JULIE HOOD: Hi, I am Dr. Julie Hood, an AP Physics
teacher from MAST Academy in Miami, Florida.
In today's video, we're going to work
through a rotational motion free response question that
has no numbers, just variables today.
If you want to work through this problem with me,
click on the link over the video now.
And you can download the PDF of this question featured
in the session.
So what I've got here is a uniform rod
of mass m and length l.
And it's attached to a pivot as shown below.
And the pivot is located l over 3 from the left end of the rod.
And we're expected to answer everything
in terms of the given quantities and fundamental constants.
Things like the acceleration of gravity
g for the universal gravitational constant, big G.
That's what they mean by fundamental constants.
So here, we're expected today to calculate
the rotational inertia of the rod about the pivot.
And then when the rods released from rest
in the horizontal position, it slides, it rotates on down.
And we're to calculate the linear speed
at the bottom end of the rod when the rod passes
through the vertical.
And finally, the rod is brought to rest
in the vertical position.
And it hangs freely.
And then we displace it slightly from that position
and calculate the period of oscillation as it swings.
So here we go.
The purpose of this question.
Well, we've got to calculate the moment of inertia
about an axis perpendicular to a rod,
at a point not at the center of mass
or even at the end of the rod.
We have to apply principles of conservation of energy
to the rod when it's in motion.
And we're going to find the period
of small oscillations of the rod when it's
displaced from the vertical.
What we have there is a physical pendulum.
So this is a complicated problem, but one that I love.
So here's our rod.
And the first thing we need to do
is find the moment of inertia with respect to that pivot.
Now, right off your equation sheet,
there's the moment of inertia.
We can't use the summation of every little mass element
at its respective r squared.
Well, we can, but we have to use calculus to do so.
And here's-- so here's our situation to find the moment
of inertia.
We've got r as the variable of integration.
But we've got little mass elements.
And we need to do a change of variables.
So it's a great idea to look at the ratio
of the mass of the rod to the length of the rod.
And that ratio is going to be equivalent to every little mass
element at its respective, teeny, tiny dr thickness.
So now we've converted the dm into a function of dr.
So we've got an integral that we could actually solve.
So let's do that.
We're going to integrate from the origin, which
I put at the pivot.
We're starting at negative l over 3,
integrating through the origin, and all the way up
to 2l over 3.
So those are our limits of integration.
You know how to integrate our r-square,
dr. It's just r cubed over 3.
And then we have to evaluate that at the limits.
Do a little algebra and get the moment of inertia,
1/9 ml squared.
You should practice this a few times.
That's our answer.
Now, you might have thought, why don't we do it
with the parallel axis theorem?
And perhaps, you can.
If they tell you to use integral calculus,
the parallel axis theorem is not going to get you points.
So remember now, the parallel axis theorem,
if we know what the moment of inertia
is at the center of mass, then we just
need to add a bit more to compensate for the fact
that the pivot has been shifted from the center of mass.
First, we have to know what the moment of inertia
at the center of mass is.
Well, I can use my handy-dandy moment of inertia integral.
And I'm going from the origin, which is the center of mass
now out to l over 2.
And I'm doing that twice.
And when I do the calculus-- and I'll let you practice that,
make sure you get 112 ml squared.
And you may even have that memorized.
But memorizing won't do you much good
if you're not using the parallel axis theorem.
So learn how to find it using integral calculus.
But just for practice, there's our 112.
And we're going to add m times the distance
the pivot is moved from the center of mass squared.
And when you do the math, you'll get
the same answer for the moment of inertia at the pivot.
Now the next part of this question,
we release the rod from the horizontal position.
And then we want to calculate the linear speed
at the bottom of--
the bottom end of the rod when it passes through the vertical.
So right when the rod is in that vertical position,
we want to find the speed of that blue dot.
Well, that's where we're headed and that's where we started.
There's our pivot.
And here's our center of mass.
And we're pivoted l over 6 from the center of mass.
Then when the ball swings down to the vertical,
there's our center of mass again.
And this is the speed at the bottom of the bar
that we want to find.
Now, this is a job for energy conservation.
Whenever you have objects bouncing
and moving like this, rotating, you
think about energy conservation.
So what we started with, initially,
was potential energy.
We let the bar go from the horizontal
and it started at rest.
And as it fell, all that potential energy
got converted to kinetic rotational energy.
So we have mgh equal 1/2 I omega squared.
And h is how far the center of mass fell.
It fell l over 6.
Now omega and v are related by how far
the pivot is from the point that you want
to calculate the linear speed.
And that r is 2/3 the length of the bar.
And when you substitute that in, that's
going to be your value for omega.
And now I'm going to substitute everything I know in.
Do some algebra.
Clean it up.
And solve for the velocity.
And there you have it.
2 root gl over 3.
And the last part of this problem, we've
got the bar hanging vertically.
We displace it slightly from that position
just a little bit.
Small theta.
And then we want to calculate the period of oscillation
as it swings.
Now let me remind you what the period of oscillation
for a simple pendulum is.
When theta is really small, it's fairly straightforward.
The length of that string and the amount of acceleration
due to gravity.
And there's your period of a simple pendulum.
Doesn't even matter what the mass is.
Now, we've got a physical pendulum.
We've got an extended mass.
Not when the mass can-- we're not going to follow just
the center of mass, we're going to follow--
well, here's the equation for a physical pendulum.
Again, not on your equation sheet.
But here's how I like to remember it.
It's going to have the same general form, 2 pi root.
And I need g down below.
And I need a length unit in the top.
But all I have is a moment of inertia.
Well, moment of inertia is proportional to mass
and distance squared.
So if I just put a mass and one single distance
in the denominator, I'll have the right units.
And that is the formula for physical pendulum.
So now, you just have to remember
that the d represents how far the center of mass
is from the pivot point.
Well, that was l over 6.
And we found i in part A. It was 1/9 ml squared.
And when you substitute all that in.
And then you do a little algebra.
Clean it up.
You get 2 pi root 2l over 3g.
That's the period of this oscillating system.
So I got to tell you, the average score on this
was only 2.9 points out of 15.
So if I wanted to tell you what the students did correctly,
well, it's really difficult in this problem
to pick out the most common errors
because very little was done correctly.
But let me help you.
Let me help you.
That's why this video.
If you're asked to find a moment of inertia,
write down this equation for starters.
You might get a point-- you'd get a point just for knowing,
oh, I need to use this formula.
Don't just do a shotgun and write momentum and conservation
of angular momentum.
Those have nothing to do with this problem.
You've got to write the equation that relates to the problem
to get the point.
And then remember to make the change in variable, which it's
just ratios, mass to length.
And substitute that in.
And you'll be good to go, if you can
add the limits of integration, you'll get another point.
And if you then follow through and do the integration,
you'll get as many points as they
allocate for that particular part of the problem.
That part A was worth 4 out of 15 points.
You can do it.
And I mentioned, if you're asked to find the speed of an object
in a situation, where you have objects that are bouncing,
falling, oscillating, rotating, it's usually a good idea
to approach it with a conservation of energy
statement like we did in part B. And take it from there.
By the way, if the situation is static, use forces.
And remember, there's a great connection
between simple pendulums and physical pendulums.
So go back and review this.
And go for the five.
And have a great day.