### Video Transcript

A uniform rod π΄π΅πΆ of length 46
centimeters was bent at its midpoint π΅ then suspended freely from π΄. Given that π΅πΆ is horizontal when
the rod is hanging in its equilibrium position, determine the distance between the
center of gravity of the rod and π΄.

In this question, we are asked to
find the distance between the center of gravity of an object and a certain point
without being told what the precise shape of the object is. We are told that the uniform rod is
bent at its midpoint, but we are not given the angle of the bend. The clues we are given are that the
rod is suspended freely from π΄ and that when the rod is hanging in its equilibrium
position, π΅πΆ is horizontal. The rodβs given length of 46
centimeters is somewhat trivial since the problem procedure would be the same
regardless of the length of the rod. So, letβs call the length πΏ to
simplify the algebra.

Letβs draw a rough diagram of the
scenario. We have a uniform rod, π΄, π΅, and
πΆ, which is bent at its midpoint π΅. When suspended freely from π΄, its
equilibrium position is such that π΅πΆ is a horizontal line. Since π΅ is the midpoint of the
rod, π΄π΅ and π΅πΆ are both the same length, half the total length of the rod, πΏ
over two. Since π΅πΆ is horizontal, the
vertical line dropped from π΄ intersects it at a right angle.

We can think of this bent rod as
two separate rods, π΄π΅ and π΅πΆ. Recall that the center of mass of a
uniform rod is at its midpoint. So if we have a uniform rod of
length πΏ over two, the center of mass is πΏ over four from either end. Going back to our diagram then, the
centers of mass of the rods π΄π΅ and π΅πΆ are πΏ over four from either of their
ends. The center of mass of the complete
rod, which we will call π, will be at the midpoint of the line between the two
center of masses of the two separate rods. When hanging in its equilibrium
position, the center of mass will also be directly below the suspension point
π΄. π therefore lies at the
intersection of the line between the two midpoints of the two separate rods and the
vertical line dropped from π΄.

The distance we need to find is the
distance between π΄ and π. There are a few ways we can go
about this. The ultimate aim is to find a right
triangle for which π΄π is one of its sides and we know the other two sides from
which we can use the Pythagorean theorem to find the length of π΄π. However, currently there are no
right triangles in this diagram whose sides we know. We therefore need to construct one
somehow.

One method is to define a new line
between π΄ and πΆ. We now have an isosceles triangle
with two identical sides of πΏ over two. If we draw a new line from π΅
through the point π, it intersects the line π΄πΆ at its midpoint πΈ, splitting the
isosceles triangle in two.

We do not yet know the length of
the side π΄πΆ. Letβs call it two π. So, the length of π΄πΈ is π, and
the length of πΆπΈ is also π. Since π is the midpoint of the two
centers of mass of the two separate rods, it is also the midpoint of π΅πΈ. Therefore, the lines π΅π and πΈπ
also have the same length, which weβll call β. Looking at the diagram again, we
now have a right triangle π΄πΈπ, where π΄π is one of the sides and the other two
sides are π and β. If we can find the lengths of π
and β, then we can find the length of π΄π.

Letβs call what we need to find the
length of π΄π π. Our end goal is to find π using
the Pythagorean theorem with the values of π and β. π is equal to the square root of
π squared plus β squared. We therefore first need to find the
values of π and β in terms of the length of the rod πΏ. Consider the triangle π΄π΅πΈ. It is a right triangle with a
hypotenuse of πΏ over two, and the other two sides are π and two β.

Using the Pythagorean theorem then,
π squared plus two β all squared is equal to πΏ over two all squared. Simplifying and rearranging for π
squared gives π squared equals πΏ squared over four minus four β squared. We can therefore substitute this
expression of π squared into our final equation so that π is equal to the square
root of πΏ squared over four minus three β squared.

We therefore need only find the
value of β in terms of the length of the rod πΏ, and we will have our answer. Finding the value of β however is
not yet trivial since we do not yet have any right triangles where β is one of its
sides and the other two sides we know. Looking at the diagram again, we
have two right triangles here, each with a side β. But we do not yet know the values
of the other sides. The line π΅πΈ bisects the isosceles
triangle π΄π΅πΆ. Letβs call the midpoints of the two
separate rods π one and π two, respectively.

