1:52
And for us on this problem, that's relatively straightforward.
They gave us those values.
One-half the mass of this object, 3 kilograms, and the linear speed,
2 meters per second, you need to make sure it's a square of that term.
So the translational kinetic energy,
the fact that this object is actually translating, moving from one location to
another, looks to me to be about 6 joules, that's a scalar.
It is positive energy, you can't have negative kinetic energy.
But since this object is rotating, it's rolling along the ground,
there's also some energy in that rotation.
So to solve for that then, our equation looks very similar.
And so there's a one-half, there's a term that represents inertia, like mass,
except it's our rotational moment of inertia,
which we'll have to know what that equation is for, specifically, a sphere.
4:43
I'm looking for this vertical height, h, so that by the time it reaches this
point down here, it's moving at 2 meters per second, like we said earlier.
That looks to me like a conservation of energy problem.
We know that it's going to have potential, gravitational potential energy at
the beginning, and then it's going to have two forms of kinetic
energy at the bottom, a rotational and a translational.
So my fundamental principle here to solve a problem like that is that the energy,
all of the energy added up before,
will equal all of the energy added up at the end of the problem.
So like we said, there's a gravitational potential energy at the beginning,
a translational final at the end.
We're setting our 0 for gravitational potential energy down here at the ground,
so there's no more gravitational potential energy, just these two.
In fact, on the right hand side, I already know these numbers.
We just calculated those.
That's our 6 joules and our 2.4 joules.
So I already know that the right hand side of this equation will be 8.4 joules.
What I'm solving for now is the height.
So that would be our mgh, which we already know the mass of the object, as well.
And g, for us, in this scenario is 10, so 3, 10,
solving for h, the height from which we would have to release this ball,
so that it would reach that speed, not very tall,
0.28 meters, about 28 centimeters.
12:14
All right, so in this question you have a planet that is moving in
a circular orbit around a star.
So that's important because it's not moving around its own axis,
it's moving around the star.
So when I need to solve for the moment of inertia for this,
I'm going to use my equation for a thin hoop which is MR squared.
All right, let's see what else they give us.
They tell us that this planet is on average 1 times 10 to the 11th
meters from the star.
Well that's the dis, distance over here from between the planet and the star.
That's my R, my radius.
[SOUND] So I'm going to make a note of that as well.
Then they tell me that it takes 300 Earth days to complete one revolution.
Well the time and seconds that it takes complete one revolution,
that would be the period.
So we're going to see how we'll use that in just a moment here.
They want us to solve for the angular momentum of
the planet around the star if the mass of the star
is 4 times 10 to the 24th kilogram.
All right, so let's go ahead and look at what we have over here.
Well the angular displacement represents the distance that's
traveled around the circumference.
14:06
So think back to this as your basic in
kinematics your velocity being your distance over time.
Your speed being your distance over time.
So now I can go ahead and plug in values, 2pi over,
remember, I need to use this time in seconds, so
I'm going to have to convert the 300 days.
There are 24 hours in one day,
[SOUND] and 3,600 seconds in one hour.
So, when I go ahead and solve for all of this,
I end up with my angular velocity of 2.42
times 10 to negative 7 radians per second.
Now this is important because I'm trying to solve for angular momentum,
and that is, angular momentum equals your moment of inertia multiplied by omega.
And earlier, we talked about what we would use for moment of inertia.
We're going to use our thin hoop equation, the MR squared.
So, and then now we can go ahead and
plug in values over here [SOUND].
The mass, remember they gave you in the question, that's 4 times 10 to the 24th.
[INAUDIBLE] clear.
[SOUND] Multiply
this by omega,
which is 2.42
times 10 to
the negative 7.
And when you go ahead and plug this into your calculator, you
should get a value of your annual momentum to be 9.68 times 10 to the 39th power.
And the units I'm using here are Newtons, times meters, times second.
[SOUND] And so that's my final answer there.
Now, that's part A.
The next part wants me to solve for the frequency.
Well, if I've figured out the amount of seconds that it took for
that one revolution.
Then I know that frequency is the inverse of the period,
1 over T, so I can go ahead and solve for it.
And again, I'll just show you the steps again.
It would be the 1 revolution over 300 days.
24 hours in one day and
3,600 seconds in one hour.
And when you solve for this,
you can go ahead and get your frequency,
which is 3.86 times 10 to the negative 8th power.
Remember frequency is in hertz.
And that's our final
answer for part B.
>> It's important to remember that momentum is conserved
when there are no outside forces affecting a system.
And so, when we're talking about an asteroid orbiting a sun or a star,
in this case.
Well this will be a more elliptical orbit than, let's say, a planet,
going around a star.
So let's say that this is our asteroid, this can do something like this.
We know from Kepler's laws, that we have here.
You're farther away, it should go slower.
18:06
When it's over here it's closer, it should be going faster.
The area swept out by this asteroid should be equal in periods of time.
That's what one of our Kepler's laws told us.
Well, angular momentum is conserved in this scenario because there aren't any
measurable or significant forces acting from the outside.
