### Video Transcript

A stone that has a mass of 1.6
kilograms is swung in a vertical circle at a constant angular velocity of 6.1
radians per second. The stone is attached to a uniform
rope of length 0.33 meters, as shown in the diagram. The length of the rope is the same
as the radius of the circle throughout the motion of the stone. What is the ratio of the maximum
force to the minimum force that the rope can apply to the stone? Give your answer to one decimal
place.

In our diagram, we see this stone
being swung by a rope around in a vertical circle; that is, the stone only moves in
a plane that is entirely vertical. The first part of our question
talks about a ratio of forces on the stone. When we think about forces on the
stone, at any given moment, there are two.

First, thereβs the tension force β
weβll call it capital π β that pulls the stone in towards the center of the
circle. The second force always acting on
the stone is its weight force; itβs mass times the acceleration due to gravity. While the weight force π times π
is always the same regardless of where the stone is along its circular path, the
tension force is not like that. Rather, the tension force varies as
the stone moves throughout this arc. We could think of the tension force
as the force that enforces circular motion on this stone. It always pulls on the stone just
hard enough so that the stone does indeed follow a vertical circle.

Part one of our question talks
about a maximum and a minimum force that the rope can apply to the stone. That force is the force weβve
called capital π, the tension force. Because the tension force always
pulls the stone towards the center of its circular arc, we say that it is a
centripetal or center-seeking force. In general, the centripetal force
πΉ sub c acting on a mass π moving in a circle of radius π is equal to π times π
times the angular speed of that object squared.

Notice in our problem statement
that weβre told that our stone moves at a constant angular velocity. In other words, all throughout its
vertical path, π is the same. In addition, the mass of the stone
is constant, and so is the radius π of its circular path. This means that for any position of
the stone anywhere along this circular path, the centripetal force πΉ sub c acting
on the stone is the same.

But, and here we must be careful,
that doesnβt mean that the tension force capital π is always the same. In fact, the tension force varies
all throughout the stoneβs circular path. This happens so that the
combination of the tension force π and the weight force π times π does yield a
constant overall centripetal force.

Since our question asks about the
maximum and minimum force that the rope can apply to the stone, we want to think
about those places in the stoneβs path where π, the tension force, has a maximum
and a minimum value.

Letβs consider the moment where the
stone is at its maximum height. As always, the weight force acts
directly downward on the stone. And here we see that the weight
force is perfectly in line with the tension force. We could say that in this position
the weight force on the stone applies a significant center-seeking force. Therefore, at this moment, the
tension force π β that is the force of the rope on the stone β does not need to be
very great in order for the overall centripetal force to equal the stoneβs mass
times its radius of revolution times its angular speed squared.

In fact, since the tension force
and the weight force are perfectly aligned at this moment, when the stone is
vertically above its rotation center, the tension in the rope is at a minimum. This force that weβve called π sub
min is the minimum force that the rope can apply to the stone.

Next, letβs think about the forces
involved when the stone is at its lowest position. As always, the tension force in the
rope pulls the stone towards the center of the circle. But now the weight force π times
π on the stone points in the opposite direction as the tension force. But we recall that the total
centripetal force acting on the stone must be constant.

Here, not only must the tension
force in the rope keep the stone moving in a circular path, it must also overcome
the weight force which is opposed to the tension force. For this reason, the tension in the
rope is highest when the stone is at its lowest position. What weβve called π sub max is the
maximum force of the rope on the stone.

To start comparing π sub max and
π sub min, letβs clear a bit of space on screen. If the centripetal force acting on
the stone is πΉ sub c, weβve said that πΉ sub c must be constant all throughout the
stoneβs motion. The reason for this is neither the
stoneβs mass nor its radius of revolution nor its angular speed vary.

At any given moment, πΉ sub c
equals the sum of the tension force in the rope and the weight force on the
stone. If we consider downward-directed
forces to be forces in the positive direction, then at the stoneβs highest point the
centripetal force is equal to π sub min, the minimum tension in the rope, plus π
times π. Here, we add these two forces
together because they both point in the same direction.

If we then turn to consider the
stone at its lowest elevation, we can say that that same centripetal force is equal
to π times π β thatβs positive because itβs directed downward β minus π sub
max. What weβre finding then is that π
sub min plus ππ is equal to ππ minus π sub max. What we basically have then is two
equations, one for π sub min and one for π sub max.

