After developing kinematics of rotational motion, the rotational motion dynamics were developed. In this the parallel concepts are applied to linear dynamics which contains some unique concepts and equations. The main important concept in rotational dynamics is a torque. It also involves the study of work, power, and kinetic energy like in linear motion, moment of inertia, **centre of mass** and equations for a description of the combination of linear and angular motion.

Here we are discussing about the centre of mass. The terms centre of mass or gravity are used in a constant field of gravity which shows a specific point in the system. It is used to describe a response of the system towards the external force and torque. If we see in on dimension plane then it is similar to seesaw at a pivot point. It contains the total mass of the body at one point. Let’s discuss more about the centre of mass and the process of finding it.

**This clock uses the principles of center of mass to keep balance on a finger.**

Here we are discussing about the centre of mass. The terms centre of mass or gravity are used in a constant field of gravity which shows a specific point in the system. It is used to describe a response of the system towards the external force and torque. If we see in on dimension plane then it is similar to seesaw at a pivot point. It contains the total mass of the body at one point. Let’s discuss more about the centre of mass and the process of finding it.

Related Calculators | |

Center of Mass Calculations | Mass Calculator |

Mass to Energy | Atomic Mass Calculator |

- The point can be real or imaginary, for example, in the case of a hollow or empty box the mass is physically not located at the center of mass point. This mass is supposed to be located at the center of mass in order to simplify calculations.
- The motion of the center of mass characterizes the motion of the entire object. The center of mass may or may not be the same to the geometric center if a rigid body is considered. It is considered as a reference point for many other calculations of mechanics.

- Shape of the body
- Distribution of mass in the body

Consider two particles A and B of masses m_{1} and m_{2}_{,} respectively. Take the line joining A and B as the X-axis. Let the coordinates of the particles at time 't' be x_{1} and x_{2}. Suppose no external force acts on the system. The particles A and B, however, exert forces on each other and these particles accelerate along the line joining them. Suppose the particles are initially at rest and the force between them is attractive.

The center of mass at time t is situated at X = $\frac{m_{1}x_{1}+m_{2}x_{2}}{m_{1}+m_{2}}$

As time passes, x_{1} and x_{2} change and hence, X changes and the center of mass moves along the X-axis. Velocity of the center of mass at time t is,

The acceleration of the center of mass is

Suppose the magnitude of the forces between the particles is F. As the only force acting on A towards B, is F, its acceleration is

The force on B is (-F) and hence,

Substituting this in equation (1)

This means, the velocity of the center of mass does not change with time. But as we assumed initially, the particles are at rest. Thus,$\vec{v_{1}}$ = $\vec{v_{2}}$ = 0 then V_{cm} has to be zero. Hence, the center of mass remains fixed and does not change with time.

Thus, if no external force acts on a two-particle system and its center of mass is at rest, initially it remains fixed even when the particles individually move and accelerate.

If the external forces do not add up to zero, the center of mass is accelerated and is given by,

If we have a single particle of mass m on which a force $\vec{F_{ext}}$ acts, its acceleration would be the same as $\frac{\vec{F_{ext}}}{m}$. Thus, the motion of the center of mass of a system is identical to the motion of a single particle of mass equal to the mass of given system, acted upon by the same external forces that act on the system.

**The formula for center of mass in three dimension**

where,

R represents the center of mass

'M' the mass of the body, m (i) = masses, r(i) = positions

This can be customized for x and y coordinates also. Then the equations would be,

The center of mass at time t is situated at X = $\frac{m_{1}x_{1}+m_{2}x_{2}}{m_{1}+m_{2}}$

As time passes, x

$\vec{V_{cm}}$ = $\frac{dx}{dt}$ = $\frac{m_{1}\vec{v_{1}}+m_{2}\vec{v_{2}}}{m_{1}+m_{2}}$

The acceleration of the center of mass is

$\vec{a_{cm}}$ = $\frac{d\vec{V_{cm}}}{dt}$ = $\frac{m_{1}\vec{a_{1}}+m_{2}\vec{a_{2}}}{m_{1}+m_{2}}$.........(1)

Suppose the magnitude of the forces between the particles is F. As the only force acting on A towards B, is F, its acceleration is

$a_{1}$ =$\frac{\vec{F}}{m_{1}}$

The force on B is (-F) and hence,

$a_{2}$ = $-\frac{\vec{F}}{m_{2}}$

Substituting this in equation (1)

$\vec{a_{cm}}$ = $\frac{m_{1}(\frac{\vec{F}}{m_{1}})+m_{2}(-\frac{\vec{F}}{m_{2}})}{m_{1}+m_{2}}$
= 0

This means, the velocity of the center of mass does not change with time. But as we assumed initially, the particles are at rest. Thus,$\vec{v_{1}}$ = $\vec{v_{2}}$ = 0 then V

Thus, if no external force acts on a two-particle system and its center of mass is at rest, initially it remains fixed even when the particles individually move and accelerate.

