A thin rod of length 'L' lies along the x-axi | vedclass

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

Question

(A) The center of mass $x_{CM}$ of a rod with linear mass density $\lambda(x)$ is given by:$x_{CM} = \frac{\int_0^L x \lambda(x) dx}{\int_0^L \lambda(

Vedclass · Accepted Answer

Vedclass · Answer

(A) The center of mass $x_{CM}$ of a rod with linear mass density $\lambda(x)$ is given by:$x_{CM} = \frac{\int_0^L x \lambda(x) dx}{\int_0^L \lambda(x) dx}$Substituting $\lambda(x) = k(\frac{x}{L})^n$:$x_{CM} = \frac{\int_0^L x \cdot k(\frac{x}{L})^n dx}{\int_0^L k(\frac{x}{L})^n dx} = \frac{\frac{k}{L^n} \int_0^L x^{n+1} dx}{\frac{k}{L^n} \int_0^L x^n dx} = \frac{[\frac{x^{n+2}}{n+2}]_0^L}{[\frac{x^{n+1}}{n+1}]_0^L} = \frac{L^{n+2}/(n+2)}{L^{n+1}/(n+1)} = L \frac{n+1}{n+2}$Analyzing the function $f(n) = L \frac{n+1}{n+2} = L (1 - \frac{1}{n+2})$:$1$. For $n = 0$,$x_{CM} = L(1 - 1/2) = L/2$.$2$. As $n 	o \infty$,$x_{CM} 	o L$.$3$. The derivative $\frac{dx_{CM}}{dn} = L \frac{(n+2) - (n+1)}{(n+2)^2} = \frac{L}{(n+2)^2} > 0$,so $x_{CM}$ is an increasing function.$4$. The second derivative $\frac{d^2x_{CM}}{dn^2} = -\frac{2L}{(n+2)^3} These properties match the graph in option $(A)$.

Similar Questions

$A$ metal sphere of radius $r$ and specific heat $S$ is rotated about an axis passing through its centre at a speed of $n$ rotations per second. It is suddenly stopped and $50 \%$ of its energy is used in increasing its temperature. Then, the rise in temperature of the sphere is

One end of a straight uniform $1\; m$ long bar is pivoted on a horizontal table. It is released from rest when it makes an angle $30^{\circ}$ with the horizontal (see figure). Its angular speed when it hits the table is given as $\sqrt{n}\; s^{-1},$ where $n$ is an integer. The value of $n$ is

$A$ body with a moment of inertia of $3\ kg\cdot m^2$ rotating with an angular velocity of $2\ rad/s$ has the same kinetic energy as a mass of $12\ kg$ moving with a velocity of .......... $m/s$.

$A$ particle falls freely near the surface of the earth. Consider a fixed point $O$ (not vertically below the particle) on the ground.

Four spheres,each of mass $M$ and diameter $2a$,are placed at the corners of a square of side $b$. The moment of inertia of this system about an axis along one of the sides of the square is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module