DEE DEE MESSER: Hi, welcome to another session of AP physics
daily practice sessions for mechanics.
I'm DeeDee Messer, AP Physics C teacher at William Mason High
School in Mason, Ohio.
In this session, we're going to look at a free response
question with rotation.
If you'd like to click the link to get a paper copy or a PDF
file of the question and work ahead of time, please do so.
Or feel free to pause the video every time
I show you a new part to the question
to work it by yourself.
And then resume the video and check your answers with mine.
All right, let's get started.
A triangular rod has a length l, mass m,
and a non-uniform linear mass density given by the equation,
lambda equals 2m over l squared times x,
where x is the distance from one end of the rod.
So there is my initial condition with my graphic and my stem.
And then each part of the question
will refer back to those examples.
So for part A, using integral calculus,
show that the rotational inertia of the rod about its left end
is 1/2 ml squared.
Remember, this is mechanics calc-based.
So don't get frustrated or nervous
when you see integral calculus.
Look at your equation sheet.
Find the equation with an integral sign
and I for rotational inertia as a starting place.
Once I write the equation, I can see that I cannot integrate
the function of r squared with respect to mass.
So I need to make a substitution.
I'm going to go back to the fact that it
has a non-uniform linear mass density
and start with the equation for lambda.
I'm going to substitute in the function they were given to me
and make that equation set equal to dm.
Now, I'm going to use that equation
and substitute into my integral for the rotational inertia.
Right now, it might look really ugly.
But let's pull out all of the constants in front
of the integral sign.
Now, we can see that we're just integrating
x cubed with respect to x.
And I want to make sure that I integrate
the entire length of the rock, so my bounds
are going to go from 0 to capital L. Do the integration.
Evaluate it with respect to the bounds.
And I get an answer of 1/2 ml squared.
A key thing to remember when you get
an expression as a final answer and you're trying to decide,
does that look right?
The basic equation for rotational inertia
is mr squared.
So does your final answer follow that same format,
mass times a distance squared?
And in this case, yes, it does.
So double check your final answer before moving on.
And make sure that it seems reasonable.
Now, let's check the scoring and see how many points we
would have earned for part A. Part A was worth three points.
The first one is for relating x to r
in an integral for rotational inertia.
So on my equation sheet, the integral is r squared dm.
I needed to show that I understand
r can be any distance variable that is represented
in the question as shown.
So I'm going to substitute in x for r.
And I'm going to integrate with respect to x.
The second point is for correctly
using the linear mass density and the correct expression.
So I needed to show how I correctly
find the expression for dm, using the linear mass density
equation given in the STEM and substitute that
into the integral equation for rotational inertia.
Finally, I needed to have the correct balance.
I needed to make sure that I show
that I'm going to integrate the entire length of the rod
from 0 to big L. And those are the three points for part A.
Let's move on.
I now introduce a hoop with a mass m, radius l,
and a rotational inertia of ml squared.
The three rods that are identical to the rod from part
A are now fastened to the thin hoop.
And I have the figure shown in number two.
A hoop with the three rods attached all fixed together
at the middle.
For question B, derive an expression
for the rotational inertia of the hoop rods system
about the center of the hoop.
Express your answer in terms of m, l, and physical constants as
appropriate.
The as appropriate lets you know that you
don't need to have m, l, and little g, physical constant,
in your final answer.
It's just reminding you that those are the only options
for your final answer.
And the verb, derived, means equation, substitute,
solve with variables.
So make sure that you show all of your work.
So I have a system that contains one hoop and three rods.
So to get the total rotational inertia,
I'm going to add all four objects together.
The rotational inertia of the hoop was given in the problem.
And the rotational inertia of the rod
was my answer to part A. Gives me a total of 5/2 ml squared.
All right, let's check the answers.
This was worth two points.
The first one was showing very clearly
that I'm adding the hoop and three rods together
to get the total.
This is not a parallax theorem.
This is not an integration.
So make sure that you are clearly stating
what your equation represents.
Finally, it was for an answer consistent with your answer
to part A. So if you had solved an incorrect equation
for the rotational inertia of the rod
and you use that in part B, you are not going
to lose any additional points.
We are just looking for you to substituting
into the correct equation, not whether or not
your final answers are correct.
Just consistent.
All right, now let's move on.
The hoop-rod system is initially at rest.
But it's fixed in place.
And a constant force of f is exerted tangentially
to the edge of the hoop for some time, delta T.
So now, the hoop is spinning but fixed in place.
For part C, derive an expression for the final angular
speed of the hoop rod system.
Express your answers in terms of m, l, f,
delta T, and physical constants as appropriate.
Pay close attention that my physical constants
are the same as little g.
But the other variables have changed
from what was allowed in part B. So
be careful with what your final expression consists of
and that it matches that particular question.
Read carefully.
