Good morning. Let’s analyze the motion of a rubber dodgeball from the moment it rebounds
off the ground, until the moment right before it strikes the ground again. So, from right
there, … to there. Of course, that happens pretty fast in real time, so I’ve slowed it down to 40
times slower than real speed. And we are analyzing from the moment it leaves the ground right here,
until right before it strikes the ground again, which will happen in this video, eventually.
Today we are going to include the drag force, so realize this ball is not in free fall. … Bobby,
what is the drag force equation we have been using up to this point.
♪ Flipping Physics ♪
Okay, uh. The force of drag equals one half the drag coefficient
times the density of the medium the object is moving through times the cross-sectional
area of the object times velocity squared. Correct Bobby. Today we are going to use a
much simpler drag force equation. The equation is the drag force equals the negative of b,
which is called the proportionality constant, times the velocity of the object.
Billy, tell me what you see in this equation. Absolutely. Well. The previous drag force
equation has velocity squared in it, whereas this equation just has velocity. That means there
is a linear relationship between drag force and velocity in the newer equation. … Also,
the previous drag force equation was a scalar and this drag force equation is
a vector. … The negative in the equation means the drag force is opposite the direction of the
velocity of the object. … The proportionality constant … What is a proportionality constant?
A proportionality constant is just a constant value which represents the ratio
of two proportional quantities. If we solve the equation for the proportionality constant,
you can see the proportionality constant is a number which represents the ratio of the
drag force to the velocity of the object. Okay, yeah. I already mentioned the linear
relationship between drag force and velocity. That is what the proportionality constant represents. …
The proportionality constant then has units of … newtons divided by meters per second.
Newtons are kilograms times meters divided by seconds squared. So, the units work out
to be … kilograms per second. Why do we have a different,
simpler equation for the force of drag? And how do we know when to use which equation?
Those are valid questions. This new proportionality constant drag force equation is
valid for small objects moving at slow speeds. (But, what does that even mean?)
Right. In order to truly determine which equation to use for the drag force in any situation,
you would need to do experiments to test the equations. Something we are not going
to do right now. As far as the AP exam is concerned, in the past, the College Board
has given the specific equation they want you to use for drag force. In other words,
do not memorize either of these equations. … Okay, Let’s look at the ball as it moves
upward. Bo, what forces act on the ball as it is moving upward?
Sure. … The force of gravity is down,
and the force of drag is … opposite the direction the ball is moving, so down.
(And what does that mean for the acceleration of the ball?)
Well. And, as the ball moves upward, the net force is down, causing a downward acceleration, which
means the speed of the ball will decrease. That means the drag force will decrease in magnitude,
that means, as the ball moves upward. The magnitude of the net force on the ball will
decrease, and therefore, the magnitude of the acceleration of the ball also decreases.
And at the very top of its path, we know the velocity of the ball is zero, so the drag
force is zero, so, at that instant in time, the dodgeball has an acceleration equal to negative g,
the acceleration due to gravity. Right.
Actually, we also know that the net force is always greater than the force of gravity,
therefore, the acceleration always has a magnitude greater than g, the acceleration due to gravity.
Right. But we knew that already because the magnitude of the acceleration
decreases to g on the way up, therefore, it must always be greater than g on the way up.
Yeah. Correct … everybody. … I want to point
out that I have exaggerated the relative magnitude of the drag force arrow in the free body diagram.
If I did not do that, because the drag force is so small, the drag force arrow would be
too small to see in this diagram. … Bo, please also discuss what happens to the acceleration
of the ball as the ball moves downward. … As the ball moves downward, the force of
gravity is still down, however, the force of drag is now up. This means, as the ball moves downward,
the net force is down, so its speed increases, causing its drag force to increase, however, the
magnitude of the net force on the ball decreases, therefore, the magnitude of the acceleration of
the ball also decreases. … That means on the way up the magnitude of the acceleration of
the ball decreases to the acceleration due to gravity … and on the way down the magnitude of
the acceleration continues to decrease. Very nice Bo. Now, what about the time
while moving up compared to the time while moving down? We know the displacement in the
y-direction has the same magnitude because it starts and ends at the same height,
however, what about the change in time? Bobby? … Well … The average acceleration on the way
up has a larger magnitude than the average acceleration on the way down. … As you said,
the magnitudes of the displacements in the y-direction is the same. …
Both going up and coming down have an initial or a final velocity equal to zero,
therefore, a larger magnitude acceleration means a smaller change in time.
In other words, it should take less time on the way up than it does on the way down.
That is correct Bobby. For the dodgeball in our example, the difference amounts to
only roughly 6 thousandths of a second. The Physics Works! Uh huh … right?