The only real forces that we have to worry about are the ones in between
the asteroid and the sun itself, those equal and opposite forces.
So, angular momentum is a great way to solve a problem like this.
This is similar to the issue of when you have the skater
bringing her arms in in a rotation.
She will speed up in her spin and her angular velocity,
because of the conservation of the angular momentum.
And so, solving a problem like this,
I'm going to start off with our fundamental principle.
Our angular momentum before, all of those angular momentums added up,
will equal a final angular momentum in the whole system.
Now, we're talking about this asteroid orbiting the sun.
So the only thing that's really moving in our reference frame here is going to be
that asteroid.
So we have a certain moment of inertia and angular speed at the beginning.
23:46
We're going to say that it has an angular momentum.
Because instantaneously,
it has a speed moving to the right which we can consider an angular velocity.
When we have a conversion between those two and
it's a certain distance from that radius.
So follow me here if you're a little lost and see if it makes more sense to you.
There's a moment of inertia of that point mass around the axis we've chosen and
it has an angular velocity.
So, I times omega.
And so, I'm going to call those for the first objects and their initial speeds.
Then after the problem, the two stick together.
So what have now is in a moment of inertia for
the sphere and the small mass that are stuck together.
So they have to have the same final angular velocity.
In fact,
it might even be easier then if I'm going to adjust my subscripts to match.
Let's make that an initial angular velocity.
This is a point mass at the beginning that's moving, so
I'm going to use m r squared.
So, it's an m1, r squared.
Now that r is, since it hits the edge of the disk or in this case sphere,
it's a distance r from the axis of rotation.
So that is the radius of the sphere that we're using there.
Again, I'm going to use our bridge equation.
V equals r omega, which comes up a lot in this unit.
So, v over r.
Now, after the collision, you have the point mass and the sphere moving together.
So, I need to sub those in.
Again, you have your mass 1 for your point mass, at a distance r squared,
because it's at the radius of the sphere plus two-fifths.
Now we have our mass of the sphere for the first time.
And again, the radius of the sphere and we're looking for
this final angular speed.
You use the same symbols here, so brackets.
I'm going to scroll down.
So let's take a look at what we have and what we don't have.
I know the mass of both, I know the radius of that sphere.
It looks to me like we're ready to sum the numbers and solve, so let's see how we do.
And we have a mass of 1 kilograms 0.3.
Now notice that the r's here can cancel out, so I only need to sub that in once.
I don't have to square it.
A speed of 4 meters per second on its way in, a linear speed.
Then on the right-hand side mass of 1 kilogram,
radius of 0.3 squared plus two-fifths.
The mass of the bigger sphere was 3, radius 0.3 squared and
then solving here for the final angular speed.
When I put all that into my calculator,
I divided by that huge term in the brackets to move it to the left-hand side.
I get a final angular velocity of 6.06 radians per second.
That's how fast the system is spinning after the collision.
Now we're not quite done, although we're getting there.
This question also asks how much energy was lost?
And typically in any real collision, energy is lost to sound,
to heat, especially an collisions like this.
So we have to calculate how much energy we had before the collision and
only one thing was moving.
We have the kinetic energy of that small point mass.
And you can do this as, in terms of rotation or in terms of linear.
It will come out the same.
I'm going to do linear, because we have those numbers and
I don't have to worry about using a bridged equation.
Just a one-half mv squared.
So one-half, 1 kilogram at speed of 4 squared.
Initially, at the beginning of this problem, we had 8 joules of energy.
Now, I need to find out how many energy do we have remaining in the problem
after the collision.
So for this, I think its a little easier to actually use rotations since we know
the final speeds.
So, after again, keeping track of my work carefully.
My total kinetic energy,
they will be one-half I omega squared plus one-half I omega squared.
Remember, we have the small point mass and the whole disk, that our sphere,
I'm sorry, that's actually rotating.
And this is a sphere that it hit, I apologize for mixing up the words.
So, one-half, now we're talking about that point mass, mr squared.
How far it is away from the center?
And we know the final angular speed, so that can just stay in.
Plus, since I'm not going to use a bridge equation one-half,
we're talking about the sphere, two-fifths mr squared.
And that's the mass of the big sphere this time.
So be careful not to mix them up.
Let's use different subscripts just to make sure and
then our final angular speed.
And they have the same final angular speed, because they're connected.
I can sub in my numbers, one-half the first mass.
Well, the mass of 1 kilogram.
The radius, 0.3 squared and
the final angular velocity that we solved,
4.606 and one-half, two-fifths.
The mass this time of that larger object, the sphere, 3.
The radius again, 0.3 and
then the 6.06 squared, the final angular speed.
When I finally solved them for the total,
final kinetic energy, I got 3.636 joules.
Notice that we have lost energy,
it's definitely less energy than the 8 joules we started out with.
It specifically asked how much was lost.
So, if I want to know how much was lost,
I've got to subtract the 8 joules minus the 3.636 joules.
All right. Find that we lost 4.36
joules in this equation.