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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Physics
Chapter
Center of Mass, Momentum and Collision

NTA Abhyas 2020 · Physics · Center of Mass, Momentum and Collision

Find the x coordinate of the centre of mass of the non-uniform rod...

Find the xx coordinate of the centre of mass of the non-uniform rod of length LL given below. The origin is taken at the left end of the rod. The density of the rod as a function of its xx-coordinate is ρ=ax2+bx+cρ = {ax}^{2} + bx + c, where aa, bb and cc are constants.

Find the x coordinate of the centre of mass of the non-uniform rod of length L given below. The origin is taken at the

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Options

  1. A2aL2+3bL2+6cL2(3aL2+4bL+8c)\dfrac{2{aL}^{2} + 3{bL}^{2} + 6cL}{2( 3{aL}^{2} + 4bL + 8c )}
  2. B4aL3+3bL2+2cL2(3aL2+2bL+c)\dfrac{4{aL}^{3} + 3{bL}^{2} + 2cL}{2( 3{aL}^{2} + 2bL + c )}
  3. C3aL2+4bL2+6cL4aL2+6bL+8c\dfrac{3{aL}^{2} + 4{bL}^{2} + 6cL}{4{aL}^{2} + 6bL + 8c}
  4. D3aL3+4bL2+6cL2(2aL2+3bL+6c)\dfrac{3{aL}^{3} + 4{bL}^{2} + 6cL}{2( 2{aL}^{2} + 3bL + 6c )}

Correct answer

D. 3aL3+4bL2+6cL2(2aL2+3bL+6c)\dfrac{3{aL}^{3} + 4{bL}^{2} + 6cL}{2( 2{aL}^{2} + 3bL + 6c )}

Step-by-step solution

\dfrac{{∫}_{0}^{L}( {ax}^{2} + bx + c )\text{x dx}}{{∫}_{0}^{L}( {ax}^{2} + bx + c )\text{dx}}</p><p></p> <p>\dfrac{{∫}_{0}^{L}( {ax}^{3} + {bx}^{2} + cx )\text{dx}}{{∫}_{0}^{L}( {ax}^{2} + bx + c )\text{dx}}</p><br/></p><br />= {\dfrac{ax4}{4} + \dfrac{bx3}{3} + \dfrac{cx2}{2}}_{0}^{L}{\dfrac{ax3}{3} + \dfrac{bx2}{2} + cx}_{0}^{L}<br/><br/><br /><br />= \dfrac{3{aL}^{4} + 4{bL}^{3} + 6{cL}^{2}12}{2{aL}^{2} + 3{bL}^{2} + 6cL6}<br/><br/><br /><br />= \dfrac{3{aL}^{3} + 4{bL}^{2} + 6cL}{2\left(2{aL}^{2} + 3bL + 6c\right)}

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