Find the centre of mass of a uniform $L$-shaped lamina (a thin flat plate) with dimensions as shown. The mass of the lamina is $3 \; kg$.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) Choosing the $X$ and $Y$ axes as shown in the figure,we have the coordinates of the vertices of the $L$-shaped lamina. We can think of the $L$-shape as consisting of $3$ squares,each of side length $1 \; m$. Since the lamina is uniform,the mass of each square is $1 \; kg$. The centres of mass $C_{1}$,$C_{2}$,and $C_{3}$ of the squares are,by symmetry,their geometric centres. Their coordinates are $(0.5, 0.5) \; m$,$(1.5, 0.5) \; m$,and $(0.5, 1.5) \; m$ respectively. We take the masses of the squares to be concentrated at these points. The centre of mass $(X, Y)$ of the whole $L$-shape is the centre of mass of these mass points.
Hence,
$X = \frac{[1(0.5) + 1(1.5) + 1(0.5)] \; kg \cdot m}{(1 + 1 + 1) \; kg} = \frac{2.5}{3} \; m = \frac{5}{6} \; m$
$Y = \frac{[1(0.5) + 1(0.5) + 1(1.5)] \; kg \cdot m}{(1 + 1 + 1) \; kg} = \frac{2.5}{3} \; m = \frac{5}{6} \; m$

Explore More

Similar Questions

Define the position vector of the centre of mass.

Two identical uniform rectangular blocks (with longest side $L$) and a solid sphere of radius $R$ are to be balanced at the edge of a heavy table such that the centre of the sphere remains at the maximum possible horizontal distance from the vertical edge of the table without toppling as indicated in the figure. If the mass of each block is $M$ and of the sphere is $M/2$,then the maximum distance $x$ that can be achieved is

The linear mass density of a rod of length $L$ varies as $\lambda = kx^2$,where $k$ is a constant and $x$ is the distance from one end. The position of the centre of mass of the rod is:

Difficult
View Solution

Three identical spheres,each of mass $2 M$,are placed at the corners of a right-angled triangle with mutually perpendicular sides equal to $4 \ m$ each. Taking the point of intersection of these two sides as the origin,the magnitude of the position vector of the centre of mass of the system is $\frac{4 \sqrt{2}}{x}$,where the value of $x$ is . . . . . .

Four particles of masses $m_1 = 2m$,$m_2 = 4m$,$m_3 = m$,and $m_4$ are placed at the four corners of a square of side $a$. Let the corners be $(0,0)$,$(a,0)$,$(a,a)$,and $(0,a)$ respectively. What should be the value of $m_4$ so that the centre of mass of the system is at the centre of the square,i.e.,at $(\frac{a}{2}, \frac{a}{2})$?

Difficult
View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo