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A circular portion of diameter r is cut out from a uniform circular disc of mass m and radius r.The moment of inertia of the remaining portion of the disc about an axis passing through the center o of the disc and perpendicular to its plane is?
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The moment of inertia of the remaining portion of the disc after cutting out a circular portion can be calculated using the parallel axis theorem and the properties of moment of inertia.
Initial Moment of Inertia of the Disc
- The moment of inertia (I) of a uniform circular disc about its center is given by:
- I = (1/2) * m * r²
- where m is the mass of the disc and r is its radius.
Mass of the Cut-out Portion
- The mass of the circular portion cut out can be calculated as:
- m_cut = (m / (π * r²)) * (π * (r/2)²)
- This simplifies to m_cut = (m / 4), since the diameter of the cut-out is r.
Moment of Inertia of the Cut-out Portion
- The moment of inertia of the cut-out portion about the center O is:
- I_cut = (1/2) * m_cut * (r/2)²
- This results in I_cut = (1/2) * (m/4) * (r²/4) = m * r² / 32.
Remaining Portion's Moment of Inertia
- To find the moment of inertia of the remaining portion, subtract the moment of inertia of the cut-out from that of the whole disc:
- I_remaining = I - I_cut
- I_remaining = (1/2) * m * r² - (m * r² / 32)
Final Calculation
- Simplifying the above:
- I_remaining = (16/32) * m * r² - (1/32) * m * r²
- I_remaining = (15/32) * m * r².
Conclusion
- The moment of inertia of the remaining portion of the disc about the center O and perpendicular to its plane is:
- I_remaining = (15/32) * m * r².
Initial Moment of Inertia of the Disc
- The moment of inertia (I) of a uniform circular disc about its center is given by:
- I = (1/2) * m * r²
- where m is the mass of the disc and r is its radius.
Mass of the Cut-out Portion
- The mass of the circular portion cut out can be calculated as:
- m_cut = (m / (π * r²)) * (π * (r/2)²)
- This simplifies to m_cut = (m / 4), since the diameter of the cut-out is r.
Moment of Inertia of the Cut-out Portion
- The moment of inertia of the cut-out portion about the center O is:
- I_cut = (1/2) * m_cut * (r/2)²
- This results in I_cut = (1/2) * (m/4) * (r²/4) = m * r² / 32.
Remaining Portion's Moment of Inertia
- To find the moment of inertia of the remaining portion, subtract the moment of inertia of the cut-out from that of the whole disc:
- I_remaining = I - I_cut
- I_remaining = (1/2) * m * r² - (m * r² / 32)
Final Calculation
- Simplifying the above:
- I_remaining = (16/32) * m * r² - (1/32) * m * r²
- I_remaining = (15/32) * m * r².
Conclusion
- The moment of inertia of the remaining portion of the disc about the center O and perpendicular to its plane is:
- I_remaining = (15/32) * m * r².
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A circular portion of diameter r is cut out from a uniform circular disc of mass m and radius r.The moment of inertia of the remaining portion of the disc about an axis passing through the center o of the disc and perpendicular to its plane is?
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A circular portion of diameter r is cut out from a uniform circular disc of mass m and radius r.The moment of inertia of the remaining portion of the disc about an axis passing through the center o of the disc and perpendicular to its plane is? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A circular portion of diameter r is cut out from a uniform circular disc of mass m and radius r.The moment of inertia of the remaining portion of the disc about an axis passing through the center o of the disc and perpendicular to its plane is? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A circular portion of diameter r is cut out from a uniform circular disc of mass m and radius r.The moment of inertia of the remaining portion of the disc about an axis passing through the center o of the disc and perpendicular to its plane is?.
A circular portion of diameter r is cut out from a uniform circular disc of mass m and radius r.The moment of inertia of the remaining portion of the disc about an axis passing through the center o of the disc and perpendicular to its plane is? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A circular portion of diameter r is cut out from a uniform circular disc of mass m and radius r.The moment of inertia of the remaining portion of the disc about an axis passing through the center o of the disc and perpendicular to its plane is? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A circular portion of diameter r is cut out from a uniform circular disc of mass m and radius r.The moment of inertia of the remaining portion of the disc about an axis passing through the center o of the disc and perpendicular to its plane is?.
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