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After cutting off a circular portion of radius R/2 from the centre of a uniform circular disc of radius R, the moment of inertia about an axis passing through its centre and perpendicular to its plane become I. The moment of inertia of the original disc about the same axis is xl. The value of x is
Correct answer is '1.06'. Can you explain this answer?
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After cutting off a circular portion of radius R/2 from the centre of ...
= 1.06
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After cutting off a circular portion of radius R/2 from the centre of ...
Given:
- Radius of the original circular disc = R
- Radius of the cut-off circular portion = R/2
- Moment of inertia of the cut-off portion about an axis passing through its center and perpendicular to its plane = I
- Moment of inertia of the original disc about the same axis = xl
To find:
The value of x
Formula:
The moment of inertia of a circular disc about an axis passing through its center and perpendicular to its plane is given by the formula:
I = (1/4) * m * R^2
Where:
- I is the moment of inertia
- m is the mass of the disc
- R is the radius of the disc
Analysis:
When a circular portion is cut off from the disc, the mass of the cut-off portion becomes a fraction of the total mass of the disc. Let's assume the mass of the cut-off portion is m_c and the mass of the remaining portion is m_r.
Since the cut-off portion is taken from the center, the mass distribution is symmetrical and the center of mass of the remaining portion does not change. Therefore, the moment of inertia of the remaining portion remains the same.
The moment of inertia of the cut-off portion about the given axis can be calculated using the formula mentioned above. Since the cut-off portion is a circular disc, its moment of inertia about its own axis passing through its center and perpendicular to its plane is given by:
I_c = (1/4) * m_c * (R/2)^2 = (1/16) * m_c * R^2
The total moment of inertia of the original disc is the sum of the moment of inertia of the cut-off portion and the moment of inertia of the remaining portion:
xl = I_c + I_r
Since the moment of inertia of the remaining portion is equal to xl, we can write:
xl = (1/16) * m_c * R^2 + I_r
xl = (1/16) * m_c * R^2 + xl
xl - xl = (1/16) * m_c * R^2
0 = (1/16) * m_c * R^2
m_c = 0
This implies that the mass of the cut-off portion is zero, which means the cut-off portion does not exist. Therefore, the original disc remains intact and the moment of inertia of the original disc about the given axis is xl, which implies x = 1.
Conclusion:
The value of x is 1.
- Radius of the original circular disc = R
- Radius of the cut-off circular portion = R/2
- Moment of inertia of the cut-off portion about an axis passing through its center and perpendicular to its plane = I
- Moment of inertia of the original disc about the same axis = xl
To find:
The value of x
Formula:
The moment of inertia of a circular disc about an axis passing through its center and perpendicular to its plane is given by the formula:
I = (1/4) * m * R^2
Where:
- I is the moment of inertia
- m is the mass of the disc
- R is the radius of the disc
Analysis:
When a circular portion is cut off from the disc, the mass of the cut-off portion becomes a fraction of the total mass of the disc. Let's assume the mass of the cut-off portion is m_c and the mass of the remaining portion is m_r.
Since the cut-off portion is taken from the center, the mass distribution is symmetrical and the center of mass of the remaining portion does not change. Therefore, the moment of inertia of the remaining portion remains the same.
The moment of inertia of the cut-off portion about the given axis can be calculated using the formula mentioned above. Since the cut-off portion is a circular disc, its moment of inertia about its own axis passing through its center and perpendicular to its plane is given by:
I_c = (1/4) * m_c * (R/2)^2 = (1/16) * m_c * R^2
The total moment of inertia of the original disc is the sum of the moment of inertia of the cut-off portion and the moment of inertia of the remaining portion:
xl = I_c + I_r
Since the moment of inertia of the remaining portion is equal to xl, we can write:
xl = (1/16) * m_c * R^2 + I_r
xl = (1/16) * m_c * R^2 + xl
xl - xl = (1/16) * m_c * R^2
0 = (1/16) * m_c * R^2
m_c = 0
This implies that the mass of the cut-off portion is zero, which means the cut-off portion does not exist. Therefore, the original disc remains intact and the moment of inertia of the original disc about the given axis is xl, which implies x = 1.
Conclusion:
The value of x is 1.
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After cutting off a circular portion of radius R/2 from the centre of a uniformcircular disc of radius R, the moment of inertia about an axis passing through its centre and perpendicular to its plane become I. The moment of inertia of the original disc about the same axis is xl. The value of x is Correct answer is '1.06'. Can you explain this answer?
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After cutting off a circular portion of radius R/2 from the centre of a uniformcircular disc of radius R, the moment of inertia about an axis passing through its centre and perpendicular to its plane become I. The moment of inertia of the original disc about the same axis is xl. The value of x is Correct answer is '1.06'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about After cutting off a circular portion of radius R/2 from the centre of a uniformcircular disc of radius R, the moment of inertia about an axis passing through its centre and perpendicular to its plane become I. The moment of inertia of the original disc about the same axis is xl. The value of x is Correct answer is '1.06'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for After cutting off a circular portion of radius R/2 from the centre of a uniformcircular disc of radius R, the moment of inertia about an axis passing through its centre and perpendicular to its plane become I. The moment of inertia of the original disc about the same axis is xl. The value of x is Correct answer is '1.06'. Can you explain this answer?.
After cutting off a circular portion of radius R/2 from the centre of a uniformcircular disc of radius R, the moment of inertia about an axis passing through its centre and perpendicular to its plane become I. The moment of inertia of the original disc about the same axis is xl. The value of x is Correct answer is '1.06'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about After cutting off a circular portion of radius R/2 from the centre of a uniformcircular disc of radius R, the moment of inertia about an axis passing through its centre and perpendicular to its plane become I. The moment of inertia of the original disc about the same axis is xl. The value of x is Correct answer is '1.06'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for After cutting off a circular portion of radius R/2 from the centre of a uniformcircular disc of radius R, the moment of inertia about an axis passing through its centre and perpendicular to its plane become I. The moment of inertia of the original disc about the same axis is xl. The value of x is Correct answer is '1.06'. Can you explain this answer?.
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After cutting off a circular portion of radius R/2 from the centre of a uniformcircular disc of radius R, the moment of inertia about an axis passing through its centre and perpendicular to its plane become I. The moment of inertia of the original disc about the same axis is xl. The value of x is Correct answer is '1.06'. Can you explain this answer?, a detailed solution for After cutting off a circular portion of radius R/2 from the centre of a uniformcircular disc of radius R, the moment of inertia about an axis passing through its centre and perpendicular to its plane become I. The moment of inertia of the original disc about the same axis is xl. The value of x is Correct answer is '1.06'. Can you explain this answer? has been provided alongside types of After cutting off a circular portion of radius R/2 from the centre of a uniformcircular disc of radius R, the moment of inertia about an axis passing through its centre and perpendicular to its plane become I. The moment of inertia of the original disc about the same axis is xl. The value of x is Correct answer is '1.06'. Can you explain this answer? theory, EduRev gives you an
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