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### Course: Class 11 Physics (India)>Unit 11

Lesson 1: Centre of mass

# What is center of mass?

Learn the definition of center of mass and learn how to calculate it.

# What is the center of mass?

The center of mass is a position defined relative to an object or system of objects. It is the average position of all the parts of the system, weighted according to their masses.
For simple rigid objects with uniform density, the center of mass is located at the centroid. For example, the center of mass of a uniform disc shape would be at its center. Sometimes the center of mass doesn't fall anywhere on the object. The center of mass of a ring for example is located at its center, where there isn't any material.
For more complicated shapes, we need a more general mathematical definition of the center of mass : it is the unique position at which the weighted position vectors of all the parts of a system sum up to zero.

# What is useful about the center of mass?

The interesting thing about the center of mass of an object or system is that it is the point where any uniform force on the object acts. This is useful because it makes it easy to solve mechanics problems where we have to describe the motion of oddly-shaped objects and complicated systems.
For the purposes of calculation, we can treat an oddly-shaped object as if all its mass is concentrated in a tiny object located at the center of mass. We sometimes call this imaginary object a point mass.
If we push on a rigid object at its center of mass, then the object will always move as if it is a point mass. It will not rotate about any axis, regardless of its actual shape. If the object is subjected to an unbalanced force at some other point, then it will begin rotating about the center of mass.

# How can we find the center of mass of any object or system?

In general the center of mass can be found by vector addition of the weighted position vectors which point to the center of mass of each object in a system. One quick technique which lets us avoid the use of vector arithmetic is finding the center of mass separately for components along each axis. I.e:
For object positions along the x axis:
${\mathrm{COM}}_{x}=\frac{{m}_{1}\cdot {x}_{1}+{m}_{2}\cdot {x}_{2}+{m}_{3}\cdot {x}_{3}+\dots }{{m}_{1}+{m}_{2}+{m}_{3}+\dots }$
And similarly for the y axis:
${\mathrm{COM}}_{y}=\frac{{m}_{1}\cdot {y}_{1}+{m}_{2}\cdot {y}_{2}+{m}_{3}\cdot {y}_{3}+\dots }{{m}_{1}+{m}_{2}+{m}_{3}+\dots }$
Together, these give the full coordinates $\left({\mathrm{COM}}_{x},{\mathrm{COM}}_{y}\right)$ of the center of mass of the system. For example, consider the system of three flat objects of uniform density shown in Figure 2.
The location of the center of mass in the $x$ direction is:
$\frac{1\cdot 4+1\cdot 6+2\cdot 12}{1+1+2}=8.5$
and in the $y$ direction:
$\frac{1\cdot 5+1\cdot 12+2\cdot 8.5}{1+1+2}=8.5$
Complex objects can often be represented as collections of simple shapes, each with uniform mass. We can then represent each component shape as a point mass located at the centroid. Voids within objects can even be accounted for by representing them as shapes with negative mass.
Consider the irregularly-shaped flat, uniform density object shown in Figure 3a.
We can break this object up into four rectangles and one circle as shown in figure 3b. Here we are only interested in the position of the center of mass in the relative units shown in the figure. The material has uniform density so the mass is proportional to the area. For simplicity we can represent the mass of each section in units of 'squares' as shown in the diagram.
In the $x$ direction, the center of mass is at:
$\frac{16\cdot 10+52\cdot 4+12\cdot 7.5+16\cdot 10+\left(-7.1\right)\cdot 4.5}{16+52+12+16–7.1}=6.6$
Note that the area of the circular void is $\pi \cdot {1.5}^{2}\simeq 7.1$. This is accounted for as a negative mass.
In the $y$ direction:
$\frac{16\cdot 13+52\cdot 7.5+12\cdot 7.0+16\cdot 2+\left(-7.1\right)\cdot 7.5}{16+52+12+16–7.1}=7.4$

# What is the center of gravity?

The center of gravity is the point through which the force of gravity acts on an object or system. In most mechanics problems the gravitational field is assumed to be uniform. The center of gravity is then in exactly the same position as the center of mass. The terms center of gravity and center of mass tend to often be used interchangeably since they are often at the same location.

