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JIPMER 2015
Exam
JIPMER
Year
2015
Subject
Physics
Chapter
Center of Mass, Momentum and Collision

JIPMER 2015 · Physics · Center of Mass, Momentum and Collision

The density of a rod having length l varies as =c+d x , where x is...

The density of a rod having length ll varies as ρ=c+dx\rho=c+d x, where xx is the distance from the left end. The centre of mass is The density of a rod having length l varies as =c+d x , where x is the distance from the left end. The centre of mass is

Options

  1. A3cl+2Dl23(2c+Dl)\frac{3 c l+2 D l^2}{3(2 c+D l)}
  2. B2cl+3Dl22(4c+8l)\frac{2 c l+3 D l^2}{2(4 c+8 l)}
  3. C2cl+3Dl23(2c+l)\frac{2 c l+3 D l^2}{3(2 c+l)}
  4. Dcl+Dl23(2c+Dl)\frac{c l+D l^2}{3(2 c+D l)}

Correct answer

A. 3cl+2Dl23(2c+Dl)\frac{3 c l+2 D l^2}{3(2 c+D l)}

Step-by-step solution

Let the cross-sectional area is α\alpha. The mass of an element of length dxd x located at a distance xx away from the left end is (C+Dx)αdx(C+D x) \alpha d x. The xx-coordinate of the centre of mass is given by Xcm=∫xdm∫dmX_{c m}=\frac{\int x d m}{\int d m}=∫01x(C+Dx)αdx∫01(C+Dx)αdx=Cl22+Dl33Cl+Dl22=3Cl+2Dl23(2C+Dl)\begin{aligned}= & \frac{\int_0^1 x(C+D x) \alpha d x}{\int_0^1(C+D x) \alpha d x} \\ = & \frac{C \frac{l^2}{2}+\frac{D l^3}{3}}{C l+\frac{D l^2}{2}}=\frac{3 C l+2 D l^2}{3(2 C+D l)}\end{aligned}$$
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