Centre of Mass Notes | EduRev

Mechanics & General Properties of Matter

IIT JAM : Centre of Mass Notes | EduRev

The document Centre of Mass Notes | EduRev is a part of the IIT JAM Course Mechanics & General Properties of Matter.
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Centre of Mass

Consider a system composed of N particles with each particle's mass described by mi, where i is an index from i = 1 to i = N . The total mass of the system is denoted by M,
Centre of Mass Notes | EduRev
where the summation over i runs from i = 1 to i = N.
Such a system is displayed in figure given. If the vector connecting the origin with the ith particle is ri then the vector defining the position of the system's center of mass is Centre of Mass Notes | EduRev
Centre of Mass Notes | EduRevThe position of the centre of mass is defined by RCMCentre of Mass Notes | EduRev
For N number of mass elements, if ris the position of the jth element, and mj is its mass, then the center of mass is defined as Centre of Mass Notes | EduRev


Example 1: The position vector of three particles of mass m= 1kg, m2 = 2 kg and m3 = 3kg are Centre of Mass Notes | EduRev 

respectively. Find the position vector of their centre of mass.

The position vector of COM of the three particles will be given by

Centre of Mass Notes | EduRev
Substituting the values, we get
Centre of Mass Notes | EduRev


Example 2: Four particles of mass 1 kg, 2kg, 3kg and 4kg are placed at the four vertices A,B, C and D of a square of side 1 m. Find the position of centre of mass of the particles.

Assuming D as the origin, DC as x - axis and DA as y - axis, we have
Centre of Mass Notes | EduRevm1 = 1kg, x1, y1) = ( 0,1 m)
m2 = 2kg, x2, y2) = (1m,1m)
m3 = 3kg, x3, y3) = (1 m, 0)
m4 = 4kg, x4, y4) = (0, 0)
and co-ordinates of their COM are
Centre of Mass Notes | EduRev
Similarly,
Centre of Mass Notes | EduRev
∴ (xCOMyCOM = (0.5,0.3)

Rigid Body Dynamics

A rigid body is defined as system of particles in which the distance between any two particles remains fixed throughout the motion.

Degree of Freedom of Rigid Body

To define rigid body, there must be minimum 3 non-collinear points.
Let P1 (x1, y1, z1), P2 (x2, y2 , z2) , P3 (x3, y3, z3) are three non-collinear points.
So, equation of constrained is
Centre of Mass Notes | EduRevr12 = (x1 - x2)2 + (y1 - y2)2 + (z1 - z2)2 = c1
r23 = (x2 - x3)2 + (y2 - y3)2 + (z2 - z3)2 = c2
r13 = (x1 - x3)2 + (y1 - y3)2 + (z1 - z3)2 = c3
Dof = 3N - K
3 x 3 - 3 = 6
So there is six degree of freedom for rigid body

Center of Mass of Continuous System

The result is not rigorous, since the mass elements are not true particles. However, in the limit where N approaches infinity, the size of each element approaches zero and the approximation becomes exact.
Centre of Mass Notes | EduRevThis limiting process defines an integral. 

FormallyCentre of Mass Notes | EduRevwhere dm is a differential mass element. Then,
Centre of Mass Notes | EduRev
To visualize this integral, think of dm as the mass in an element of volume dV located at position r. If the mass density at the element is ρ, then dm = ρdV and Centre of Mass Notes | EduRev. This 

integral is called a volume integral.

Example 3: A rod of length L is placed along the x - axis between x = 0 and x = L. The linear density (mass/length) λ of the rod varies with the distance x from the origin as λ = Rx . Here, R is a positive constant. Find the position of centre of mass of this rod.

Mass of element dx situated at x = x is
dm = λdx = Rx dx
The COM of the element has coordinates (x,0,0).
Therefore, x -coordinate of COM of the rod will be

Centre of Mass Notes | EduRev
Centre of Mass Notes | EduRev
The y -coordinate of COM of the rod is yCOM Centre of Mass Notes | EduRev(as y = 0 )
Similarly, zCOM = 0 Hence, the centre of mass of the rod lies at Centre of Mass Notes | EduRev


Example 4: Find the center of mass of a thin rectangular plate with sides of length α and b, whose mass per unit area σ varies in the following fashion:
σ = σ0(xy/αb) where σ0 is a constant.

Centre of Mass Notes | EduRev

We find M , the mass of the plate, as follows:
Centre of Mass Notes | EduRev
Centre of Mass Notes | EduRevWe first integrate over x , treating y as a constant.
Centre of Mass Notes | EduRev
The x component of R is
Centre of Mass Notes | EduRevCentre of Mass Notes | EduRev

Centre of Mass Notes | EduRev
Similarly,Centre of Mass Notes | EduRev
Centre of Mass Notes | EduRev
⇒ y = 2/3 b So center of mass is Centre of Mass Notes | EduRev

Example 5 : Find the centre of mass of semicircular disc of mass M and radius R.

From the circular  symmetry x = r cos θ, y = r sin θ

Centre of Mass Notes | EduRev
Centre of Mass Notes | EduRevCentre of Mass Notes | EduRev


Example 6: Find the centre of mass of a hollow hemisphere of radius R and mass M.

Consider a ring at angle θ from the base area of ring This problem can be solved in spherical symmetry
Centre of Mass Notes | EduRev

x = R sinθ cos∅ , y = R sinθ sin∅ , z = R cos θ

Centre of Mass Notes | EduRev
XC.M = ∫xdm
Centre of Mass Notes | EduRevR sin θ cos ∅.sin θdθd∅Centre of Mass Notes | EduRev
Centre of Mass Notes | EduRev

Second method
The integral is over three dimensions, but the symmetry of the situations lets us treat it as a one dimensional integral. We mentally subdivide the hemisphere into a pile of thin disks. Consider the circular disk of radius r and thickness dz . Its volume is dV = πr2dz , and its mass is
dM = ρdV= (M/V) dV, where V = 2/3 πR3.

Centre of Mass Notes | EduRev
Hence, Centre of Mass Notes | EduRev

To evaluate the integral, we need to find r in terms of z. Since, r2 = R- z2, we have
Centre of Mass Notes | EduRevCentre of Mass Notes | EduRev

Example 7: Find the centre of mass triangular plate of Mass M .  Where l and h are given parameter shown in figure.
Centre of Mass Notes | EduRev 

Centre of Mass Notes | EduRev
Centre of Mass Notes | EduRevCentre of Mass Notes | EduRev
lf some mass M1 is removed from a rigid body of mass M , then the position of centre of mass of the remaining portion is obtained from the following formulae:
Centre of Mass Notes | EduRev

Here, Centre of Mass Notes | EduRevthe values mass and center of mass of  the whole mass while areCentre of Mass Notes | EduRevthe values for the mass which has been removed.


Example 8: Find the position of centre of mass of the remaining plate when a circle is cut from right half radius of circle is l/8. Centre of circle is in centre of the right part of the plate.
Centre of Mass Notes | EduRev

Area Centre of Mass Notes | EduRev 
A1 l2/4 l/4 l/2
A2 l2/4 3l/8 l/2
A3 πl2/64 3l/8 l/2
Taking origin at corner O
Centre of Mass Notes | EduRev
Centre of Mass Notes | EduRev
Centre of Mass Notes | EduRev
Centre of mass is (0.28l, 0.5l).

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