# What about determining the center of mass for a real object?

There are a couple of useful experimental tests that can be done to determine the center of mass of rigid physical objects.
The table edge method (Figure 4) can be used to find the center of mass of small rigid objects with at least one flat side. The object is pushed slowly without rotating along the surface of a table towards an edge. At the point where the object is just about to fall, a line is drawn parallel to the table edge. The procedure is repeated with the object rotated 90°. The intersection point of the two lines gives the location of the center of mass in the plane of the table.
The plumb line method (Figure 5) is also useful for objects which can be suspended freely about a point of rotation. An irregularly shaped piece of cardboard suspended on a pin-board is a good example of this. The cardboard pivots freely around the pin under gravity and reaches a stable point. A plumb line is hung from the pin and used to mark a line on the object. The pin is moved to another location and the procedure repeated. The center of mass then lies beneath the intersection point of the two lines.

# Center of mass and toppling stability

One useful application of the center of mass is determining the maximum angle that an object can be tilted before it will topple over.
Figure 6a shows a cross section of a truck. The truck has been poorly loaded with many heavy items loaded on the left-hand side. The center of mass is shown as a red dot. A red line extends down from the center of mass, representing the force of gravity. Gravity acts on all the weight of the truck through this line.
If the truck is tipped at an angle ${\theta }_{\mathrm{t}}$ (as shown in figure 6b) then all the weight of the truck will be supported by the left-most edge of the left wheel. Should the angle be further increased then the point of support will move outside of any point of contact with the road and the truck is guaranteed to topple over. The angle ${\theta }_{\mathrm{t}}$ is the topple limit.
Exercise 1: Determine the topple limit for the uniform density object shown in figure 7 as it is tipped to the right.

# Center of mass reference frame

When the term reference frame is used in physics it refers to the coordinate system used for calculations. A reference frame has a set of axes and an origin (zero point). In most problems the reference frame is fixed relative to the laboratory and a convenient (but arbitrary) origin point is chosen. This is known as the laboratory reference frame. However, in classical physics it is also possible to use any other reference frame and expect the laws of physics to hold within it. This includes reference frames which are moving relative to the laboratory.
One very useful property of the center of mass is that it can be used to define the origin of a moving reference frame for a system. This reference frame is sometimes called the COM frame. The COM frame is particularly useful in collision problems. It turns out that the momentum of a fully-defined system measured in the COM frame is always zero. This means that calculations can often be much simpler when done in the COM frame compared to the laboratory reference frame. Let us consider a simple example:
Consider two trolleys running along a track in the same direction as shown in Figure 9. The left trolley is traveling faster so there will inevitably be a collision. Let's assume the collision is elastic. What are the velocities after the collision?

## Want to join the conversation?

• Why is it that we define centre of mass as the sum of masses times their perpendicular distances from a reference point divided by the total mass of the body?.. How exactly does this give us the location of the centre of mass..?
(18 votes)
• It's defined that way because the point you get from this calculation has a lot of interesting properties, like:
1) If you hang the body somehowand let it in equilibrium, a line parallel to gravity will connect the suspension point and the center of mass.
2) If you apply a force on the center of mass, the body will feel no tendency to spin, just translate.
If you don't apply a force on the center of mass, the rotation axis in the body will cross the center of mass.
3) If a body is both translating and spinning due to previous action of a single force, you can find the speed of each of its points by adding vectorialy the center of mass' velocity and the body's rotation relative to the center of mass.
4)It can greatly simplify calculations of elastic collisions
These are characteristics satisfied by the point defined as you said. Since they are very special indeed, the point was given a name and is often studied.
(18 votes)
• i'm not understanding The COM reference frame and the Toppling Stability. Can someone help me?
(8 votes)
• For toppling stability (let's take the Truck example):
First, it'd be important to take note of its coordinate reference.
There are two important values here: The x-coordinate of the center of mass, and the x-coordinate of the part of the object that touches the ground; it has to be the most extreme point at the direction of 'toppling'.

