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NEET PHYSICSRotational motionMedium

Question

Point masses m1m_1 and m2m_2 are placed at the opposite ends of a rigid rod of length LL and negligible mass. The rod is to be set rotating about an axis perpendicular to it. The position of point P on this rod through which the axis should pass so that the work required to set the rod rotating with angular velocity ω0\omega_0 is minimum, is given by:

A

x=m1m2Lx = \frac{m_1}{m_2}L

B

x=m2m1Lx = \frac{m_2}{m_1}L

C

x=m2Lm1+m2x = \frac{m_2 L}{m_1 + m_2}

D

x=m1Lm1+m2x = \frac{m_1 L}{m_1 + m_2}

Step-by-Step Solution

The work required to set the rod rotating with angular velocity ω0\omega_0 is equal to its rotational kinetic energy, W=ΔK=12Iω02W = \Delta K = \frac{1}{2}I\omega_0^2. For the work to be minimum, the moment of inertia II must be minimum. Let the axis of rotation pass through a point at a distance xx from mass m1m_1. Then its distance from mass m2m_2 is (L−x)(L - x). The moment of inertia of the system about this axis is: I=m1x2+m2(L−x)2I = m_1x^2 + m_2(L - x)^2 To find the condition for minimum moment of inertia, we differentiate II with respect to xx and set it to zero: dIdx=ddx[m1x2+m2(L−x)2]=0\frac{dI}{dx} = \frac{d}{dx}[m_1x^2 + m_2(L - x)^2] = 0 2m1x+2m2(L−x)(−1)=02m_1x + 2m_2(L - x)(-1) = 0 m1x−m2(L−x)=0m_1x - m_2(L - x) = 0 m1x=m2L−m2xm_1x = m_2L - m_2x (m1+m2)x=m2L(m_1 + m_2)x = m_2L x=m2Lm1+m2x = \frac{m_2L}{m_1 + m_2} This implies that the axis must pass through the centre of mass of the two-particle system for the moment of inertia, and thus the work done, to be minimum.

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 motionmassesplacedoppositelengthnegligible

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