Video Transcript
Center of Gravity of Particles
In this video, we will learn how to
find the position of the center of gravity of a set of particles arranged in a
two-dimensional plane.
We know that the weight of a
particle on Earth acts from its center directly downwards towards the center of the
Earth. This makes calculations with
particles or individual point masses very easy to work with. But what happens when we have a
collection or system of particles and we want to consider all of these particles in
one go? This is where the center of gravity
comes in handy. The center of gravity is
essentially the point in space, which is the weighted average of all the particles
in the system.
Let’s now consider two particles in
the 𝑥𝑦-plane.
Here, we have our two
particles. Let’s suppose that the weight of
the particle on the left is 𝑤 one and that the weight of the particle on the right
is 𝑤 two. We can also add in the position
vectors of 𝑤 one and 𝑤 two. We can label the position vector of
𝑤 one as 𝐫 one, and the position vector of 𝑤 two is 𝐫 two. Now, the position of the center of
gravity of the system containing these two weights will lie somewhere along the line
joining center of the two particles. Now, if the weight of the two
particles are equal — that is, that 𝑤 one is equal to 𝑤 two — then the center of
gravity will be directly in the middle of the line segment joining the center of the
two particles. However, in the cases where the two
weights are not equal, the center of gravity will shift along this line towards the
larger weight.
If 𝑤 one is larger than 𝑤 two,
then the center of gravity will shift along this line towards 𝑤 one. Similarly, if 𝑤 two is larger than
𝑤 one, then the center of gravity will shift along this line towards 𝑤 two. In order to actually find the
position of this center of gravity, we′ll need to use a formula. This formula tells us that the
position of the center of gravity 𝐑 is equal to 𝑤 one 𝐫 one plus 𝑤 two 𝐫 two
over 𝑤 one plus 𝑤 two. Now, in most systems where we’re
trying to find the position of the center of gravity, we′ll be considering more than
two particles. Hence, we will need to expand this
formula in order to cater for any number of particles.
Here we have the definition for the
center of gravity of a system of particles in two dimensions. Suppose we have a system of 𝑛
particles with weights of 𝑤 one, 𝑤 two, 𝑤 three, and so on up to 𝑤 𝑛, which are
located at the position vectors 𝐫 one, 𝐫 two, 𝐫 three, and so on up to 𝐫 𝑛,
respectively. Then, the center of gravity of the
system can be found using the formula 𝐫 is equal to 𝑤 one 𝐫 one plus 𝑤 two 𝐫
two plus 𝑤 three 𝐫 three and so on all the way up to 𝑤 𝑛 𝐫 𝑛 over 𝑤 one plus
𝑤 two plus 𝑤 three plus all the way up to 𝑤 𝑛.
Using this formula, we′re able to
find the center of gravity of a system of particles in two dimensions when we’re
given the weights and respective position vectors of each of the particles. We can also rewrite this formula
using summation notation. We can see that these two formulas
are identical. However, the one using summation
notation is a lot more compact. We have that 𝐑 is equal to the sum
from 𝑖 equals one to 𝑛 of 𝑤 𝑖 𝐫 𝑖 over the sum from 𝑖 equals one to 𝑛 of 𝑤
𝑖.
Next, we will be rearranging this
formula slightly so we can see how we can find the center of gravity of a system
when we’re given the masses of the particles instead of the weight.
We know that the weight of any
particle is equal to its mass 𝑚 multiplied by the acceleration due to gravity
𝑔. Now, in a uniform gravitational
field, 𝑔 is a constant. On Earth, 𝑔 is roughly equal to
9.8 meters per second squared.
Using this, we can rewrite 𝑤 one
as 𝑚 one 𝑔, 𝑤 two as 𝑚 two 𝑔, all the way up to 𝑤 𝑛 as 𝑚 𝑛 𝑔. Now, if we’re in a uniform
gravitational field, the value of 𝑔 in each of these weights will be the same. Therefore, we can say that for any
of our weights, 𝑤 𝑖 is equal to 𝑚 𝑖 𝑔. We can substitute this equation for
𝑤 𝑖 into our formula for finding the center of gravity. We have that 𝐑 is equal to the sum
from 𝑖 equals one to 𝑛 of 𝑚 𝑖 𝑔𝐫 𝑖 over the sum from 𝑖 equals one to 𝑛 of
𝑚 𝑖 𝑔. Since 𝑔 is just a constant in both
the numerator and denominator, we can take it out to the front in both
summations. Next, we can cancel this factor of
𝑔. Here, we have found another formula
for finding the center of gravity of a system of particles. It tells us that 𝐑 is equal to the
sum from 𝑖 equals one to 𝑛 of 𝑚 𝑖 𝐫 𝑖 over the sum from 𝑖 equals one to 𝑛 of
𝑚 𝑖.