If the external forces do not add up to zero, the center of mass is accelerated and is given by,

$\vec{a_{cm}}$=$\frac{\vec{F_{ext}}}{m}$

If we have a single particle of mass m on which a force $\vec{F_{ext}}$ acts, its acceleration would be the same as $\frac{\vec{F_{ext}}}{m}$. Thus, the motion of the center of mass of a system is identical to the motion of a single particle of mass equal to the mass of given system, acted upon by the same external forces that act on the system.

R_{cm}= $\frac{\sum m(i)r{i}}{M}$

where,

R represents the center of mass

'M' the mass of the body, m (i) = masses, r(i) = positions

This can be customized for x and y coordinates also. Then the equations would be,

X_{cm} = $\frac{\sum m(i)x{i}}{M}$

Y_{cm} = $\frac{\sum m(i)y{i}}{M}$

Z_{cm}=$\frac{\sum m(i)z{i}}{M}$

Y

Z

- By using line of symmetry
- By using integration

The bodies which have a uniform density through the entire body are called uniform bodies. Now if the body has a line of symmetry then the center of mass will lie along this line.

For example: In case of a circle the center of mass is present in the center.

When the center of mass of a body cannot be found using the axis/axes of symmetry, it can be found by integration. Also, from the mathematical definition of the center of mass, it is basically the net resultant of the masses or weights of the distinct and individual particles that make up the body. So, moments at any point can be taken,

Moment of the Total Body = The Total of Moments

Hence we can say that, the distance of center of mass from P multiplied by the weight of body will be equal to the sum of the mass of individual particle multiplied by the distance of each particle from P. We can replace the sum of part with integration sign.For example: To calculate the center of mass position of a semi-circular lamina (uniform) with radius r. It is known that the center of mass lies on the semi-circle’s axis of symmetry but we are not aware where exactly on it. So, the semi-circle is divided into many small strips, each with thickness of dx.

Let us label this line with x-axis and introduce a y-axis as follows:

The mass of each strip = volume $\times$ density

The volume of each strip is its area only, as it has no depth being a lamina.

The area of each strip = d x $\times$ 2 y

We know that the body is uniform, which interprets that the density of it is K (constant). Let the density be called r. Hence the mass of each strip = $2y r d x$

Now the total area of the semicircle = $\frac{1}{2} pr^{2}$

So, the total mass is $\frac{1}{2} pr^{2}r$

(Distance of center of mass from O) × (weight of body) = $\sum$ ( the mass of each particle) × (the distance of each particle from O)

Hence center of mass (c o m) ×$\frac{1}{2} pr^{2}$ = $\sum 2y x r d x$

The submission is from x = 0 to x = r

center of mass ×$\frac{1}{2} pr^{2}$ = $\int 2y x r dx$

The integral is with respect to the term x, so, we must replace the term y by a function of x.

By using similar triangles concept, we can interpret that $\frac{y}{r}$ =$\frac{x}{r}$, so y = x

(C.O.M) × $\frac{1}{2}pr^{2}r$ = $\int 2x^{2}rdx$

(C.O.M) × $\frac{1}{2}pr^{2}r$ = $\frac{2r^{3}r}{3}$

Hence center of mass is present at a distance of $\frac{4r}{3p}$ from O.

$\frac{Total\ mass\ of\ triangle}{Total\ area\ of\ triangle}$ = $\frac{Mass\ of\ rectangle}{Area\ of\ rectangle}$

Or

$\frac{M}{A }$ = $\frac{dm}{dA}$

So $dm$ = $\frac{M}{A}$$dA$

=$\left [\frac{M}{\frac{1}{2}ab}\right ]( y dx )$

Hence, the x – coordinate for center of mass

X = $\frac{1}{M\int(xdm)}$

= $\frac{1}{M\int(x(\frac{2M}{ab})ydx)}$

Solving this we get,

X = $\frac{2}{3}$a

Similarly,

Y = $\frac{1}{3}$b

Hence coordinate C O M = ($\frac{2}{3}$a , $\frac{1}{3}$b)

where,

a = base length and

b = height

Here are some points stating the difference between the center of mass and center of gravity:

**Example**: If we consider a ring then the center of mass and center of
gravity both lie at or near the middle of the ring, which lies outside
the body of the ring. As gravity decreases according to the inverse
square law, center of gravity depends on how the ring is oriented relative to the gravity field. If the ring is in the same plane as the Earth's radius is, the center of mass lies slightly below the center of gravity. If the plane of the ring is perpendicular to the Earth's radius, the center of mass would lie above the center of mass.

- When we are talking about a uniform gravitational field then both the center of mass as well as the center of gravity will be the same point.
- But if we consider a non-uniform field it is not always true. An object can rotate due to the torque produced by a non-uniform gravitational field.

- The center of mass of a car is closer to the ground rather than in the geometric center of the car so that the car can be better balanced.
- The technique of a high jumper is another example. A high jumper bends his
body in a certain way so that the center of mass does not clear the bar,
but the body clears it.

Related Topics | |

Physics Help | Physics Tutor |