A thin bar of length L has a mass per unit le | vedclass

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

Question

(C) Let the mass per unit length be $\lambda(x) = \lambda_0 + kx$.Total mass $M = \int_{0}^{L} (\lambda_0 + kx) dx = \lambda_0 L + \frac{kL^2}{2}$.Fro

Vedclass · Accepted Answer

$\frac{2L}{3} - \frac{\lambda_0 L^2}{6M}$

Vedclass · Answer

(C) Let the mass per unit length be $\lambda(x) = \lambda_0 + kx$.Total mass $M = \int_{0}^{L} (\lambda_0 + kx) dx = \lambda_0 L + \frac{kL^2}{2}$.From this,$k = \frac{2(M - \lambda_0 L)}{L^2} = \frac{2M}{L^2} - \frac{2\lambda_0}{L}$.The center of mass $x_{cm}$ is given by $x_{cm} = \frac{1}{M} \int_{0}^{L} x dm = \frac{1}{M} \int_{0}^{L} x (\lambda_0 + kx) dx$.$x_{cm} = \frac{1}{M} [\frac{\lambda_0 x^2}{2} + \frac{kx^3}{3}]_{0}^{L} = \frac{1}{M} (\frac{\lambda_0 L^2}{2} + \frac{kL^3}{3})$.Substituting $k = \frac{2M}{L^2} - \frac{2\lambda_0}{L}$ into the equation:$x_{cm} = \frac{1}{M} [\frac{\lambda_0 L^2}{2} + \frac{L^3}{3} (\frac{2M}{L^2} - \frac{2\lambda_0}{L})] = \frac{1}{M} [\frac{\lambda_0 L^2}{2} + \frac{2ML}{3} - \frac{2\lambda_0 L^2}{3}]$.$x_{cm} = \frac{2L}{3} + \frac{\lambda_0 L^2}{M} (\frac{1}{2} - \frac{2}{3}) = \frac{2L}{3} - \frac{\lambda_0 L^2}{6M}$.

$A$ thin bar of length $L$ has a mass per unit length $\lambda$ that increases linearly with distance $x$ from one end. If its total mass is $M$ and its mass per unit length at the lighter end $(x=0)$ is $\lambda_0$,then the distance of the centre of mass from the lighter end is:

Similar Questions

The position vectors of three particles of masses $1\, kg, 2\, kg$,and $3\, kg$ are $\vec{r_1} = (\hat{i} + 4\hat{j} + \hat{k})\,m$,$\vec{r_2} = (\hat{i} + \hat{j} + \hat{k})\,m$,and $\vec{r_3} = (2\hat{i} - \hat{j} - 2\hat{k})\,m$ respectively. The position vector of their centre of mass is:

The centre of mass of a non-uniform rod of length $L$ whose mass per unit length $\lambda$ varies as $\lambda = \frac{k x^3}{L^3}$ (where $k$ is a constant and $x$ is the distance of any point on the rod from one end) is at a distance from the same end equal to:

Write the meaning of homogeneous bodies.

Vedclass Test Series

Exam Paper Generator

Online Exam Module