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Two rings of the same radius and mass are placed such that their centres are at a common point and their planes are perpendicular to each other. The moment of inertia of the system about an axis passing through the centre and perpendicular to the plane of one of the rings is (mass of the ring = m and radius= r)
- a)1/2 mr2
- b)mr2
- c)3/2 mr2
- d)2mr2
Correct answer is option 'C'. Can you explain this answer?
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Solution:
Given, two rings of same mass and radius, placed in such a way that their centres are at a common point and their planes are perpendicular to each other.
Let the moment of inertia of each ring about an axis perpendicular to its plane and passing through its centre be I1.
The moment of inertia of the system about the axis passing through the centre and perpendicular to the plane of one of the rings can be found by using the perpendicular axis theorem.
According to this theorem, the moment of inertia of a planar body about an axis perpendicular to its plane is equal to the sum of its moments of inertia about two perpendicular axes lying in its plane and intersecting each other at the point where the perpendicular axis passes through the plane.
In this case, let us consider the moment of inertia of the system about an axis passing through the centre and perpendicular to the plane of the first ring. This axis intersects the plane of the second ring at the common centre.
The moments of inertia of the two rings about two perpendicular axes lying in their plane and passing through their common centre are given by:
I1 = 1/2 mr^2 (for each ring)
The moment of inertia of the system about the perpendicular axis passing through the common centre can be found by using the perpendicular axis theorem as:
I = 2I1 = 2 x 1/2 mr^2 = mr^2
Therefore, the moment of inertia of the system about an axis passing through the centre and perpendicular to the plane of one of the rings is mr^2, which is option (c).
Given, two rings of same mass and radius, placed in such a way that their centres are at a common point and their planes are perpendicular to each other.
Let the moment of inertia of each ring about an axis perpendicular to its plane and passing through its centre be I1.
The moment of inertia of the system about the axis passing through the centre and perpendicular to the plane of one of the rings can be found by using the perpendicular axis theorem.
According to this theorem, the moment of inertia of a planar body about an axis perpendicular to its plane is equal to the sum of its moments of inertia about two perpendicular axes lying in its plane and intersecting each other at the point where the perpendicular axis passes through the plane.
In this case, let us consider the moment of inertia of the system about an axis passing through the centre and perpendicular to the plane of the first ring. This axis intersects the plane of the second ring at the common centre.
The moments of inertia of the two rings about two perpendicular axes lying in their plane and passing through their common centre are given by:
I1 = 1/2 mr^2 (for each ring)
The moment of inertia of the system about the perpendicular axis passing through the common centre can be found by using the perpendicular axis theorem as:
I = 2I1 = 2 x 1/2 mr^2 = mr^2
Therefore, the moment of inertia of the system about an axis passing through the centre and perpendicular to the plane of one of the rings is mr^2, which is option (c).
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Two rings of the same radius and mass are placed such that their centres are at a common point and their planes are perpendicular to each other. The moment of inertia of the system about an axis passing through the centre and perpendicular to the plane of one of the rings is (mass of the ring = m and radius= r)a)1/2 mr2b)mr2c)3/2 mr2d)2mr2Correct answer is option 'C'. Can you explain this answer?
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