Mr.p: Good morning. Bobby, please read the problem and Billy, please translate.
"Flipping Physics" Intro
Bobby: As shown, a 5.3 g dart is moving vertically at 16.5 m/s just before it collides with and sticks
to a 33.9 centimeter long thin piece of cardboard.
If the dart hits the 71.8 g piece of cardboard 28.7 centimeters from its fixed end,
to what maximum angle does the cardboard rise?
The rotational inertia of a long, thin object about its end equals one third mass times length squared.
Billy: Well, to start with, we need all of our variables in base SI units.
So the mass of the dart equals 0.0053 kilograms,
the initial velocity of the dart equals 16.5 meters per second.
the mass of the cardboard equals 0.0718 kilograms.
The distance, let's let's call it "x", from the stationery end of the cardboard to where the dart embeds itself
in the cardboard equals 0.287 meters.
The length of the thin piece of cardboard equals 0.339 meters. Theta final equals question mark.
And the rotational inertia of a long, thin rod, like this piece of cardboard,
equals one-third times mass times length squared.
You know what? Let's actually just get a number for that.
So, one third times times 0.0718 times 0.339 squared equals 0.0027504 kilograms times meters squared.
Mr.p: Bo, please walk us briefly through why angular momentum is conserved in this situation
and begin solving the problem.
Bo: We just did this last time,
so, we know angular momentum is conserved because there is no net torque acting on the system.
The system is the dart and the cardboard, and the axis of rotation is at the stationery end of the cardboard.
Net torque equals change in angular momentum over change in time,
and when that equals zero, the change in angular momentum equals zero,
so the angular momentum is not changing and angular momentum is conserved.
Before the collision
we have the initial angular momentum of the dart plus the initial angular momentum of the cardboard
and after the collision
we have the final angular momentum of the dart plus the final angular momentum of the cardboard.
Bobby: Actually, the collision is just the first part of the problem.
After the collision the dart and cardboard rise together
and we know mechanical energy is conserved while the cardboard rotates because there is no mechanical
energy added to or removed from the system via work done by a force of friction or force applied.
Bo: Right. So part 1, the collision, we can use conservation of angular momentum,
and part 2, while the cardboard rotates,
we can use conservation of mechanical energy. Thanks.
Mr.p: Exactly. This problem has two parts.
Part 1 is a collision where we use conservation of angular momentum.
And then part 2 is while the dart and cardboard rise up and we use conservation of mechanical energy.
Let's start with part 1.
Billy, remind me, what are the two general equations we have for the angular momentum of objects?
Billy: Again, the angular momentum of a point particle equals
r, the vector from the axis of rotation to the center of mass of the point particle,
times its mass, times its velocity, times the sine of the angle between the r vector and its velocity.
And the angular momentum of a rigid object with shape equals rotational inertia times the angular velocity.
For the dart before the collision we use the angular momentum of a point particle equation.
r for the dart for part one initial times mass of the dart times the velocity of the dart part one initial
times the sine of the angle for the dart part one initial.
Before the collision, the cardboard is not moving, so it does not have any angular momentum.
For the dart after the collision, we have already shown
that we can use either of those two equations for the angular momentum of the dart or point particle,
because it is moving in a circle.
So, let's use rotational inertia for the dart for part one final times angular velocity of the dart for part one final.
For the cardboard after the collision,
we have to use the equation for the angular momentum of a rigid object with shape,
so rotational inertia of the cardboard times the angular velocity of the cardboard for part one final.
Oh, and you did this last time.
Looking specifically at the equation
for the initial angular momentum of the dart as it moves toward the cardboard before the collision,
we showed previously that,
while the values of r and theta change as the point particle moves towards the rigid object with shape,
the value of "x",
which equals r for the dart part 1 final initial times the sine of theta for the dart for part one initial,
does not change. So we can substitute the value of "x" in
for r for the dart part 1 initial times the sine of theta for the dart for part 1 initial!
Oh, and we might as well also substitute in mass of the dart times the square of r for the dart part 1 final
in for the rotational inertia for the dart for part 1 final.
Mr.p: Very nice Billy!
Also notice r for the dart for part 1 final equals x
because after the collision the dart is still a distance x from the axis of rotation.
And both of the angular velocities for part 1 final are the same.
The dart and the cardboard have the same angular velocity after the collision
because they are moving together.
And the final angular velocity for part 1 is the initial angular velocity for part 2
because the end of part 1 is the beginning of part 2.
