Mr.p: Good morning. Bobby, please read the problem and Bo, please translate.
"Flipping Physics" Intro
Bobby: 0.100 kilogram and 0.200 kilogram masses hang from either side of a frictionless pulley
with a rotational inertia of 0.0137 kilograms times meters squared and radius of 0.0385 meters.
Bo: Mass 1 equals 0.100 kilograms. Mass 2 equals 0.200 kilograms.
The rotational inertia of the pulley equals 0.0137 kilograms times meters squared.
The radius of the pulley equals 0.0385 meters.
Bobby: (a)What is the angular acceleration of the pulley? (b) What is the force of tension in each string?
Bo: Part (a) angular acceleration equals question mark.
Part (b) force of tension 1 and force of tension 2 both equal question mark.
Billy: If the rotational inertia of the pulley is 0.0137 kilograms times meters squared,
what about those spiky things sticking out of the pulley?
Should we not include the rotational inertia of those as well?
Bo: You were here last time, right?
0.0137 kilograms times meters squared is the rotational inertia of the whole thing with the spokes.
Bobby: Yeah, we determined that last time. Are you okay?
Yeah, I remember, now.
Mr.p: Billy, please begin solving the problem.
Billy: Let's draw free body diagrams.
The pulley has a force of gravity acting down at its center of mass
and a force normal up at the same location.
The pulley also has two forces of tension acting down on it, one on each side.
Let's have force of tension 1 point towards mass 1 and force of tension 2 point towards mass 2.
Mass 1 has a force of gravity 1 down and an upward force of tension 1,
which has the same magnitude as the force of tension 1 which acts down on the pulley.
Mass 2 has the equivalently named forces only with subscripts of 2 instead of 1.
Bo: Hold up, shouldn't the two forces of tension on either side of the pulley be the same?
It is just a single string, so the force of tension in any part of the string should be the same, right?
Mr.p: Bo, I do understand that it is the same string,
however, because the pulley has mass and therefore rotational inertia,
it takes a net torque to angularly accelerate the pulley
and therefore these two tension forces on either side of the pulley cannot be the same.
In the past, we have done pulley problems
where we assumed the pulley had negligible friction and negligible mass.
In that simplified case, the two forces of tension on either side of the pulley would be the same.
But that is not the case here. Billy, please keep going.
Billy: Okay, let's start by summing the torques on the pulley with its axle as the axis of rotation.
Because mass 2 is greater than mass 1, mass 2 should apply a larger torque on the pulley
and cause the pulley to rotate in the clockwise, or into the board, direction.
Therefore, let's define clockwise or into the board as positive.
Oh, because both forces act on the axis of rotation,
neither the force normal nor the force of gravity acting on the pulley will cause a torque on the pulley.
Therefore the net torque equals
the torque caused by force of tension 2 minus the torque caused by force of tension 1.
And the net torque also equals the rotational inertia of the pulley times the angular acceleration of the pulley.
We can substitute the equation for torque, r times force of tension times sine of theta
in for each torque equation with corresponding 1 and 2 subscripts.
Both r values are equal to the radius of the pulley.
Both angles are 90 degrees because both r values are horizontal and both forces of tension are down.
The sine of 90 degrees equals 1.
So we now have pulley radius times force of tension 2 minus pulley radius times force of tension 1
equals rotational inertia times angular acceleration.
In part a, we are solving for the angular acceleration of the pulley,
and in part b, we are solving for both forces of tension.
So everything we are solving for is an unknown in this equation.
So let's put this equation in our equation holster.
Mr.p: Bobby, please continue from here.
Bobby: Well, we've sum the torques. So now let's sum the forces.
Let's start with mass 1.
The net force on mass 1 in the y-direction equals force of tension 1 minus force of gravity 1,
which equals mass 1 times acceleration 1.
We can bring force of gravity 1 over to the other side
and substitute in mass 1 times acceleration due to gravity
and we now have an equation for the force of tension 1, which we can put in our equation holster.
Now, let's sum of the forces mass 2 in the y-direction...
Actually, it's the same as what I did with mass 1 only replace every 1 subscript with a 2
because we are summing the forces on mass 2.
Why is mr.p just staring at me? He should be writing on the board.
Mr.p: You know, I really don't think you want me to write that on the board yet.
Bobby: I don't?
Billy: Oh, it's the direction. I defined the direction of positive torque as clockwise, or into the board,
so down on the side of the pulley where mass 2 is is actually positive, not negative.
Bobby: Yeah, I actually made this same mistake the last time we did this. So, okay.
Then let the net force on mass 2 equal the negative of force of tension 2 plus force of gravity 2,
which equals mass 2 times acceleration 2. Therefore, force of tension 2 equals
mass 2 times acceleration due to gravity minus mass 2 times acceleration 2.
That makes more sense. Thanks, Billy.
Billy: You are welcome, Bobby.
Mr.p: Notice I have better identified the direction in which we are summing the forces.
We are actually not summing the forces in the y-direction, but rather what I call the "positive direction".
The positive direction is the direction which Billy defined earlier when he defined clockwise,
or into the board, as positive.
This means that to the left of the pulley, up is positive and to the right of the pulley, down is positive.
By defining this positive direction, it helps remind us to be careful of direction.