This formula will be useful for
when we’re given the masses of the particles rather than their weight. It is worth noting that this
formula for the center of gravity, which uses masses instead of weight, is sometimes
called the center of mass of a system. However, as we can see from our
calculations, it is equivalent to the center of gravity of a system of particles
when the particles are in a uniform gravitational field.
In this video, we’ll only be
considering systems which meet this condition of a uniform gravitational field. Therefore, we′ll be able to use
these formulas interchangeably for finding the center of gravity. Now, when we’re actually finding
the center of gravity of a system, we may find it easier to break down this formula
even further. We will now consider the 𝑥- and
𝑦-coordinates of 𝐑.
Since 𝐑 is a vector in two
dimensions, we can say that 𝐑 is equal to 𝑥𝐢 plus 𝑦𝐣, where 𝐢 and 𝐣 are the
fundamental unit vectors. If we were to draw the vector 𝐑 on
a grid, then the 𝑥- and 𝑦-coordinates of the endpoint of 𝐑 can be found by
looking at the coefficients of 𝐢 and 𝐣. So that is what we have labeled 𝑥
and 𝑦 here. We can also rewrite the position
vectors for each of the particles in our system. We’ll have that the vector 𝐫 𝑖 is
equal to 𝑥 𝑖 𝐢 plus 𝑦 𝑖 𝐣.
Now, when we look back at the
formula for the center of gravity of a system, we can consider just the horizontal
or just the vertical part. Considering just the horizontal
parts of these vectors, we can say that the 𝑥-coordinate of the position of the
center of gravity is equal to the sum from 𝑖 equals one to 𝑛 of 𝑚 𝑖 𝑥 𝑖 over
the sum from 𝑖 equals one to 𝑛 of 𝑚 𝑖. Similarly, the 𝑦-coordinate is
equal to the sum from 𝑖 equals one to 𝑛 of 𝑚 𝑖 𝑦 𝑖 over the sum from 𝑖 equals
one to 𝑛 of 𝑚 𝑖.
It’s important to note that these
formulas also work when we’re given the weights of the particles instead of their
masses. We can obtain these formulas by
multiplying the numerator and denominator of the original formulas by 𝑔 and then
substituting 𝑤 𝑖 for 𝑚 𝑖 𝑔. And this will give us the second
line of formulas for finding the center of gravity when we′re given the weights of
the particles instead of their masses.
Let’s now move on to look at an
example.
In the given figure, three weights
of magnitudes two newtons, five newtons, and three newtons are placed on the
vertices of an equilateral triangle of side length eight centimeters. Find the center of gravity of the
system.
In order to solve this problem,
we’re first going to need to find the 𝑥- and the 𝑦-coordinates of each of the
weights. We have been given that the weights
lie on the vertices of an equilateral triangle. Hence, each of these side lengths
will also be eight centimeters. We can already see that the weight
of two newtons lies at the origin. So its coordinates will be zero,
zero. The weight of five newtons lies on
the 𝑦-axis and is eight centimeters from the weight of two newtons. So, its coordinates will be zero,
eight.
In order to find the position of
the weight of three newtons, we can add in this horizontal line to help us. This line will make a right angle
with the vertical axis. Also, since this is an equilateral
triangle, this line will bisect the vertical side. So we can see that this length will
be four centimeters. So this is the 𝑦-coordinate of the
weight of three newtons.
In order to find the 𝑥-coordinate,
we’re going to use the Pythagorean theorem. The Pythagorean theorem tells us
that this 𝑥-coordinate is equal to the square root of eight squared minus four
squared, which is equal to the square root of 64 minus 16, which is also equal to
the square root of 48. And this simplifies to give four
root three. So the coordinate of our weight of
three newtons is four root three four.
We can now write a table which will
show us the positions of each of the weights. We’ll also need the formulas for
finding the 𝑥- and 𝑦-coordinates of the center of gravity. Let’s start by finding the
𝑥-coordinate. In the numerator, we need to sum
the products of the weight and their 𝑥-coordinates. So that’s two multiplied by zero
plus five multiplied by zero plus three multiplied by four root three. And then in the denominator, we
simply need to sum the weights. This gives us two multiplied by
zero plus five multiplied by zero plus three multiplied by four root three over two
plus five plus three. The first two terms in the
numerator are multiplied by zero. So we can ignore these. Simplifying the rest of the
fraction, we find that 𝑥 is equal to 12 root three over 10 centimeters.