I guess so. 6 thousandths of a second is not very long.
Right. Remember the drag force on this dodgeball is quite small. Because of that,
it only causes a very small increase in the time on the way down relative to on the way
up. … I also tried this with a large beach ball. The beach ball has a much smaller mass and a
much larger diameter and therefore a much larger relative drag force. … However, the difference for
the beach ball is still quite small at roughly 20 thousands or 2 hundredths of a second. So,
the physics does work, however, realize at small speeds and distances like this, the decrease in
the time from going up to down is quite small. See, The Physics Works!
Yeah? Eh.
In a bit we are going to look at graphs of motion, however, first, Billy, please determine
the terminal velocity of the dodgeball. Okay. Determine the terminal velocity of
the ball. … Uh, on the way up or the way down? … At terminal velocity the acceleration of the
ball equals zero. So, according to what we determined before about acceleration in this
situation, it has to be on the way down. Correct Bobby. And I just want to make it
absolutely clear, that, in the dodgeball and beach ball examples I show here,
neither of these demonstrations have the ball getting close to
terminal velocity. … Okay. Billy? Okay. The net force in the y-direction
equals the drag force minus the force of gravity and the net force equals mass times acceleration
in the y-direction. As Bobby just mentioned, at terminal velocity the acceleration of the ball
equals zero, so the drag force equals the force of gravity. … Plugging in equations gives us … the
proportionality constant times terminal velocity equals mass times acceleration due to gravity. So,
terminal velocity equals mass times acceleration due to gravity divided
by the proportionality constant. … What happened to the negative
in the drag force equation? … Oh, that negative gives the
direction of the drag force. By drawing the free body diagram and identifying the drag force to be
in the positive y-direction, we have already identified the direction of the drag force
and do not need to include that negative. Oh right. Just like the negative in Hooke’s Law.
Yep, just like the force of the spring equals negative spring constant times displacement
from equilibrium position. Yeah, that equation.
I was just clarifying. Of course you were.
That means this equation actually only gives us the magnitude of the terminal velocity or the
terminal speed. What?
We know the ball is moving down, so the terminal velocity is negative.
Ah. Oh.
Now let’s look at graphs of position, velocity, and acceleration as a function of time for the
dodgeball. Let’s compare the graphs of Free Fall, in other words, without a drag force,
to graphs with a drag force. Let’s start with the graph of position as a function
of time for the dodgeball assuming there is no drag force. Notice it is a symmetrical
parabola because the time going up is the same as the time coming down. … Bobby, how does the
graph change when we add a drag force? … Uh, Okay. Well, first off, we know it
is not a symmetrical parabola because the time going up is less than the time going down. So,
just slide the apex of the curve to a bit earlier in time. And, assuming the same initial velocity,
the ball will reach a smaller maximum height. (You know what Bobby. Let’s also assume the
ball reaches terminal velocity so we can see what that looks like on the graphs.)
Then … that means the final slope of the graph equals the terminal velocity of the ball or
negative mass times acceleration due to gravity over the proportionality constant.
Very nice Bobby. … Now let’s look at velocity as a function of time.
In free fall, without the drag force, the slope of the line is a constant
negative 9.81 meters per second squared because the acceleration is constant at negative g. The
line crosses the time axis at half the total time, when the ball is at the top of its path.
And the initial and final velocities have the same magnitude. … Bo, how does adding
the drag force change the velocity graph? … Well, initially the slope has a magnitude
larger than 9.81 meters per second squared because the magnitude of the acceleration is larger than
g. … The magnitude of the slope will get smaller as time goes by and cross the time axis
before half the total time, again this is when the ball is at the top of its path. … The slope at the
top of its path equals negative g. … The final slope is zero because the acceleration is zero
at its terminal velocity. … Oh, and the value of the final velocity is the terminal velocity.
Again, negative mass times acceleration over the proportionality constant.
Thanks Bo. … Lastly, let’s look at acceleration as a function of time.
In free fall, the acceleration is constant near the Earth at roughly negative 9.81 meters per
second squared. So that is a horizontal line, or a line with zero slope, at negative 9.81
meters per second squared. Billy, what does that look like when we include a drag force?
Okay. The initial acceleration has a magnitude larger than 9.81 meters per second squared.
Because it is negative, that means it is more negative than negative
9.81 meters per second squared. … At the top of its path, the ball will have an
acceleration equal to negative g. … And the final acceleration approaches zero
because it will reach terminal velocity. Thanks Billy. And very nice everybody.
Thank you very much for learning with me today, I enjoyed learning with you.