Mr.p: Good morning. Bo, please read the problem and Billy, please translate.
"Flipping Physics" Intro
Bo: Mass 1 and mass 2 hang from either side of a frictionless pulley with rotational inertia, I, and radius, R.
“frictionless pulley” means the axle of the pulley is frictionless,
not that the surface of the pulley is frictionless, right mr.p?
Mr.p: That is correct Bo. A frictionless pulley has an axle with negligible friction. Please keep reading.
Bo: What is the angular acceleration of the pulley?
Assume the system starts at rest and mass 2 is greater than mass 1.
Billy: Knowns are mass 1, mass 2, pulley rotational inertia capital I and pulley radius capital R.
Angular acceleration equals question mark.
Initial angular velocity is zero and mass 2 is greater than mass 1.
Bo: This looks eerily similar to the problem we just did, does it not?
Bobby: It does. Billy: Yeah.
Mr.p: You are correct that this is similar to the problem we just did, I have simply removed the numbers.
This time we are going to solve the problem using conservation of mechanical energy
instead of net force and net torque.
First off, Bobby, how do you know mechanical energy is conserved in this situation?
Bobby: Well, there is no force applied and no force of friction,
so no forces are adding or removing energy from the system, therefore mechanical energy is conserved.
Billy: Actually, there is a force of friction between the string and the pulley,
otherwise the pulley would not rotate.
But the force of friction is internal to the pulley, string, and 2 masses system,
so it does not add or remove energy from the system.
Mr.p: Great. And Bobby, please begin solving the problem.
Bobby: Well, let’s set the initial point where the objects are at rest at the start.
And the final point … uh, do the masses stop at some point?
Mr.p: Well, they do stop at some point, however, the final point of the problem has to be before that happens.
How about the final point is after both masses have travelled a distance y?
Bobby: Sure.
And to make it easier, let’s make it so the final point of mass 2
is at the same height as the initial point for mass 1.
That way we can set the zero line at the initial point for mass 1
which will be the same as the final point for mass 2.
And both masses will have zero gravitational potential energies
at their respective points.
Mr.p: Good idea Bobby. Now please identify all the
mechanical energies initial and final for all the objects.
Bobby: Okay. Everything starts at rest, so no initia kinetic energies.
There are no springs at all, so no elastic potential energies at all in the problem.
Mass 1 has a height initial of zero, so it has no gravitational potential energy initial.
So the only initial mechanical energy is the initial gravitational potential energy of mass 2.
As far as mechanical energies finial are concerned.
Mass 1 is above the zero line at the final point so it has final gravitational potential energy.
Both masses are moving so they both have final kinetic energies.
Bo: The pulley is above the zero line the whole time so
the pulley should have gravitational potential energy both initial and final, right?
Bobby: Oh, the pulley starts at the same height above the zero line
and ends at the same height above the zero line.
So, its gravitational potential energy does not change.
So, it would be the same on both sides of the equation and would cancel out.
So there is no need to add identical initial and final gravitational potential energies of the pulley
to the conservation of mechanical energy equation.
Billy: But the pulley does have rotational kinetic energy at the final point, because it is rotating, right?
Bobby: Right. That's right. I always forget rotational kinetic energy.
That means the two kinetic energies we got before of the two masses are translational kinetic energies.
Thanks Billy.
Billy: You are welcome Bobby.
Mr.p: Billy, please substitute in equations and continue from here.
Billy: Sure. Gravitational potential energy initial of mass 2
equals mass 2 times acceleration due to gravity times height 2 initial.
Gravitational potential energy final of mass 1
equals mass 1 times acceleration due to gravity times height 1 final.
Both final translational kinetic energies equal
one half mass time velocity final squared with their respective subscripts.
And the final rotational kinetic energy of the pulley equals
one half rotation inertia times the final angular velocity of the pulley squared.
And we know the initial height of mass 2 equals the final height of mass 1
because they both equal the distance y.
Just like we showed with the bike tire in a previous lesson,
we know the linear velocities of the two hanging masses have the same magnitude
as the tangential velocity of the rim of the bike wheel or, in this case, the rim of the pulley.
And tangential velocity equals radius times angular velocity.
So, velocity 1 final equals velocity 2 final
which equals the radius of the pulley times the final angular velocity of the pulley.
And we can substitute that into our equation
and multiply the whole equation by 2 to get rid of the one half’s.
Now let’s square both radius times angular velocity expressions.
Factor out the square of the pulley’s angular velocity final.
And we can solve for the square of the final angular velocity of the pulley.
It equals the quantity 2 times mass 2 times acceleration due to gravity times y
minus 2 times mass 1 times acceleration due to gravity times y
all divided by the quantity mass 1 times radius squared plus mass 2 times radius squared
plus rotational inertia.
That’s all I see to do right now.
Mr.p: Billy, this is great so far.
Okay, we already showed that the tangential velocity of the rim of the pulley
and the linear velocities of the masses have the same magnitude.
Do you also see how the linear distance travelled by the masses
has the same magnitude as the linear distance travelled by the rim of the pulley or the arc length?
Bo: Oh right. Arc length equals radius times angular displacement
so y equals the radius of the pulley times the angular displacement of the pulley.
So change in theta equals y divided by radius.
And we know the net torque is constant, so the angular acceleration of the pulley is constant,
so we can use the uniformly angularly accelerated motion equations.
We can use angular velocity final squared
equals angular velocity initial squared plus two times angular acceleration times angular displacement.
The initial angular velocity equals zero,
so angular acceleration equals
final angular velocity squared over the quantity 2 times angular displacement.
But we can substitute y over pulley radius in for angular displacement.
Bobby: And we can substitute in that giant expression we got for the
final angular velocity of the pulley squared.
Billy: And y cancels out
and we can factor out the square of the pulley radius from two expressions in the denominator.
And we get that the angular acceleration of the pulley equals
pulley radius times acceleration due to gravity times the quantity mass 2 minus mass 1
all divided by the quantity pulley radius squared times the quantity mass 1 plus mass 2
and then plus the rotational inertia of the pulley, still in the denominator.
Bo: Which is exactly what we got last time. That’s cool.
Bobby: Agreed. Billy: Yes. Cool.
Mr.p: Yes. This is the same equation we got last time for the angular acceleration of the pulley.
Turns out we can solve this problem using either conservation of mechanical energy
or the translational and rotational forms of Newton’s Second Law.
Thank you very much for learning with me today, I enjoyed learning with you.