These two triangles π΅π one π and
π΅π two π are similar triangles, meaning that they have all the same angles. They have a right angle. And since π΅πΈ bisects the
isosceles triangle, they have the same angle here, π. Recall that similar triangles have
the property that the ratios of all of their sides are the same. So, for example, the ratio of the
hypotenuse to the second longest side is the same in both triangles. In the first purple triangle, the
length of the hypotenuse is πΏ over four, and the length of the adjacent side is
β.

In the second green triangle, the
length of the hypotenuse is β and the length of the adjacent side is the length of
π΅π two, which weβll call π₯. So, for the first triangle, we have
cos π equals β over πΏ over four. And for the second triangle, we
have cos π equals π₯ over β. π is the same angle for both
triangles. So, cos π has the same value. Therefore, β over πΏ over four is
equal to π₯ over β. Rearranging this equation gives us
β squared equals π₯πΏ over four. We can substitute this expression
for β squared into our final equation for π, giving us π equals the square root of
πΏ squared over four minus three π₯πΏ over four.

So, the final step is to find the
value of π₯ in terms of πΏ. Going back to our main diagram, π₯
is the length of the side of this green triangle here, which is the distance between
the point π΅ and the point where the vertical line from π΄ meets the line π΅πΆ. Letβs call this point of
intersection πΉ. Before going any further, letβs
redraw the main isosceles triangle. Letβs rotate the triangle so that
π΄πΆ is now horizontal and π΅ is at the top. The point π is halfway between π΄
and πΆ and halfway up the triangle. The line from π΄, which passes
through the point π, meets the line π΅πΆ at the point πΉ.

What we want to find is the length
π₯ of the line π΅πΉ. This is equivalent to finding how
far along the line π΅πΆ the point πΉ is. Consider completing the rectangle
of this triangle by drawing a vertical line upwards from π΄ and πΆ to the altitude
of the triangle and then a horizontal line between their two endpoints. The point π is now the centroid of
this rectangle since it was halfway up the triangle and halfway along the line
π΄πΆ. If we extend the line π΄πΉ, it
therefore meets the rectangle at its corner. This creates two new triangles, one
and two. Since they are formed by the
intersection of two straight lines and two horizontal lines, these two triangles are
similar.

Recall from the original diagram
that the length of π΄πΆ is two π. Since the point π΅ is at the
midpoint of the top side of the rectangle, this length here is π. The line π΅πΆ meets the horizontal
line containing π at halfway to this midpoint. The length of this side of triangle
two is therefore π over two. Letβs now take a closer look at
these two triangles.

These two triangles are formed by
the intersection of two straight lines and two horizontal lines of length π and π
over two. These two triangles are therefore
similar and share a Z angle here of π. The ratio of the larger triangleβs
top side π to its adjacent side π₯ is therefore equal to the ratio of the smaller
triangleβs bottom side π over two to its adjacent side. Letβs call it π¦. So, we have π over π₯ equals π
over two all over π¦. The πβs on both sides will
cancel. And rearranging gives us π₯ is
equal to two π¦.

Recall from the original diagram
that the length of the side π΅πΆ is equal to πΏ over two. The two sides of these triangles,
π₯ and π¦, therefore add together to give half the length of this side, πΏ over
four. Since π₯ is equal to two π¦, π¦ is
equal to π₯ over two. So, we have π₯ plus π₯ over two
equals πΏ over four. Rearranging for π₯ will give us π₯
equals πΏ over six. We now have π₯ exclusively in terms
of πΏ and therefore everything we need to find π.

Substituting in the expression π₯
equals πΏ over six, we get π equals the square root of πΏ squared over four minus
three πΏ squared over 24. Placing over the common denominator
of 24 gives us the square root of six πΏ squared minus three πΏ squared over 24. Simplifying gives us πΏ root three
over root 24. Simplifying further gives us πΏ
root two over four. Substituting the original length of
the rod, 46 centimeters, gives us 46 root two over four. And simplifying once more gives us
the final answer. The distance between the center of
gravity of the rod and π΄ π equals 23 root two over two centimeters.