = 2b)x = 0c)x = 1d)x = 3Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A rigid body can be hinged about any point on the x-axis. When it is hinged such that the hinge is at x, the moment of inertia is given byI = 2x2-12x + 27 The x-coordinate of centre of mass isa)x = 2b)x = 0c)x = 1d)x = 3Correct answer is option 'D'. Can you explain this answer?.
Solutions for A rigid body can be hinged about any point on the x-axis. When it is hinged such that the hinge is at x, the moment of inertia is given byI = 2x2-12x + 27 The x-coordinate of centre of mass isa)x = 2b)x = 0c)x = 1d)x = 3Correct answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 11.
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A rigid body can be hinged about any point on the x-axis. When it is hinged such that the hinge is at x, the moment of inertia is given byI = 2x2-12x + 27 The x-coordinate of centre of mass isa)x = 2b)x = 0c)x = 1d)x = 3Correct answer is option 'D'. Can you explain this answer?, a detailed solution for A rigid body can be hinged about any point on the x-axis. When it is hinged such that the hinge is at x, the moment of inertia is given byI = 2x2-12x + 27 The x-coordinate of centre of mass isa)x = 2b)x = 0c)x = 1d)x = 3Correct answer is option 'D'. Can you explain this answer? has been provided alongside types of A rigid body can be hinged about any point on the x-axis. When it is hinged such that the hinge is at x, the moment of inertia is given byI = 2x2-12x + 27 The x-coordinate of centre of mass isa)x = 2b)x = 0c)x = 1d)x = 3Correct answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
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