In our equation for π sub min, if
we subtract π times π from both sides, then on the right-hand side, π times π
minus π times π cancels out. And we find that negative π times
π plus π times π times π squared equals π sub min. Factoring out an π and rearranging
this slightly, we find that π sub min is equal to π times the quantity π times π
squared minus π.

Now, letβs find a similar
expression for π sub max. If we add π sub max to both sides
of the second equation, then π sub max cancels on the right. And next, we subtract π times π
times π squared from both sides, canceling these two terms on the left. Finally, we can factor out an π on
the right-hand side. And we find that π sub max equals
π times the quantity ππ squared plus π. Notice then that the only
difference between π sub min and π sub max in these equations is the sign in front
of the acceleration due to gravity π.

Anyway, we want to find the ratio
of these forces. That is, we want to solve for π
sub max divided by π sub min. Notice that when we write out this
fraction, the mass π of our stone is common to numerator and denominator, and so it
cancels out. Our result will be the same
regardless of the mass of our stone. Now, the radius π of our stoneβs
path is 0.33 meters. π, the magnitude of our stoneβs
angular velocity, is given a 6.1 radians per second. And π, the acceleration due to
gravity, is 9.8 meters per second squared.

With all these values substituted
in, weβre now ready to calculate our ratio. The answer, to one decimal place,
is 8.9. Note that there are no units in
this result because itβs a ratio of two values that have the same units. The ratio then of the maximum force
to the minimum force that the rope can apply to the stone is 8.9.

Letβs look now at part two of our
question.

What is the force that the rope
applies to the stone when the rope makes an angle π equals 33 degrees above the
horizontal? Give your answer to the nearest
newton.

We can now think of the stone being
in this position so that we want to solve for the tension π along the rope at this
angle π of 33 degrees. To help us do this, letβs consider
this up-close sketch. The tension in the rope acts
towards the center of the stoneβs rotation, thatβs here, and the weight force as
always acts straight downward. Weβve said that for the stone to
follow a circular arc, it must experience a constant center-seeking force. At all points in the stoneβs
motion, this force is equal to some combination of the tension in the rope and the
weight force π times π. At this particular position of our
stone, we want to see how these two forces interact.

Notice that the weight force on the
stone can be broken up into components that are parallel and perpendicular to the
rope. This parallel component, because it
points in the same direction as the tension in the rope, contributes to the overall
centripetal force on the stone. If we call this component of the
weight force πΉ sub π€, we can write that this component added to the tension in the
rope is equal to the total centripetal force on the stone. The reason we add these two forces
together is because we see they point in the same direction, towards the center of
the stoneβs rotation.

In this part of our question, we
want to solve for what weβve called π in this equation, the tension in the
rope. Rearranging this equation a bit, we
see that π equals π times π times π squared minus πΉ sub π€.

From what weβre told in our problem
statement, we know the mass of the stone, the radius of its circular arc, and the
angular speed with which it moves. What we donβt yet know is πΉ sub
π€. If we consider the pink triangle of
which πΉ sub π€ is one side, we note that that right triangle is similar to this
right triangle weβve sketched in in orange. Therefore, this angle here in the
orange triangle, 33 degrees, is the same as this angle here in the pink one.

Now, if we have a right
triangle,and one of the other interior angles is called π, then the side of the
triangle opposite π β weβve called it π β is equal to the hypotenuse of the right
triangle β times the sin on the angle π. In our pink triangle, πΉ sub π€ is
the side opposite this known angle of 33 degrees. For this right triangle, the
hypotenuse is π times π, the mass of the stone times the acceleration due to
gravity.

We can write then that πΉ sub π€,
the component of the weight force that acts in the same direction as the tension
force, equals the mass of the stone times the acceleration due to gravity times the
sin of 33 degrees. We can substitute this result for
πΉ sub π€ into our equation for the tension π. If we do that and then also factor
out the mass of our stone, common to both terms on the right-hand side, we find that
the tension in the rope in this position is equal to the mass of the stone
multiplied by the quantity ππ squared minus π times the sin of 33 degrees.

Clearing a bit of space so we can
substitute in for these variables, we know that the mass of our stone is 1.6
kilograms. The radius of the stone circle is
0.33 meters. And the angular speed of the stone
is 6.1 radians per second. And π is 9.8 meters per second
squared.

Calculating this entire expression,
rounded to the nearest newton, we get a result of 11 newtons. This is the force that the rope
applies to the stone when the rope makes an angle of 33 degrees above the
horizontal.