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A thin bar of length L has a mass per unit  length\lambda, that increases linearly with distance from one end.If its total mass is M and its mass per unit length at the lighter end is \lambda _{0}, then the distance of the centre of mass from the lighter end is :

  • Option 1)

    \frac{L}{2}-\frac{\lambda _{0}L^{2}}{4M}

  • Option 2)

    \frac{L}{3}+\frac{\lambda _{0}L^{2}}{8M}

  • Option 3)

    \frac{L}{3}+\frac{\lambda _{0}L^{2}}{4M}

  • Option 4)

    \frac{2L}{3}-\frac{\lambda _{0}L^{2}}{6M}

 

Answers (1)

best_answer

As we have learned

Centre of Mass of a continuous Distribution -

x_{cm}=\frac {\int xdm}{\int dm}, \; \;y_{cm}=\frac{\int ydm}{\int dm}, \;z_{cm}=\frac{\int zdm}{\int dm}

- wherein

dm is mass of small element. x, y, z are coordinates of dm part.

 

 

\lambda = \lambda _0 +kx 

k is some constant 

Total mass = \int dm = \int_{0}^{L}\lambda dx

M = \int_{0}^{L}(\lambda_0 +k x )dx = \lambda _0 L + \frac{k L^2}{2}.......(1)

x_c_m = \frac{\int xdm}{\int dm } = \frac{\int_{0}^{L}x (\lambda dx)}{M}

= \frac{\int_{0}^{L}(\lambda _0+kx )\cdot xdx }{M}= \frac{\lambda _0\cdot \frac{L^2}{2}+ k \frac{L^3}{3}}{M}

x_c_m = \frac{L^2\left [ \frac{\lambda _0}{2}+\frac{KL}{3} \right ]}{M}

From equation (1) KL = \left ( \frac{M}{L}-\lambda _0 \right )\cdot 2

x_c_m = \frac{L^2}{M} \left [ \frac{\lambda _0}{2}+\frac{2M}{3L} - \frac{2 \lambda_0}{3}\right ]

= \frac{L^2}{M} \left [ \frac{3\lambda _0L+2M-4\lambda _0L}{6L}\right ]

= \frac{L}{6M}\cdot (2M - \lambda _0L)= L/3 - \frac{\lambda _0L^2}{6M}

 

 

 

 

 

 

 

 


Option 1)

\frac{L}{2}-\frac{\lambda _{0}L^{2}}{4M}

Option 2)

\frac{L}{3}+\frac{\lambda _{0}L^{2}}{8M}

Option 3)

\frac{L}{3}+\frac{\lambda _{0}L^{2}}{4M}

Option 4)

\frac{2L}{3}-\frac{\lambda _{0}L^{2}}{6M}

Posted by

SudhirSol

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