Video Transcript
Four particles of masses nine
kilograms, 10 kilograms, four kilograms, and seven kilograms are placed on the
𝑥-axis at the points four, zero; three, zero; eight, zero; and one, zero,
respectively. What is the position of the center
of mass of the four particles?
In this question, we have four
particles which all lie on the 𝑥-axis. We are told that the first mass of
nine kilograms is at the point four, zero. The second of 10 kilograms is at
three, zero. And the third and fourth masses of
four kilograms and seven kilograms are at eight, zero and one, zero,
respectively. In order to calculate the center of
mass of these four masses, we could use the formula which will produce the position
vector of the center of mass. This position vector, which we can
denote as vector 𝐑, is found by calculating the mass times the position vector of
every mass, adding them, and then multiplying by one over 𝑀, where 𝑀 is the total
mass of all the particles.
In the context of this question, we
will take each mass, multiply it by its position vector, add them up, and then
divide by the total of the four masses. We might notice that we are given
coordinates instead of position vectors. So the first thing we’ll do is
change each coordinate into a position vector. So the coordinate one, zero is one
𝐢 plus zero 𝐣. The coordinate three, zero is three
𝐢 plus zero 𝐣. Four, zero is four 𝐢 plus zero
𝐣. And eight, zero is eight 𝐢 plus
zero 𝐣. We can also work out the total
mass. It will be nine plus 10 plus four
plus seven, which is 30 kilograms.
Now let’s substitute these values
into the formula. The mass of each object multiplied
by its position vector can be added in any order. Here we’ve written the order of
seven, 10, nine, and four kilograms. However, the order given in the
question of nine, 10, four, and seven multiplied by each of their position vectors
could also be used. So we would have that the position
vector 𝐑 is equal to one over 30 times seven times one 𝐢 plus zero 𝐣 plus 10
times three 𝐢 plus zero 𝐣 plus nine times four 𝐢 plus zero 𝐣 plus four times
eight 𝐢 plus zero 𝐣. Of course, we don’t need to include
all of these zero 𝐣’s, but it’s useful to see how this formula can be applied in a
different context when the masses aren’t on the 𝑥-axis.
But the next step here would be to
simplify the masses times their position vectors. When we do this, the right-hand
side of our equation simplifies to one over 30 times seven 𝐢 plus 30𝐢 plus 36𝐢
plus 32𝐢. This is equivalent to one over 30
times 105𝐢. Dividing 105𝐢 by 30 gives us seven
over two 𝐢, or 3.5𝐢. In this question, we were asked for
the position of the center of mass, and what we have here is a position vector. As a coordinate then, we can write
the position vector of 3.5𝐢 as the coordinate 3.5, zero because we know that 3.5𝐢
is the same as 3.5𝐢 plus zero 𝐣. Therefore, the coordinates of the
center of mass of these four masses at their given positions is 3.5, zero.