We can now substitute those variables into our equation
and rearrange it to solve for the final angular velocity for part 1.
With numbers that works out to be
0.287 times 0.0053 times 16.5 all divided by the quantity 0.0053 times the square of 0.287
plus 0.0027504, which equals 7.897517 radians per second.
So now we have the final angular velocity of the system for part 1,
which is also the initial angular velocity of the system for part 2.
Bobby, please work on part two.
Bobby: Okay, part 2 is conservation of mechanical energy.
The initial point is right after the collision, but before the cardboard starts to rise.
Let's set the final point at the maximum angle the system rotates to.
And let's put the horizontal zero line at the initial point for part 2.
There are no springs, so no elastic potential energy in the problem.
Initially the dart and cardboard are at the horizontal zero line,
so neither object has gravitational potential energy initial.
Then the only type of initial mechanical energy is kinetic energy and, because the system is rotating,
it is the combined rotational kinetic energy initial of both the dart and the cardboard.
At the final point, the dart and cardboard are not moving, so they have no final kinetic energy.
At the final point, the dart and cardboard are above the zero line,
so both the dart and the cardboard have final gravitational potential energy.
The rotational inertia of the system equals
the rotational inertia of the dart plus the rotational inertia of the cardboard.
We use the rotational inertia equation for a point particle for the dart,
so the rotational inertia of the system equals the mass of the dart times
the square of r for the dart for part 2 final plus the rotational inertia of the cardboard.
Oh, but r for the dart for part 2 final equals the distance x
because the dart is always the distance x from the axis of rotation after the collision.
We can substitute x into the rotational inertia equation for the system
and then substitute that equation back into our conservation of mechanical energy equation
and we have
one half times the quantity mass of the dart times x squared plus the rotational inertia of the cardboard
all times the square of the initial angular velocity of the system for part 2 equals
the mass of the dart times acceleration due to gravity times height for the dart for part 2 final plus
the mass of the cardboard times acceleration due to gravity
times the height for the cardboard for part 2 final
But we have two unknowns in that equation. We do not know either of the two final heights.
We cannot solve one equation with two unknowns.
Mr.p: This is great so far. Now, in order to reduce this equation to one unknown,
we need to draw triangles and use trigonometry to find relationships between the angle theta,
which is between the cardboard and the horizontal, and both final heights.
Before we do so, realize the height for Part two final of the cardboard is to the center of mass of the cardboard.
We will consider the cardboard to be of uniform density,
so the center of mass of the cardboard is in the middle of the cardboard.
Sine of theta for part 2 final equals opposite over hypotenuse.
For the dart, that equals height for the dart for part two final divided by x.
For the cardboard
that equals height for the cardboard for part 2 final divided by the length of the cardboard divided by 2
because it is the distance from the axis of rotation to the center of mass of the cardboard.
We can solve both equations for height 2 final
and substitute them back into our conservation of mechanical energy equation.
Which means we now have one equation with one unknown.
We can factor out sine of theta 2 final,
divide both sides by everything in the parenthesis in front of the sine of theta 2 final,
and then take the inverse sine of the whole equation to solve for theta 2 final.
And we get theta 2 final equals the inverse sine of the quantity....
Actually, you know what?
That would be a bit overkill for me to read and you can always pause the video to read it to yourself.
Billy: Wait, plus what video? Bo: Hold up, I got it.
Billy: What are you doing? Mr.p: Sorry, just kidding.
Mr.p: We can now plug in numbers and we get 48 degrees with two significant digits.
So, 48 degrees is our predicted final angle of the cardboard.
Now, let's see what the measured final angle is. As you can see it is roughly 46 degrees.
That means the percentage difference between our predicted and our measured final angle
for the cardboard works out to be roughly 4.5%, which is pretty close.
Thank you very much. Billy: Wait, the physics worked, did it not?
Mr.p: Uh huh, it was pretty close.
Bo: Actually what I want to know is,
why did we solve for the final angular velocity for part 1?
I mean, if you really wanted us to let go of our numbers dependency,
we should have left the final angular velocity for part 1 as variables. Am I right or am I right?
Billy: I don't like that.
Mr.p: Okay, sure, Bo, we could have done that. And it would have looked like this.
And I sort of felt that that was pushing it a bit too far.
You can disagree with me if you would like to and you can solve it that way,
but this is what I chose to do this time.
Perhaps next time, Bo, I will take your advice.
Thank you very much for learning with me today, I enjoyed learning with you.