Let’s now move on to find the
𝑦-coordinate. We follow a similar process to
finding the 𝑥-coordinate, except this time in the numerator, we need to sum the
products of the weights with their 𝑦-coordinates. So that′s two multiplied by zero
plus five multiplied by eight plus three multiplied by four. Then, we need to divide this by the
sum of the weights, giving us that 𝑦 is equal to two multiplied by zero plus five
multiplied by eight plus three multiplied by four over two plus five plus three. Since the first term in the
numerator is multiplied by zero, we can again ignore this term. Simplifying the rest of the
fraction, and we′re left with 𝑦 is equal to 52 over 10 centimeters. Now that we have both the 𝑥- and
𝑦-coordinates, we found the center of gravity of the system. And it’s located at the coordinates
12 root three over 10, 52 over 10.
We’ll now move on to our second
example, where we’ll see how we can find the center of gravity of a system when
we’re given the masses of the particles instead of their weights.
Suppose three masses of one
kilogram, four kilograms, and six kilograms are located at points whose positions
are negative six 𝐢 minus 𝐣, two 𝐢 minus nine 𝐣, and seven 𝐢 plus eight 𝐣. Determine the position vector of
the center of gravity of the system of masses.
We can start by forming a table of
the three masses and each of their 𝑥- and 𝑦-coordinates of their positions. Next, we need to recall the
formulas for finding the 𝑥- and 𝑦-coordinates of the center of gravity of a system
in terms of their masses. We have that 𝑥 is equal to the sum
from 𝑖 equals one to 𝑛 of 𝑚 𝑖 𝑥 𝑖 over the sum from 𝑖 equals one to 𝑛 of 𝑚
𝑖. And 𝑦 is equal to the sum from 𝑖
equals one to 𝑛 of 𝑚 𝑖 𝑦 𝑖 over the sum from 𝑖 equals one to 𝑛 of 𝑚 𝑖.
We will start by finding the
𝑥-coordinate of the center of gravity. For the numerator, we need to sum
the products of the masses with their 𝑥-coordinates. And in the denominator, we simply
have the sum of the masses. This gives us one multiplied by
negative six plus four multiplied by two plus six multiplied by seven over one plus
four plus six. Simplifying both the numerator and
denominator, we have that 𝑥 is equal to 44 over 11. Since 44 is a multiple of 11, we
can cancel through by this factor. Hence, we have that 𝑥 is equal to
four.
Next, we’ll find the 𝑦-coordinate
of the center of gravity. Using the formula, we have that 𝑦
is equal to one multiplied by negative one plus four multiplied by negative nine
plus six multiplied by eight over one plus four plus six. Simplifying this fraction, we have
that 𝑦 is equal to 11 over 11. And of course, this simplifies to
one. So we found that the 𝑦-coordinate
of the center of gravity is equal to one.
So we’ve now found the coordinates
of the center of gravity. However, the question has asked us
to find the position vector of the center of gravity. The 𝑥-coordinate of the center of
gravity will be the coefficient of the 𝐢 in the position vector. Similarly, the 𝑦-coordinate will
be the coefficient of 𝐣. Here, we reach our solution, which
is that the position vector of the center of gravity of the system of masses is four
𝐢 plus 𝐣.
In the next example, we’ll see how
the center of gravity can be used to find a missing mass in a system of masses.
The points zero, six; zero, nine;
and zero, four on the 𝑦-axis are occupied by three solid masses of nine kilograms,
six kilograms, and 𝑀 kilograms, respectively. Determine the value of 𝑀 given the
center of gravity of the system is at the point zero, seven.
The first thing that we notice
about this question is that all of the masses along with the center of mass are
located on the 𝑦-axis. Hence, each of the 𝑥-coordinates
are equal to zero. If we were to try and perform any
calculations with the 𝑥-coordinate, we would end up with every term being
multiplied by zero. And so, this would not tell us
anything useful. Hence, we should focus on the
𝑦-coordinates here. Let’s recall the formula for
finding the 𝑦-coordinate of the center of gravity of a system of masses. We have that 𝑦 is equal to the sum
from 𝑖 equals one to 𝑛 of 𝑚 𝑖 𝑦 𝑖 over the sum from 𝑖 equals one to 𝑛 of 𝑚
𝑖. In order to make our lives a little
easier, let’s form a table consisting of the masses and their relative
𝑦-coordinates.
Now, we are ready to use the
formula. We know that the 𝑦 in our formula
is the 𝑦-coordinate of the center of gravity, which we’ve been given as seven. Using our formula, we have that
seven is equal to nine multiplied by six plus six multiplied by nine plus four
multiplied by 𝑀 over nine plus six plus 𝑀. Simplifying both the numerator and
denominator, we have that seven is equal to 108 plus four 𝑀 over 15 plus 𝑀. Next, we can multiply through on
the left and right by 15 plus 𝑀. We have that seven multiplied by 15
plus 𝑀 is equal to 108 plus four 𝑀.
Next, we can expand the parentheses
on the left-hand side, giving us 105 plus seven 𝑀 is equal to 108 plus four 𝑀. Next, we can subtract 105 from both
sides and four 𝑀 from both sides. We are left with three 𝑀 is equal
to three. Finally, we divide both sides by
three to find our solution, which is that 𝑀 is equal to one. Therefore, the mass of our final
solid is one kilogram.
For the final example of this
video, we’ll see how we can find the center of gravity of a system where the masses
are placed on the vertices of a square.
A square 𝐴𝐵𝐶𝐷 has side length
𝐿. Three masses of 610 grams are
placed at 𝐴, 𝐵, and 𝐷. Find the coordinates of the center
of gravity of the system.
In order to solve this problem,
we′ll need to use the formulas for finding the 𝑥- and 𝑦-coordinates of the center
of gravity of a system of masses. We have that the 𝑥-coordinate of
the center of gravity is equal to the sum from 𝑖 equals one to 𝑛 of 𝑚 𝑖 𝑥 𝑖
over the sum from 𝑖 equals one to 𝑛 of 𝑚 𝑖. And the 𝑦-coordinate is equal to
the sum from 𝑖 equals one to 𝑛 of 𝑚 𝑖 𝑦 𝑖 over the sum from 𝑖 equals one to
𝑛 of 𝑚 𝑖. We can now form a table to show the
masses and their corresponding coordinates. Noting that the side length of the
square is 𝐿, we have that the coordinates of the masses are zero, zero at 𝐴; 𝐿,
zero at 𝐵; and zero, 𝐿 at 𝐷.
We are now ready to substitute
these values into our formulas to find the coordinates of the center of gravity. For the 𝑥-coordinate, we have that
𝑥 is equal to 610 multiplied by zero plus 610 multiplied by 𝐿 plus 610 multiplied
by zero over 610 plus 610 plus 610. Now, the first and last term in the
numerator are multiplied by zero. So we can ignore these. Simplifying what we have left, we
have 610𝐿 over three multiplied by 610. So we can cancel this factor of
610. Here, we have found the
𝑥-coordinate, which is 𝐿 over three.
Next, we can find the 𝑦-coordinate
of the center of gravity in a similar fashion. We have that the 𝑦-coordinate is
equal to the sum of the masses multiplied by their respective 𝑦-coordinates divided
by the sum of the masses. We have that 𝑦 is equal to 610
multiplied by zero plus 610 multiplied by zero plus 610 multiplied by 𝐿 divided by
610 plus 610 plus 610. Since the first two terms in the
numerator are multiplied by zero, we can ignore them. We can then simplify this to 610𝐿
over three multiplied by 610. And we can simplify it further by
canceling the factor of 610. So now, we have found our
𝑦-coordinate of the center of gravity, which is 𝐿 over three. Combining this with our
𝑥-coordinate, and we have found the coordinates of the center of gravity of the
system. And this is at 𝐿 over three, 𝐿
over three.
We have now covered a variety of
examples. Let’s recap some key points of the
video.
Key Points
The center of gravity of a system
of particles is the average position of the particles weighted according to their
weights. We can find the center of gravity
of a system using 𝐑 is equal to the sum from 𝑖 equals one to 𝑛 of 𝑤 𝑖 𝐫 𝑖
over the sum from 𝑖 equals one to 𝑛 of 𝑤 𝑖. To find the 𝑥- and 𝑦-coordinates
of the center of gravity, we can use 𝑥 is equal to the sum from 𝑖 equals one to 𝑛
of 𝑤 𝑖 𝑥 𝑖 over the sum from 𝑖 equals one to 𝑛 of 𝑤 𝑖. And 𝑦 is equal to the sum from 𝑖
equals one to 𝑛 of 𝑤 𝑖 𝑦 𝑖 over the sum from 𝑖 equals one to 𝑛 of 𝑤 𝑖.
