Class 11 Exam > Class 11 Questions > A light rod of length l has two masses m1 and...
Start Learning for Free
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?
Most Upvoted Answer
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.
Attention Class 11 Students!
To make sure you are not studying endlessly, EduRev has designed Class 11 study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in Class 11.
|
Explore Courses for Class 11 exam
|
|
Similar Class 11 Doubts
Top Courses for Class 11View all
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?
Question Description
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?.
Solutions 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? in English & in Hindi are available as part of our courses for Class 11.
Download more important topics, notes, lectures and mock test series for Class 11 Exam by signing up for free.
Here you can find the meaning of 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? defined & explained in the simplest way possible. Besides giving the explanation of
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 detailed solution 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? has been provided alongside types of 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? theory, EduRev gives you an
ample number of questions to practice 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? tests, examples and also practice Class 11 tests.
|
|
Explore Courses for Class 11 exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup