Question:

Two point masses of 0.3 kg and 0.7 kg are fixed at the ends of a rod of length 1.4 m and of negligible mass. The rod is set rotating about an axis perpendicular to its length with a uniform angular speed. The point on the rod through which the axis should pass in order that the work required for rotation of the rod is minimum, is located at a distance of

Updated On: Jun 14, 2022
  • 0.42 m from mass of 0.3 kg
  • 0.70 m from mass of 0.7 kg
  • 0.98 m from mass of 0.3 kg
  • 0.98 m from mass of 0.7 kg
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The Correct Option is C

Solution and Explanation

Work done $W=\frac{1}{2}I\omega^2$
If x is the distance of mass 0.3 kg from the centre of mass, we will have,
$I=(0.3)x^2+(0.7)(1.4-x)^2$
For work to be minimum, the moment of inertia (/) should be minimum, or
or $2(0.3x)-2(0.7)(1.4-x)=0$
or $(0.3)x=(0.7)(1.4-x)$
$\Rightarrow x=\frac{(0.7)(1.4)}{0.3+0.7}=0.98m$
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Questions Asked in JEE Advanced exam

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Concepts Used:

System of Particles and Rotational Motion

  1. The system of particles refers to the extended body which is considered a rigid body most of the time for simple or easy understanding. A rigid body is a body with a perfectly definite and unchangeable shape.
  2. The distance between the pair of particles in such a body does not replace or alter. Rotational motion can be described as the motion of a rigid body originates in such a manner that all of its particles move in a circle about an axis with a common angular velocity.
  3. The few common examples of rotational motion are the motion of the blade of a windmill and periodic motion.