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NEET PHYSICSRotational motionMedium

Question

A light rod of length ll has two masses m1m_1 and m2m_2 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

m1m2m1+m2l2\frac{m_1 m_2}{m_1 + m_2}l^2

B

m1+m2m1m2l2\frac{m_1 + m_2}{\sqrt{m_1 m_2}}l^2

C

(m1+m2)l2(m_1 + m_2)l^2

D

m1m2l2\sqrt{m_1 m_2}l^2

Step-by-Step Solution

Let the center of mass be at a distance r1r_1 from mass m1m_1 and r2r_2 from mass m2m_2. The position of the center of mass is given by m1r1=m2r2m_1 r_1 = m_2 r_2. We also know that the total length of the rod is l=r1+r2l = r_1 + r_2. From these two equations, we can find the distances: r1=m2lm1+m2r_1 = \frac{m_2 l}{m_1 + m_2} and r2=m1lm1+m2r_2 = \frac{m_1 l}{m_1 + m_2} The moment of inertia of the system about the center of mass is the sum of the moments of inertia of the two masses: I=m1r12+m2r22I = m_1 r_1^2 + m_2 r_2^2 Substituting the values of r1r_1 and r2r_2: I=m1(m2lm1+m2)2+m2(m1lm1+m2)2I = m_1 \left( \frac{m_2 l}{m_1 + m_2} \right)^2 + m_2 \left( \frac{m_1 l}{m_1 + m_2} \right)^2 I=m1m22l2(m1+m2)2+m2m12l2(m1+m2)2I = \frac{m_1 m_2^2 l^2}{(m_1 + m_2)^2} + \frac{m_2 m_1^2 l^2}{(m_1 + m_2)^2} I=m1m2l2(m1+m2)(m1+m2)2I = \frac{m_1 m_2 l^2 (m_1 + m_2)}{(m_1 + m_2)^2} I=m1m2m1+m2l2I = \frac{m_1 m_2}{m_1 + m_2} l^2 This is also known as the reduced mass μ=m1m2m1+m2\mu = \frac{m_1 m_2}{m_1 + m_2}, so I=μl2I = \mu l^2.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from Rotational motion. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSRotational motionlengthmassesattachedmomentinertia

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