A rod of length L has non-uniform linear mass | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

$\mathrm{x}_{\mathrm{cm}}=\frac{\int \mathrm{xdm}}{\int \mathrm{dm}}=\frac{\int(\lambda \mathrm{dx}) \mathrm{x}}{\int \mathrm{dm}}$

$=\frac{\int_{0

Vedclass · Accepted Answer

$\frac{3}{4}\left(\frac{2 a+b}{3 a+b}\right) L$

Vedclass · Answer

$\mathrm{x}_{\mathrm{cm}}=\frac{\int \mathrm{xdm}}{\int \mathrm{dm}}=\frac{\int(\lambda \mathrm{dx}) \mathrm{x}}{\int \mathrm{dm}}$

$=\frac{\int_{0}^{L}\left(a+\frac{b x^{2}}{L^{2}}\right) x d x}{\int_{0}^{L}\left(a+\frac{b x^{2}}{L^{2}}\right) d x}$

$=\frac{\frac{a L^{2}}{2}+\frac{b}{L^{2}} \cdot \frac{L^{4}}{4}}{a L+\frac{b}{L^{2}} \cdot \frac{L^{3}}{3}}$

$=\frac{\left(\frac{4 a+2 b}{8}\right) L}{\frac{(3 a+b)}{3}}=\frac{3}{4} \frac{(2 a+b) L}{(3 a+b)}$

A rod of length $L$ has non-uniform linear mass density given by $\rho(\mathrm{x})=\mathrm{a}+\mathrm{b}\left(\frac{\mathrm{x}}{\mathrm{L}}\right)^{2},$ where $a$ and $\mathrm{b}$ are constants and $0 \leq \mathrm{x} \leq \mathrm{L}$. The value of $\mathrm{x}$ for the centre of mass of the rod is at

Similar Questions

From a circular disc of radius $R$ a triangular portion is cut (see fig.). The distance of $COM$ of the remaining disc from centre of disc $O$ is:-

From a uniform disc of radius $R$, an equilateral triangle of side $\sqrt 3 \,R$ is cut as shown. The new position of centre of mass is :

Three particles of masses $1\,kg,\,\frac {3}{2}\,kg$ , and $2\,kg$ are located at the vertices of an equilateral triangle of side $a$ . The $x, y$ coordinates of the centre of mass are

Locate the centre of mass of arrangement shown in figure. The three rods are identical in mass and length

Seven identical homogeneous bricks, each of length $L$ , are arranged as shown in figure. Each brick is displaced with respect to the one in contacts by $\frac{L}{{10}}$ . Calculate the $x$-co-ordinate of the centre of mass of this system relative to the origin $O$ as shown