A uniform rod of length L and mass M is place | vedclass

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

Question

(B) For a uniform rod of length $L$ and mass $M$ placed along the $x$-axis with one end at the origin,the linear mass density $\lambda$ is constant an

Vedclass · Accepted Answer

$L/2$

Vedclass · Answer

(B) For a uniform rod of length $L$ and mass $M$ placed along the $x$-axis with one end at the origin,the linear mass density $\lambda$ is constant and is given by $\lambda = M/L$.The position of the centre of mass $X_{cm}$ is defined as:$X_{cm} = \frac{1}{M} \int x \, dm$Since the rod is uniform,$dm = \lambda \, dx = (M/L) \, dx$.Substituting this into the integral:$X_{cm} = \frac{1}{M} \int_{0}^{L} x \left( \frac{M}{L} \right) dx$$X_{cm} = \frac{1}{L} \int_{0}^{L} x \, dx$$X_{cm} = \frac{1}{L} \left[ \frac{x^2}{2} \right]_{0}^{L}$$X_{cm} = \frac{1}{L} \left( \frac{L^2}{2} - 0 \right) = \frac{L}{2}$.Thus,the centre of mass is at $L/2$.

$A$ uniform rod of length $L$ and mass $M$ is placed along the $x$-axis with one end at the origin. Find the $x$-coordinate of the centre of mass of the rod.

Similar Questions

The variation of density of a solid cylindrical rod of cross-sectional area $\alpha$ and length $L$ is given by $\rho = \rho_0 \frac{x^2}{L^2}$, where $x$ is the distance from one end of the rod. The position of its centre of mass from that end $(x=0)$ is:

The centre of mass of two particles lies:

In the case of a system of two particles of different masses,the centre of mass lies

Find the $x$ coordinate of the centre of mass of the books shown in the figure. Every book has mass $m$.

The centre of mass of a homogeneous semi-circular plate of radius $r$ is located at $A$ as shown in the figure. The distance $OA$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module