All right, if there is a force acting tangentially
to the edge of the system, then I
know there is a torque applied.
So the definition equation for torque is r cross f.
And a network for Newton's second law of a rotating system
is equivalent to I alpha, where alpha, by definition,
is the change of angular velocity
over the change in time.
The initial condition, we're starting from rest.
So omega naught is 0.
I know that l represents the distance from the center
to the edge of the hoop.
They are tangent, so is the sine of 90.
I for the system was 5/2 and all squared.
That was my answer to part B. And now
that I have made all of my substitutions,
I can rearrange the equation and make it say omega equals.
Once again, that equation may not
look like a pretty final answer to some of you.
But nevertheless, it obeys all of the rules
with only using variables that are allowed.
So now, let's check the points.
This was worth three points.
One point for using a correct equation
to find the angular speed.
So knowing that alpha was equal to the change of omega
over the change in time was worth that first point.
For relating the lever arm to the length of the rod,
so noting that r in the torque equation
was going to be represented by capital L was the second point.
And the third point was the correct substitution
of the rotational inertia of the system from part B.
Once again, don't get frustrated,
if there is a part that you are unsure
and think that you're going to miss the rest of the problems.
As long as your answers are consistent,
you won't lose any additional points.
So if you are unsure how to answer one part,
write down any answer that seems logical,
an expression, or a number, depending upon the question.
And then keep substituting that into the other parts
of the question.
All right, now the system has changed.
The hoop rod system is no longer fixed in place,
but now is rolling without slipping along
a level horizontal surface with angular speed omega.
At time t equals 0, the system begins
rolling without slipping up a ramp as shown in the figure.
For part D, on the figure of the hoop rod system
below, draw and label the forces, not the components
that act on the system.
It's important, especially with this free body diagram
being on an incline that you understand not
to draw any components at all.
If you need to draw components later to help you analyze,
draw a separate picture somewhere else on the page.
Each force must be represented by a distinct arrow starting at
and pointing away from the point at which the force is exerted.
So a new addition to sometimes our free body diagrams.
We need to have all the forces, but we
need you to start drawing them from point of application.
So I know the Earth is pulling down with a force of g.
And that's always straight down.
The source of the normal force is the ramp,
so it starts from that bottom of the hoop.
And the source of the frictional force
is also the ramp, so it starts at the bottom of the hoop.
Now, let's look and see the points.
Part DI was worth three points.
One for the correct gravitational force.
One for the correct normal force.
And one for the correct frictional force.
Any additional forces would just give a maximum value
for this of two points.
All right, the second part of D. Justify your choice
for the direction of each force.
So I want to justify each force.
There were three forces that I drew.
So I want three short, sweet, and to the point sentences.
The gravitational force is always vertically down.
The normal force is always perpendicular to the surface.
Therefore, it's going to be a diagonal up into the left.
And the frictional force is opposite the direction
of rotation.
Therefore, it's up the ramp because the bottom of the hoop
is moving down.
If it's not slipping, then I want
it to come to rest at that contact point.
So friction has to point up.
This was worth one point.
So one point that all three justifications
were correct with all the forces that you drew in your free body
diagram.
All right, let's look at this last part.
Derive an expression for the change
in height of the center of the hoop from the moment it reaches
the bottom of the ramp until it reaches maximum height.
Express your answer in the following terms
and physical constants.
Once again, pay attention that the variables
allowed in this part are different from the others.
So when you get a final expression,
check the directions and make sure that you're only
using variables allowed.
If it is rolling without slipping,
there is no kinetic friction.
So energy is conserved.
Initially, it has both translational and rotational
kinetic energy.
And at the end, it only has gravitational potential.
I substitute in my definition equations
for those three energies.
And then I'm going to substitute in the specifics.
For example, the mass of the system is 4m.
The linear speed, v, is equal to r omega
or, in this particular case, l omega.
I total is a variable that I'm allowed to have for the system.
And finally, for the potential energy, mass once again is 4m.
Now, that I have all the variables
that I'm allowed to use, I can rearrange the equation
with my algebra.
And I get an expression for h.
So let's see how well we did according to the scoring.
Part E was worth three points.
One point for using linear, and rotational kinetic energy,
and a conservation of energy equation.
I highlighted three different lines
to let you know that you could have started with any of those.
You just needed to make it very clear
that that system had both translational and rotational
kinetic energy to start with.
The second point was for understanding
the substitution of linear speed is
equivalent to r omega or, more specifically, l omega.
And the last point for correct substitution of mass
into a conservation of energy equation.
So we needed to recognize that the entire system has
a mass of 4m.
And there is our 15 points.
And that concludes this AP practice daily session
for mechanics.
The more practice you do, the more
you're going to see the results on the day of your AP exam.
So continue practicing, continue watching.
And I'll see you next time.