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A thin uniform rod of length 4L, mass 4m is bent at the points as a W shape. What is the moment of inertia of the rod about the axis passing through point on the middle upper sharp edge & perpendicular to the plane of the paper.?
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A thin uniform rod of length 4L, mass 4m is bent at the points as a W ...
Since, the mass of the rod is 4 m and the length is 4 l
so, mass of AB = BO = OC = CD = m
and, length of AB = BO = OC = CD = l
We know, Moment of inertia of a rod about its end = (ml^2)/3
Moment of inertia of AB bout B (I 1) = (ml^2)/3
Moment of inertia of BO about O (I 2) = (ml^2)/3
Moment of inertia of OC about O (I 3)= (ml^2)/3
Moment of inertia of CD about C (I 4)= (ml^2)/3
Now, from parallel axis theorem,
Moment of inertia of AB about O (I 5)
= I 1 + ml^2 ————(parallel axis theorem)
= (ml^2)/3 + ml^2
= (4 ml^2)/3
Similarly, Moment of inertia of CD about O (I 6)
= (4 ml^2)/3
So, moment of inertia of the rod about O
= I 2 + I 3 + I 5 + I 6
= ( ml^2)/3 + ( ml^2)/3 + (4 ml^2)/3 + (4 ml^2)/3
= (10 ml^2)/3
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A thin uniform rod of length 4L, mass 4m is bent at the points as a W ...
To find the moment of inertia of the bent rod about the axis passing through the middle upper sharp edge, we can divide the rod into three segments: the middle segment (length 2L) and the two side segments (length L each).
Let's denote the moment of inertia of the middle segment as I_m, and the moment of inertia of each side segment as I_s.
The moment of inertia of a uniform rod about an axis passing through one of its ends and perpendicular to its length is given by the formula:
I = (1/3) * m * L^2,
where m is the mass of the rod and L is its length.
Since the middle segment has a length of 2L, its moment of inertia is:
I_m = (1/3) * (2m) * (2L)^2 = (8/3) * m * L^2.
Similarly, the moment of inertia of each side segment is:
I_s = (1/3) * m * L^2.
Now, to find the total moment of inertia of the bent rod about the axis passing through the middle upper sharp edge, we need to add the moments of inertia of the three segments:
I_total = I_m + I_s + I_s = (8/3) * m * L^2 + (1/3) * m * L^2 + (1/3) * m * L^2.
Simplifying this expression gives:
I_total = (8/3) * m * L^2 + (2/3) * m * L^2.
Combining the terms gives:
I_total = (10/3) * m * L^2.
Therefore, the moment of inertia of the rod about the axis passing through the middle upper sharp edge is (10/3) * m * L^2.
Let's denote the moment of inertia of the middle segment as I_m, and the moment of inertia of each side segment as I_s.
The moment of inertia of a uniform rod about an axis passing through one of its ends and perpendicular to its length is given by the formula:
I = (1/3) * m * L^2,
where m is the mass of the rod and L is its length.
Since the middle segment has a length of 2L, its moment of inertia is:
I_m = (1/3) * (2m) * (2L)^2 = (8/3) * m * L^2.
Similarly, the moment of inertia of each side segment is:
I_s = (1/3) * m * L^2.
Now, to find the total moment of inertia of the bent rod about the axis passing through the middle upper sharp edge, we need to add the moments of inertia of the three segments:
I_total = I_m + I_s + I_s = (8/3) * m * L^2 + (1/3) * m * L^2 + (1/3) * m * L^2.
Simplifying this expression gives:
I_total = (8/3) * m * L^2 + (2/3) * m * L^2.
Combining the terms gives:
I_total = (10/3) * m * L^2.
Therefore, the moment of inertia of the rod about the axis passing through the middle upper sharp edge is (10/3) * m * L^2.
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A thin uniform rod of length 4L, mass 4m is bent at the points as a W shape. What is the moment of inertia of the rod about the axis passing through point on the middle upper sharp edge & perpendicular to the plane of the paper.?
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A thin uniform rod of length 4L, mass 4m is bent at the points as a W shape. What is the moment of inertia of the rod about the axis passing through point on the middle upper sharp edge & perpendicular to the plane of the paper.? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A thin uniform rod of length 4L, mass 4m is bent at the points as a W shape. What is the moment of inertia of the rod about the axis passing through point on the middle upper sharp edge & perpendicular to the plane of the paper.? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A thin uniform rod of length 4L, mass 4m is bent at the points as a W shape. What is the moment of inertia of the rod about the axis passing through point on the middle upper sharp edge & perpendicular to the plane of the paper.?.
A thin uniform rod of length 4L, mass 4m is bent at the points as a W shape. What is the moment of inertia of the rod about the axis passing through point on the middle upper sharp edge & perpendicular to the plane of the paper.? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A thin uniform rod of length 4L, mass 4m is bent at the points as a W shape. What is the moment of inertia of the rod about the axis passing through point on the middle upper sharp edge & perpendicular to the plane of the paper.? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A thin uniform rod of length 4L, mass 4m is bent at the points as a W shape. What is the moment of inertia of the rod about the axis passing through point on the middle upper sharp edge & perpendicular to the plane of the paper.?.
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