Bo, we now have three equations in equation holsters. Please combine them all.
Bo: Actually, all we need to do is substitute the equations for force of tension 1 and 2
into the equation we got from summing the torques. That's all we can really do with that.
Mr.p: Bo, what about relating the linear and angular accelerations to one another?
Bo: Oh, yeah. That's right. So, when we did the bike wheel example a few lessons ago,
you showed us that the linear acceleration of the hanging mass
is the same as the tangential acceleration of the rim of the bike wheel, or in this case the rim of the pulley.
So, acceleration 1 equals acceleration 2 which equals
tangential acceleration or radius times angular acceleration.
Because it's the rim of the wheel, the radius is the radius of the pulley.
So we can substitute the pulley radius times angular acceleration for both linear accelerations.
Mr.p: Bo, please finish part a. Please find the angular acceleration of the pulley.
Bo: Okay. Multiply through by the radius of the pulley
and then move all the terms to the left-hand side, which have angular acceleration in them,
so we can factor out angular acceleration on the left hand side.
And we can factor out the acceleration due to gravity times radius on the right hand side.
Divide the whole equation by everything inside the parentheses on the left hand side.
I suppose we could factor out radius squared from two terms in the denominator
and we get angular acceleration of the pulley equals
acceleration due to gravity times pulley radius times the quantity of mass 2 minus mass 1
all divided by the quantity rotational inertia plus radius squared times the quantity mass 2 plus mass 1.
With numbers that is 9.81 times 0.0385 times the quantity 0.2 minus 0.1
all divided by the quantity 0.0137 plus 0.0385 squared times the quantity 0.2 plus 0.1,
which works out to be 2.67016 or 2.67 radians per second squared with 3 significant digits.
Mr.p: Let's compare this to our measured angular acceleration.
The initial angular velocity of the pulley is zero. Change in theta is two revolutions or four pi radians.
Change in time is 193 frames at 60 frames per second or 3.216 repeating seconds.
The net torque is constant, so the angular acceleration is constant,
so we can use the uniformly angularly accelerated motion equation angular displacement equals
initial angular velocity times change in time plus one half angular acceleration times change in time squared.
Initial angular velocity is zero,
so the angular acceleration equals two times an angular displacement
divided by change in time squared, or
2 times 4 pi divided by 3.216 repeating squared, which equals 2.4290 or 2.43 radians per second squared.
We can use the relative error equation to solve for our percentage difference
with the measured angular acceleration as the observed value
and the predicted angular acceleration is the accepted value.
We get a percentage difference of negative 9.03 percent.
So our measured angular acceleration was about 9 percent less than our predicted angular acceleration.
Eh, that's pretty close.
Billy, please solve part b. Please solve for the predicted forces of tension.
Billy: We can use the two equations for forces of tension which are in our equation holsters.
But we can factor out mass
and substitute in pulley radius times angular acceleration for linear acceleration in both equations.
That means force of tension one equals
mass one times the quantity acceleration due to gravity plus pulley radius times angular acceleration, or
0.1 times the quantity 9.81 plus 0.0385 times 2.67016, or 0.991280 or with three significant digits 0.991 N.
Force of tension 2 equals 0.2 times the quantity 9.81 minus 0.0385 times 2.67016,
which is 1.94144 or 1.94 newtons.
Mr.p: Now, please notice that those two forces of tension are not equal in magnitude.
Recall that is because our pulley has mass and therefore has rotational inertia
and therefore requires a net torque to angularly accelerate it.
Therefore in order to cause that net torque on the pulley,
force of tension 1 and force of tension 2 cannot have the same magnitude.
But if the pulley had negligible mass and therefore a negligible rotational inertia,
the equation we got from summing the net torque,
would actually show that the two forces of tension acting on the pulley would be the same.
Bo: Hold up mr.p. You also said the pulley has to have negligible friction.
But if the pulley has no friction,
then the string would just slide on the pulley and the pulley would never rotate.
What's up with that?
Mr.p: Interesting point Bo. So what we mean when we say the pulley has negligible friction
is that the axle of the pulley has negligible friction.
The surface of the pully still has friction to interact with the string.
Okay. I also want to point out two ways which students try to use to solve this problem,
which are both incorrect.
One is to sum of the torques on the whole system all at once with the axis of rotation at the axle of the pulley.
It works out to be a pretty long equation, but some students will try to solve it this way.
But realize, the rotational inertia and angular acceleration on the right hand side of the equation
would then refer to everything in the system, including the two hanging masses.
And hopefully you recognize that the hanging masses do not have rotational inertia
nor do they have angular acceleration. So this equation is incorrect.
Another way students try to solve this all at once
is to sum the forces on the entire system in the positive direction.
Again, this is a pretty long equation, and again it does not work.
This time it is because the mass times acceleration on the right hand side of the equation
would be for the entire system.
Now, I do understand that the rim of the pulley does have the same magnitude tangential acceleration
as the linear acceleration of the two masses.
However, the pulley itself does not have a single tangential acceleration.
Because the tangential acceleration depends on radius,
the larger the radius the larger the tangential acceleration
of whatever specific point on the pulley you are referring to. So that equation does not work either.
Thank you very much for learning with me today, I enjoyed learning with you.