Video Transcript
A uniform rod 𝐴𝐵𝐶 of length 46
centimeters was bent at its midpoint 𝐵 then suspended freely from 𝐴. Given that 𝐵𝐶 is horizontal when
the rod is hanging in its equilibrium position, determine the distance between the
center of gravity of the rod and 𝐴.
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 𝐴 and that when the rod is hanging in its equilibrium
position, 𝐵𝐶 is horizontal. The rod’s given length of 46
centimeters is somewhat trivial since the problem procedure would be the same
regardless of the length of the rod. So, let’s call the length 𝐿 to
simplify the algebra.
Let’s draw a rough diagram of the
scenario. We have a uniform rod, 𝐴, 𝐵, and
𝐶, which is bent at its midpoint 𝐵. When suspended freely from 𝐴, its
equilibrium position is such that 𝐵𝐶 is a horizontal line. Since 𝐵 is the midpoint of the
rod, 𝐴𝐵 and 𝐵𝐶 are both the same length, half the total length of the rod, 𝐿
over two. Since 𝐵𝐶 is horizontal, the
vertical line dropped from 𝐴 intersects it at a right angle.
We can think of this bent rod as
two separate rods, 𝐴𝐵 and 𝐵𝐶. Recall that the center of mass of a
uniform rod is at its midpoint. So if we have a uniform rod of
length 𝐿 over two, the center of mass is 𝐿 over four from either end. Going back to our diagram then, the
centers of mass of the rods 𝐴𝐵 and 𝐵𝐶 are 𝐿 over four from either of their
ends. The center of mass of the complete
rod, which we will call 𝑃, 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
𝐴. 𝑃 therefore lies at the
intersection of the line between the two midpoints of the two separate rods and the
vertical line dropped from 𝐴.
The distance we need to find is the
distance between 𝐴 and 𝑃. There are a few ways we can go
about this. The ultimate aim is to find a right
triangle for which 𝐴𝑃 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 𝐴𝑃. 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 𝐴 and 𝐶. We now have an isosceles triangle
with two identical sides of 𝐿 over two. If we draw a new line from 𝐵
through the point 𝑃, it intersects the line 𝐴𝐶 at its midpoint 𝐸, splitting the
isosceles triangle in two.
We do not yet know the length of
the side 𝐴𝐶. Let’s call it two 𝑏. So, the length of 𝐴𝐸 is 𝑏, and
the length of 𝐶𝐸 is also 𝑏. Since 𝑃 is the midpoint of the two
centers of mass of the two separate rods, it is also the midpoint of 𝐵𝐸. Therefore, the lines 𝐵𝑃 and 𝐸𝑃
also have the same length, which we’ll call ℎ. Looking at the diagram again, we
now have a right triangle 𝐴𝐸𝑃, where 𝐴𝑃 is one of the sides and the other two
sides are 𝑏 and ℎ. If we can find the lengths of 𝑏
and ℎ, then we can find the length of 𝐴𝑃.
Let’s call what we need to find the
length of 𝐴𝑃 𝑑. Our end goal is to find 𝑑 using
the Pythagorean theorem with the values of 𝑏 and ℎ. 𝑑 is equal to the square root of
𝑏 squared plus ℎ squared. We therefore first need to find the
values of 𝑏 and ℎ in terms of the length of the rod 𝐿. Consider the triangle 𝐴𝐵𝐸. It is a right triangle with a
hypotenuse of 𝐿 over two, and the other two sides are 𝑏 and two ℎ.
Using the Pythagorean theorem then,
𝑏 squared plus two ℎ all squared is equal to 𝐿 over two all squared. Simplifying and rearranging for 𝑏
squared gives 𝑏 squared equals 𝐿 squared over four minus four ℎ squared. We can therefore substitute this
expression of 𝑏 squared into our final equation so that 𝑑 is equal to the square
root of 𝐿 squared over four minus three ℎ squared.
We therefore need only find the
value of ℎ in terms of the length of the rod 𝐿, and we will have our answer. Finding the value of ℎ however is
not yet trivial since we do not yet have any right triangles where ℎ 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 ℎ. But we do not yet know the values
of the other sides. The line 𝐵𝐸 bisects the isosceles
triangle 𝐴𝐵𝐶. Let’s call the midpoints of the two
separate rods 𝑀 one and 𝑀 two, respectively.