For example, check this image: https://ka-perseus-images.s3.amazonaws.com/fd327f9a43e55a7a26319969805b9cce7fb51d89.svg

Notice that for (b), Fg directly intersects with the second important point I was referring to (the extreme point), Because at that angle (and at angles less that it), the wheel is still able to support the Truck.

Notice what happens when you increase the angle just a little bit.. Gf and that part won't intersect anymore and Fg will go beyond the wheel.
What that means is that there will be nothing to support the load of the truck and will eventually topple over.
(8 votes)
• If we apply a certain force on the COM then the object moves in the direction of the applied force . But , what if the COM is not located on the object ?
Does it mean that we can never make the object move in the direction of the applied force ?
(5 votes)
• If you apply a force off-center from the COM, then some of that force will create a torque on the object causing the object to have both rotational and translational motion. To determine how much force goes into rotational vs translational motion, you need to break up the applied force vector into a component that goes straight through the COM and a component perpendicular to the COM. The force component going through the COM will cause the translational motion and the component perpendicular to the COM will cause the rotational motion.
(11 votes)
• IN GENERAL OTHER THAN EQUATIONS can any one give a simple example when center of mass is same as center of gravity and when not ... and i might be a wrong thinking but to sure?? what i was think that for regular bodies like ruler etc it have both centre of mass and center of gravity SAME and for irregular body there might be Different points for both is this right or wrong
(4 votes)
• One way to look at this is that the center of mass is a point where the amount mass of an object is equal in all directions where the center of gravity is a point where the amount weight of an object is equal in all directions.

So if the gravitational field that the object is in can be considered uniform across the volume of the object then the center of mass and center of gravity are the same point, the shape of the object doesn't matter.
(9 votes)
• Is the entire mass of a system located at its center of mass ?
(1 vote)
• no

but it is sometimes AS IF it is at the centre.

for example our world has mass in the seas and mountains and all around you. but it pulls you towards the centre of its mass

ok??
(9 votes)
• I didn't understand what exactly is a weighted position vector. Also could someone please explain what was the meaning of the expression - summation of m(i)r(i)=0? Also why are we dividing by mass in the next expression (when origin is a point other than the COM)?

Do the initial and final velocities and momenta always swap signs in case of elastic collisions?
(4 votes)
• Is it impossible to balance a donut from one point? (Its center of mass is at the emptied part of it.)
(2 votes)
• If the circle of the donut is vertical rather than horizontal, it can be balanced from any point on the circumference. If the donut is oriented horizontally, no, it cannot be balanced from a single point.
(5 votes)
• For the Figure 8 problem, why is it tan inverse of (0.7/7.6)? According to SOHCAHTOA, is it not supposed to be OPPOSITE over ADJACENT? Therefore, the side opposite of alpha is 7.6 and the side adjacent to alpha is 0.7?
(2 votes)
• You are correct about alpha, but note that they are finding theta, not alpha. So while it's true that alpha = tan⁻¹(7.6/0.7) = 84.74° (the big angle), they want theta = tan⁻¹(0.7/7.6) = 5.26° (the small angle). To be honest, I'm not sure why they labeled that angle alpha, when they didn't ever want to solve for it.
(5 votes)
• can center of mass depend on gravity
(2 votes)
• No, center of mass is independent of gravity. Center of gravity can be different from center of mass if an object is not in a uniform gravitational field.
(4 votes)
• hello,
I have a problem with finding the centre of mass of a cycloide :
x = r(theta -sin(theta))
y = r(1-cos(theta))
thanks in advance
(3 votes)