Question

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 :

• A. $\frac{L}{2} - \frac{\lambda_{0}L^{2}}{4M}$
• B. $\frac{L}{3} + \frac{\lambda_{0}L^{2}}{8M}$
• C. $\frac{L}{3} + \frac{\lambda_{0}L^{2}}{4M}$
• D. $\frac{2L}{3} - \frac{\lambda_{0}L^{2}}{6M}$

Answer 1

Answer: (D)

$\frac{2L}{3} - \frac{\lambda_{0}L^{2}}{6M}$

Answer 2

Mass per unit lengh $\lambda_{0} + kx$
$M = \int\limits^{L}_{0}\left(\lambda _{0} + kx\right)dx$
$M = \lambda_{0}L + \frac{K\times L^{2}}{2}$
$\frac{2M-\lambda _{0}L }{L^{2}} = K$
$\frac{2M}{L^{2}}-\frac{\lambda _{0}}{L} = K$
$\frac{\int dm\left(r\right)}{\int dm} = \frac{\int\left(\lambda dn\right)x}{M} = \frac{ \int ^{L}_{0}\left(\lambda _{0}x + kx^{2}\right)dx}{M}$
$r_{cm} = \frac{\lambda_{0}L+\frac{kL^{2}}{2}}{M}$
substitute 'k'
$r_{cm} = \frac{2L}{3} - \frac{\lambda_{0}\ell^{2}}{6M}$