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From a circular disc of radius R and mass 9M a small disc of radius R/3 is removed as shown in figure the moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through O is?
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From a circular disc of radius R and mass 9M a small disc of radius R/...
The moment of inertia of remaining disk = Moment of inertia of whole disc - Moment of inertia of removed small disc
moment of inertia of any disk about its own central axis = 1/2 mr^2
Moment of inertia of total disk of mass 9M and radius R = (1/2) 9M R^2
Iwhole = (1/2) 9M R^2
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Let us calculate the mass of the removed disk and the location of its centre of mass
mass density = 9M / area of disk = 9M/ piR^2
mass of small disk of radius R/3 = area of small disk x mass density = pi(R/3)^2 x 9M/ piR^2
We get the mass of the removed small disk = M
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Its centre of mass is located at R - R/3 = 2R/3 from the main axis
its moment of inertia can be calculated using parallel axis theorem
moment of inertia of removed disk = moment of inertia of small disk about its own axis + moment of inerita of its centre of mass about the main axis
Iremoved = (1/2) M (R/3)^2 + M (2R/3)^2
= MR2 /18 + 4MR^2 /9
Iremoved = MR^2 /2
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hence the moment of inertia of remaining disk = Iwhole - Iremoved
= (1/2) 9M R^2 - MR^2 /2
= 4MR^2
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From a circular disc of radius R and mass 9M a small disc of radius R/...
Problem Statement:
A circular disc of radius R and mass 9M is given. A small disc of radius R/3 is removed from the larger disc. The task is to find the moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through O.
Solution:
To solve this problem, we can consider the moment of inertia of the remaining disc as the difference between the moment of inertia of the original disc and the moment of inertia of the removed disc.
Moment of Inertia of the Original Disc:
The moment of inertia of a circular disc about an axis perpendicular to its plane and passing through its center is given by the formula:
I = (1/2) * M * R^2
Here, M is the mass of the disc and R is its radius.
In this case, the mass of the original disc is 9M and its radius is R. Therefore, the moment of inertia of the original disc is:
I_original = (1/2) * (9M) * R^2 = 4.5M * R^2
Moment of Inertia of the Removed Disc:
The moment of inertia of a circular disc about an axis perpendicular to its plane and passing through its center is given by the same formula as above:
I = (1/2) * m * r^2
Here, m is the mass of the disc and r is its radius.
In this case, the mass of the removed disc is M and its radius is R/3. Therefore, the moment of inertia of the removed disc is:
I_removed = (1/2) * (M) * (R/3)^2 = (1/18) * M * R^2
Moment of Inertia of the Remaining Disc:
The moment of inertia of the remaining disc can be calculated by subtracting the moment of inertia of the removed disc from the moment of inertia of the original disc:
I_remaining = I_original - I_removed
= 4.5M * R^2 - (1/18) * M * R^2
= (81/18) * M * R^2 - (1/18) * M * R^2
= (80/18) * M * R^2
= (40/9) * M * R^2
Therefore, the moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through O is (40/9) * M * R^2.
A circular disc of radius R and mass 9M is given. A small disc of radius R/3 is removed from the larger disc. The task is to find the moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through O.
Solution:
To solve this problem, we can consider the moment of inertia of the remaining disc as the difference between the moment of inertia of the original disc and the moment of inertia of the removed disc.
Moment of Inertia of the Original Disc:
The moment of inertia of a circular disc about an axis perpendicular to its plane and passing through its center is given by the formula:
I = (1/2) * M * R^2
Here, M is the mass of the disc and R is its radius.
In this case, the mass of the original disc is 9M and its radius is R. Therefore, the moment of inertia of the original disc is:
I_original = (1/2) * (9M) * R^2 = 4.5M * R^2
Moment of Inertia of the Removed Disc:
The moment of inertia of a circular disc about an axis perpendicular to its plane and passing through its center is given by the same formula as above:
I = (1/2) * m * r^2
Here, m is the mass of the disc and r is its radius.
In this case, the mass of the removed disc is M and its radius is R/3. Therefore, the moment of inertia of the removed disc is:
I_removed = (1/2) * (M) * (R/3)^2 = (1/18) * M * R^2
Moment of Inertia of the Remaining Disc:
The moment of inertia of the remaining disc can be calculated by subtracting the moment of inertia of the removed disc from the moment of inertia of the original disc:
I_remaining = I_original - I_removed
= 4.5M * R^2 - (1/18) * M * R^2
= (81/18) * M * R^2 - (1/18) * M * R^2
= (80/18) * M * R^2
= (40/9) * M * R^2
Therefore, the moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through O is (40/9) * M * R^2.
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From a circular disc of radius R and mass 9M a small disc of radius R/3 is removed as shown in figure the moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through O is?
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From a circular disc of radius R and mass 9M a small disc of radius R/3 is removed as shown in figure the moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through O is? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about From a circular disc of radius R and mass 9M a small disc of radius R/3 is removed as shown in figure the moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through O is? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for From a circular disc of radius R and mass 9M a small disc of radius R/3 is removed as shown in figure the moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through O is?.
From a circular disc of radius R and mass 9M a small disc of radius R/3 is removed as shown in figure the moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through O is? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about From a circular disc of radius R and mass 9M a small disc of radius R/3 is removed as shown in figure the moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through O is? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for From a circular disc of radius R and mass 9M a small disc of radius R/3 is removed as shown in figure the moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through O is?.
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