These two triangles 𝐵𝑀 one 𝑃 and
𝐵𝑀 two 𝑃 are similar triangles, meaning that they have all the same angles. They have a right angle. And since 𝐵𝐸 bisects the
isosceles triangle, they have the same angle here, 𝜃. 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 𝐿 over four, and the length of the adjacent side is
ℎ.
In the second green triangle, the
length of the hypotenuse is ℎ and the length of the adjacent side is the length of
𝐵𝑀 two, which we’ll call 𝑥. So, for the first triangle, we have
cos 𝜃 equals ℎ over 𝐿 over four. And for the second triangle, we
have cos 𝜃 equals 𝑥 over ℎ. 𝜃 is the same angle for both
triangles. So, cos 𝜃 has the same value. Therefore, ℎ over 𝐿 over four is
equal to 𝑥 over ℎ. Rearranging this equation gives us
ℎ squared equals 𝑥𝐿 over four. We can substitute this expression
for ℎ squared into our final equation for 𝑑, giving us 𝑑 equals the square root of
𝐿 squared over four minus three 𝑥𝐿 over four.
So, the final step is to find the
value of 𝑥 in terms of 𝐿. Going back to our main diagram, 𝑥
is the length of the side of this green triangle here, which is the distance between
the point 𝐵 and the point where the vertical line from 𝐴 meets the line 𝐵𝐶. Let’s call this point of
intersection 𝐹. Before going any further, let’s
redraw the main isosceles triangle. Let’s rotate the triangle so that
𝐴𝐶 is now horizontal and 𝐵 is at the top. The point 𝑃 is halfway between 𝐴
and 𝐶 and halfway up the triangle. The line from 𝐴, which passes
through the point 𝑃, meets the line 𝐵𝐶 at the point 𝐹.
What we want to find is the length
𝑥 of the line 𝐵𝐹. This is equivalent to finding how
far along the line 𝐵𝐶 the point 𝐹 is. Consider completing the rectangle
of this triangle by drawing a vertical line upwards from 𝐴 and 𝐶 to the altitude
of the triangle and then a horizontal line between their two endpoints. The point 𝑃 is now the centroid of
this rectangle since it was halfway up the triangle and halfway along the line
𝐴𝐶. If we extend the line 𝐴𝐹, 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 𝐴𝐶 is two 𝑏. Since the point 𝐵 is at the
midpoint of the top side of the rectangle, this length here is 𝑏. The line 𝐵𝐶 meets the horizontal
line containing 𝑃 at halfway to this midpoint. The length of this side of triangle
two is therefore 𝑏 over two. Let’s 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 𝑏 and 𝑏
over two. These two triangles are therefore
similar and share a Z angle here of 𝜃. The ratio of the larger triangle’s
top side 𝑏 to its adjacent side 𝑥 is therefore equal to the ratio of the smaller
triangle’s bottom side 𝑏 over two to its adjacent side. Let’s call it 𝑦. So, we have 𝑏 over 𝑥 equals 𝑏
over two all over 𝑦. The 𝑏’s on both sides will
cancel. And rearranging gives us 𝑥 is
equal to two 𝑦.
Recall from the original diagram
that the length of the side 𝐵𝐶 is equal to 𝐿 over two. The two sides of these triangles,
𝑥 and 𝑦, therefore add together to give half the length of this side, 𝐿 over
four. Since 𝑥 is equal to two 𝑦, 𝑦 is
equal to 𝑥 over two. So, we have 𝑥 plus 𝑥 over two
equals 𝐿 over four. Rearranging for 𝑥 will give us 𝑥
equals 𝐿 over six. We now have 𝑥 exclusively in terms
of 𝐿 and therefore everything we need to find 𝑑.
Substituting in the expression 𝑥
equals 𝐿 over six, we get 𝑑 equals the square root of 𝐿 squared over four minus
three 𝐿 squared over 24. Placing over the common denominator
of 24 gives us the square root of six 𝐿 squared minus three 𝐿 squared over 24. Simplifying gives us 𝐿 root three
over root 24. Simplifying further gives us 𝐿
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 𝐴 𝑑 equals 23 root two over two centimeters.
