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A light rod of length l has two masses m1 and m2 attached to its two ends. The moment of inertia of the system about an axis perpendicular to the rod and passing through the centre of mass is A) _/m1m2 l^2 B) m1m2 upon m1 m2 multiplied by l^2 C) m1 m2 upon m1m2 multiplied by l^2 D) (m1 m2) l^2?
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A light rod of length l has two masses m1 and m2 attached to its two e...
To find the moment of inertia of the system about an axis perpendicular to the rod and passing through the center of mass, we can use the parallel axis theorem. The parallel axis theorem states that the moment of inertia of a system about an axis parallel to and a distance "d" away from an axis through the center of mass is given by:
I = Icm + Md^2
Where I is the moment of inertia about the parallel axis, Icm is the moment of inertia about the axis through the center of mass, M is the total mass of the system, and d is the distance between the two axes.
In this case, let's consider the axis passing through the center of mass as the x-axis. The masses m1 and m2 are attached to the ends of the rod, so the center of mass of the system is at the midpoint of the rod.
Therefore, the distance between the two axes, d, is l/2. The total mass of the system, M, is m1 + m2.
Now, let's calculate the moment of inertia about the axis through the center of mass, Icm. For a rod of length l rotating about an axis passing through its center and perpendicular to its length, the moment of inertia is given by:
Icm = (1/12) Ml^2
Substituting the values of M and d into the parallel axis theorem equation, we get:
I = (1/12) Ml^2 + M(l/2)^2
= (1/12) Ml^2 + (1/4) Ml^2
= (1/12 + 1/4) Ml^2
= (1/3) Ml^2
Since M = m1 + m2, the moment of inertia can be written as:
I = (1/3) (m1 + m2) l^2
So, the correct option is D) (m1 + m2) l^2.
To summarize:
- The moment of inertia of the system about an axis perpendicular to the rod and passing through the center of mass is given by the parallel axis theorem.
- The distance between the two axes is l/2, and the total mass of the system is m1 + m2.
- The moment of inertia about the axis through the center of mass is (1/12) Ml^2.
- Substituting the values into the parallel axis theorem equation, we get the moment of inertia as (1/3) (m1 + m2) l^2.
- Therefore, the correct option is D) (m1 + m2) l^2.
I = Icm + Md^2
Where I is the moment of inertia about the parallel axis, Icm is the moment of inertia about the axis through the center of mass, M is the total mass of the system, and d is the distance between the two axes.
In this case, let's consider the axis passing through the center of mass as the x-axis. The masses m1 and m2 are attached to the ends of the rod, so the center of mass of the system is at the midpoint of the rod.
Therefore, the distance between the two axes, d, is l/2. The total mass of the system, M, is m1 + m2.
Now, let's calculate the moment of inertia about the axis through the center of mass, Icm. For a rod of length l rotating about an axis passing through its center and perpendicular to its length, the moment of inertia is given by:
Icm = (1/12) Ml^2
Substituting the values of M and d into the parallel axis theorem equation, we get:
I = (1/12) Ml^2 + M(l/2)^2
= (1/12) Ml^2 + (1/4) Ml^2
= (1/12 + 1/4) Ml^2
= (1/3) Ml^2
Since M = m1 + m2, the moment of inertia can be written as:
I = (1/3) (m1 + m2) l^2
So, the correct option is D) (m1 + m2) l^2.
To summarize:
- The moment of inertia of the system about an axis perpendicular to the rod and passing through the center of mass is given by the parallel axis theorem.
- The distance between the two axes is l/2, and the total mass of the system is m1 + m2.
- The moment of inertia about the axis through the center of mass is (1/12) Ml^2.
- Substituting the values into the parallel axis theorem equation, we get the moment of inertia as (1/3) (m1 + m2) l^2.
- Therefore, the correct option is D) (m1 + m2) l^2.
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A light rod of length l has two masses m1 and m2 attached to its two ends. The moment of inertia of the system about an axis perpendicular to the rod and passing through the centre of mass is A) _/m1m2 l^2 B) m1m2 upon m1 m2 multiplied by l^2 C) m1 m2 upon m1m2 multiplied by l^2 D) (m1 m2) l^2?
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A light rod of length l has two masses m1 and m2 attached to its two ends. The moment of inertia of the system about an axis perpendicular to the rod and passing through the centre of mass is A) _/m1m2 l^2 B) m1m2 upon m1 m2 multiplied by l^2 C) m1 m2 upon m1m2 multiplied by l^2 D) (m1 m2) l^2? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A light rod of length l has two masses m1 and m2 attached to its two ends. The moment of inertia of the system about an axis perpendicular to the rod and passing through the centre of mass is A) _/m1m2 l^2 B) m1m2 upon m1 m2 multiplied by l^2 C) m1 m2 upon m1m2 multiplied by l^2 D) (m1 m2) l^2? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A light rod of length l has two masses m1 and m2 attached to its two ends. The moment of inertia of the system about an axis perpendicular to the rod and passing through the centre of mass is A) _/m1m2 l^2 B) m1m2 upon m1 m2 multiplied by l^2 C) m1 m2 upon m1m2 multiplied by l^2 D) (m1 m2) l^2?.
A light rod of length l has two masses m1 and m2 attached to its two ends. The moment of inertia of the system about an axis perpendicular to the rod and passing through the centre of mass is A) _/m1m2 l^2 B) m1m2 upon m1 m2 multiplied by l^2 C) m1 m2 upon m1m2 multiplied by l^2 D) (m1 m2) l^2? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A light rod of length l has two masses m1 and m2 attached to its two ends. The moment of inertia of the system about an axis perpendicular to the rod and passing through the centre of mass is A) _/m1m2 l^2 B) m1m2 upon m1 m2 multiplied by l^2 C) m1 m2 upon m1m2 multiplied by l^2 D) (m1 m2) l^2? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A light rod of length l has two masses m1 and m2 attached to its two ends. The moment of inertia of the system about an axis perpendicular to the rod and passing through the centre of mass is A) _/m1m2 l^2 B) m1m2 upon m1 m2 multiplied by l^2 C) m1 m2 upon m1m2 multiplied by l^2 D) (m1 m2) l^2?.
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