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Four spheres of diameter 2a and mass M are placed with their centres on the four corners of a square of side b. The n moment of inertia of the system about an axis about one of the sides of the square is 1) Ma² 2Mb² 2) Ma² 3) Ma² 4Mb² 4) 8/5 Ma² 2Mb²?
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Four spheres of diameter 2a and mass M are placed with their centres o...
The answer is 4.
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Four spheres of diameter 2a and mass M are placed with their centres o...
The moment of inertia of a system of particles is the sum of the moments of inertia of each individual particle. In this problem, we have four spheres placed at the corners of a square, and we need to find the moment of inertia of the system about an axis passing through one side of the square.
Let's assume that the axis of rotation is passing through the side of the square, and let's label the spheres as A, B, C, and D.
To find the moment of inertia of the system, we can use the parallel axis theorem. According to the parallel axis theorem, the moment of inertia of a system about an axis parallel to and a distance "d" away from an axis passing through the center of mass is given by:
I = I_cm + Md^2
where I_cm is the moment of inertia of the system about the axis passing through the center of mass, M is the total mass of the system, and d is the distance between the two axes.
We know that the moment of inertia of a sphere about an axis passing through its center is given by:
I_sphere = (2/5)mr^2
where m is the mass of the sphere and r is its radius.
Let's calculate the moment of inertia of the system about the axis passing through the center of mass.
- Moment of inertia of sphere A about the axis passing through its center = (2/5)Ma^2
- Moment of inertia of sphere B about the axis passing through its center = (2/5)Mb^2
- Moment of inertia of sphere C about the axis passing through its center = (2/5)Ma^2
- Moment of inertia of sphere D about the axis passing through its center = (2/5)Mb^2
Since the spheres are placed at the corners of a square, the distance between the axis passing through the center of mass and the axis passing through one side of the square is b/2.
Using the parallel axis theorem, the total moment of inertia of the system about the axis passing through one side of the square is:
I_total = I_A + I_B + I_C + I_D + M(d^2)
where d = b/2.
Substituting the values, we get:
I_total = (2/5)Ma^2 + (2/5)Mb^2 + (2/5)Ma^2 + (2/5)Mb^2 + M(b^2/4)
Simplifying, we get:
I_total = (4/5)Ma^2 + (4/5)Mb^2 + M(b^2/4)
Comparing this with the given options, we can see that the correct answer is 4) 8/5 Ma² + 2Mb².
Let's assume that the axis of rotation is passing through the side of the square, and let's label the spheres as A, B, C, and D.
To find the moment of inertia of the system, we can use the parallel axis theorem. According to the parallel axis theorem, the moment of inertia of a system about an axis parallel to and a distance "d" away from an axis passing through the center of mass is given by:
I = I_cm + Md^2
where I_cm is the moment of inertia of the system about the axis passing through the center of mass, M is the total mass of the system, and d is the distance between the two axes.
We know that the moment of inertia of a sphere about an axis passing through its center is given by:
I_sphere = (2/5)mr^2
where m is the mass of the sphere and r is its radius.
Let's calculate the moment of inertia of the system about the axis passing through the center of mass.
- Moment of inertia of sphere A about the axis passing through its center = (2/5)Ma^2
- Moment of inertia of sphere B about the axis passing through its center = (2/5)Mb^2
- Moment of inertia of sphere C about the axis passing through its center = (2/5)Ma^2
- Moment of inertia of sphere D about the axis passing through its center = (2/5)Mb^2
Since the spheres are placed at the corners of a square, the distance between the axis passing through the center of mass and the axis passing through one side of the square is b/2.
Using the parallel axis theorem, the total moment of inertia of the system about the axis passing through one side of the square is:
I_total = I_A + I_B + I_C + I_D + M(d^2)
where d = b/2.
Substituting the values, we get:
I_total = (2/5)Ma^2 + (2/5)Mb^2 + (2/5)Ma^2 + (2/5)Mb^2 + M(b^2/4)
Simplifying, we get:
I_total = (4/5)Ma^2 + (4/5)Mb^2 + M(b^2/4)
Comparing this with the given options, we can see that the correct answer is 4) 8/5 Ma² + 2Mb².
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Four spheres of diameter 2a and mass M are placed with their centres on the four corners of a square of side b. The n moment of inertia of the system about an axis about one of the sides of the square is 1) Ma² 2Mb² 2) Ma² 3) Ma² 4Mb² 4) 8/5 Ma² 2Mb²?
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Four spheres of diameter 2a and mass M are placed with their centres on the four corners of a square of side b. The n moment of inertia of the system about an axis about one of the sides of the square is 1) Ma² 2Mb² 2) Ma² 3) Ma² 4Mb² 4) 8/5 Ma² 2Mb²? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Four spheres of diameter 2a and mass M are placed with their centres on the four corners of a square of side b. The n moment of inertia of the system about an axis about one of the sides of the square is 1) Ma² 2Mb² 2) Ma² 3) Ma² 4Mb² 4) 8/5 Ma² 2Mb²? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Four spheres of diameter 2a and mass M are placed with their centres on the four corners of a square of side b. The n moment of inertia of the system about an axis about one of the sides of the square is 1) Ma² 2Mb² 2) Ma² 3) Ma² 4Mb² 4) 8/5 Ma² 2Mb²?.
Four spheres of diameter 2a and mass M are placed with their centres on the four corners of a square of side b. The n moment of inertia of the system about an axis about one of the sides of the square is 1) Ma² 2Mb² 2) Ma² 3) Ma² 4Mb² 4) 8/5 Ma² 2Mb²? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Four spheres of diameter 2a and mass M are placed with their centres on the four corners of a square of side b. The n moment of inertia of the system about an axis about one of the sides of the square is 1) Ma² 2Mb² 2) Ma² 3) Ma² 4Mb² 4) 8/5 Ma² 2Mb²? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Four spheres of diameter 2a and mass M are placed with their centres on the four corners of a square of side b. The n moment of inertia of the system about an axis about one of the sides of the square is 1) Ma² 2Mb² 2) Ma² 3) Ma² 4Mb² 4) 8/5 Ma² 2Mb²?.
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