Question:

If linear density of a rod of length 3 m varies as $ \lambda =2+x, $ then the position of the centre of gravity of the rod is:

Updated On: Jun 20, 2022
  • $ \frac{7}{3}m $
  • $ \frac{12}{7}m $
  • $ \frac{10}{7}m $
  • $ \frac{9}{7}m $
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The Correct Option is B

Solution and Explanation

Let rod is placed along $ x- $ axis. Mass of element PQ of length
$ dx $ situated at $ x=x $ is


$ dm=\lambda dx=(2+x)dx $
The COM of the element has coordinates $ (x,0,0). $ Therefore, $ x- $ coordinate of COM of the rod will be
$ {{x}_{COM}}=\frac{\int_\limits{0}^{3}{xdm}}{\int_\limits{0}^{3}{dm}} $
$ =\frac{\int_\limits{0}^{3}{x(2+x)dx}}{\int_\limits{0}^{3}{(2+x)dx}} $
$ =\frac{\int_\limits{0}^{3}{(2x+{{x}^{2}})dx}}{\int_\limits{0}^{3}{(2+x)dx}} $
$ =\frac{\left[ \frac{2{{x}^{2}}}{2}+\frac{{{x}^{3}}}{3} \right]_{0}^{3}}{\left[ 2x+\frac{{{x}^{2}}}{2} \right]_{0}^{3}} $
$ =\frac{\left[ {{(3)}^{2}}+\frac{{{(3)}^{3}}}{3} \right]}{\left[ 2\times 3+\frac{{{(3)}^{2}}}{2} \right]} $
$ =\frac{9+9}{6+9/2} $
$ =\frac{18\times 2}{21} $
$ =\frac{12}{7}m $
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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.