Sigma Percentile
JEE Main 2019, 12 Jan Shift-I
LEVELJEE Main

Animated Solution for Physics - System of Particles: The position vector of the centre of mass of an asymmetric uniform bar of negligible area of cross-section as shown in figure is

Select Answer:

Visualized Solution

  • The given asymmetric bar can be broken down into three distinct uniform straight segments.
  • Segment 1: Horizontal, length , mass .
  • Segment 2: Vertical, length , mass .
  • Segment 3: Horizontal, length , mass .

  • For a system of discrete extended bodies, we can treat each uniform segment as a point mass located at its own geometric center.

  • Segment 1 spans from to at height .
  • Mass
  • Center

  • Segment 2 spans vertically from to at .
  • Mass
  • Center

  • Segment 3 spans horizontally from to at .
  • Mass
  • Center

  • Combining the coordinates, the position vector is:

  • By decomposing complex continuous bodies into simpler geometric shapes, we can easily locate the center of mass using discrete summation.

The Sigma Insight: Centre of Mass

Solution Diagram

Breaking Down the Complex Shape

Imagine you are looking at a strangely shaped, asymmetric bar. At first glance, finding its center of mass might seem like a daunting task requiring complex integration. But let's take a breath and look closer.
This continuous bar can be elegantly sliced into three distinct, uniform straight segments. By doing this, we transform a complex continuous body problem into a much simpler discrete particle problem.

The Power of Point Masses

For any uniform symmetric body, like a straight rod, we know that its center of mass lies exactly at its geometric midpoint. This is a powerful realization! It means we can replace each of our three segments with a single "point mass" located at its respective center.
The master equation for the position vector of the center of mass of a discrete system is:
Let's map out our three new "point masses":
Segment 1: This is the top horizontal part. It spans from to at a height of . Its mass is given as . Its geometric center is exactly in the middle, so .
Segment 2: This is the vertical drop. It spans from down to at . Its length is , so its mass is . Its geometric center is halfway down, giving us .
Segment 3: This is the bottom horizontal part. It spans from to along the x-axis (). Its length is , so its mass is . Its geometric center is in the middle, at .

Calculating the Coordinates

Now, we simply plug these coordinates into our center of mass formula. Let's start with the X-coordinate:
Substituting our values:
Exactly in the same way, let's calculate the Y-coordinate:

The Final Position Vector

Finally, we combine our calculated X and Y coordinates to form the complete position vector of the center of mass for the entire asymmetric bar:
By breaking the complex shape into simpler parts, we easily navigated to the correct answer!

Similar Questions

JEE Main 2020, 9 Jan Shift-II
LEVELJEE Main

A rod of length has non-uniform linear mass density given by , where and are constants and . The value of for the centre of mass of the rod is at

(A)
(B)
(C)
(D)
JEE Main 2019, 8 April Shift-II
LEVELJEE Main

A uniform rectangular thin sheet of mass has length and breadth , as shown in the figure. If the shaded portion is cut-off, the coordinates of the centre of mass of the remaining portion will be

(A)
(B)
(C)
(D)
JEE Main 2021, 22 July Shift-II
LEVELJEE Main

The position of the centre of mass of a uniform semi-circular wire of radius placed in XY-plane with its centre at the origin and the line joining its ends as X-axis is given by . Then, the value of is ......... .

JEE Main 2020
LEVELJEE Main

The coordinates of centre of mass of a uniform flag shaped lamina (thin flat plate) of mass . (The coordinates of the same are shown in figure) are

(A)
(B)
(C)
(D)
JEE Main 2019, 12 April Shift-II
LEVELJEE Main

Three particles of masses , and are placed at the vertices of an equilateral triangle of side (as shown in the figure). The coordinates of the centre of mass will be

(A)
(B)
(C)
(D)
JEE Advanced (1980)
LEVELJEE Main

A circular plate of uniform thickness has a diameter of . A circular portion of diameter is removed from one edge of the plate as shown in figure. Find the position of the centre of mass of the remaining portion.

LEVELJEE Advanced

A thin rod of length is lying along the x-axis with its ends at and . Its linear density (mass/length) varies with as , where can be zero or any positive number. If the position of the centre of mass of the rod is plotted against , which of the following graphs best approximates the dependence of on ?

(A)
(B)
(C)
(D)
JEE Main 2020, 7 Jan Shift-I
LEVELJEE Main

Three point particles of masses 1.0 kg, 1.5 kg and 2.5 kg are placed at three corners of a right angle triangle of sides 4.0 cm, 3.0 cm and 5.0 cm as shown in the figure. The centre of mass of the system is at a point

(A)
2.0 cm right and 0.9 cm above 1 kg mass
(B)
0.6 cm right and 2.0 cm above 1 kg mass
(C)
1.5 cm right and 1.2 cm above 1 kg mass
(D)
0.9 cm right and 2.0 cm above 1 kg mass
JEE Main 2021, 17 March Shift-II
LEVELJEE Main

The disc of mass with uniform surface mass density is shown in the figure. The centre of mass of the quarter disc (the shaded area) is at the position , where is ............. (Round off to the nearest integer) ( is an area as shown in the figure)

JEE Main 2021, 24 Feb Shift-II
LEVELJEE Main

A circular hole of radius is cut out of a circular disc of radius as shown in figure. The centroid of the remaining circular portion with respect to point will be

(A)
(B)
(C)
(D)