in this video we're looking at a vertical Atwood machine with a heavy disc pulley and
this is framed in a very general way we have one hanging mass of M one hanging mass of 2m and then
the pulley itself has a mass of M and a radius of R what we're trying to find is the speed of
the system after it's fallen through a distance of D so the heavier Mass 2m is going to fall downward
the Little M is going to be pulled upward by a distance of D so we want the final speeds of those
masses after they've evolved through a distance of D and the new complication in this problem
is that some of that kinetic energy goes into the rotation of the heavy disc pulley as well so it's
going to end this process at an angular speed of Omega while the hanging masses will in the process
at a speed of V so we apply energy conservation and say e initial is equal to e final and we have
to define the vertical coordinate carefully for each of these hanging masses what I'm going to
do for the mass 2m is call its initial vertical position D and its final vertical position zero
and for the mass m I'll call its initial position 0 in its final position D so looking at all the
sources of energy in the initial State there's no kinetic energy because everything was released
from rest all I have is potential energies and with an initial y coordinate of zero the hanging
mass m has an initial potential energy of zero so the only energy source here is just the potential
energy of my larger Mass 2m that's given by the mass times g d so I end up with a 2m GD for my
initial energy in my final state if I just account for the potential energies first because my larger
Mass ends up at a coordinate of zero it has no gravitational potential energy but the mass m has
a potential energy of MGD all right so we can see why we end up with some kinetic energy in
the final State because our total gravitational potential energy has decreased so in our final
State we also have three sources of kinetic energy that's the translational motion of each Mass plus
the rotational motion of the disc so we'll just handle the biggest Mass first one half times the
mass times its final speed squared the smaller Mass one half times M times V squared and finally
the disc and that's one half I Omega squared where I is the moment of inertia and for a disk that's
one half m r squared and there's our Omega squared the final angular speed squared now in our final
State each mass is moving with a speed of V that's the same as the speed of the string and because
the string is not slipping that means it's the same as the speed at the outer edge of the disk
but we know that speed at the outer edge of the disc is related to the angular speed through the
relationship V equals R Omega and I'll post a link to the video where I first discussed that
relationship between the tangential speed and the rotational speed of an object so V is equal to R
Omega and conveniently we see an r squared Omega squared sitting in our expression for the kinetic
energy of the disk well that's just V squared where V is the translational speed along the edge
of the disk which is the same as the translational speed of our hanging masses so now every one of
our kinetic energy terms has an m v squared in it and some constants out in front and we need
to get these all combined I'll go ahead and cancel an M from every term in our energy conservation
equation before I write down the next step and I may as well subtract a g d from both sides while
I'm at it I had a 2gd on the left and one on the right so I'm left with one on the left so I have
GD is equal to let's see what our first term is one half of two gives me a One V squared and I'm
going to factor the V squared out my second term has a one-half V squared and my third
term is a one-fourth V squared so we'll combine our fractions with a common denominator of four
1 4 plus 2 4 that's the same as a half so that gives me three-fourths and
then plus the number one which is four fourths so I end up with seven fourths
V squared and now we're just about to solve for V we had to multiply both sides by 4 7 and square
root the result so I end up with a four g d over seven square rooted now in Part B we're asked to
compute the percentage of the kinetic energy that ends up in each of the moving pieces of the system
so that's each hanging mass in addition to the rotating disk and before we start I'd like to just
have a convenient expression for the total kinetic energy in the system if we go back to our original
energy equation those three terms on the right give you the total kinetic energy that was the
kinetic energy of each Mass plus the rotational kinetic energy of the disk so I'll just call that
k t for total and if I just subtract that MGD term from both sides I end up with an MGD on the left
hand side which means my total kinetic energy is just given by MGD that was the total change in
gravitational potential energy now I can just go through each term in my kinetic energy expression
and plug in the v that I found and we'll figure out what percentage of the kinetic energy is
in each piece so we start with the 2m piece k2m that's one half times two M times V squared the
one-half and the two cancel and I get an M times V squared and now I plug in my final V the squaring
undoes the square root and I get a 4G D over 7. so I end up with 4 7 MGD in other words 4 7 the total
kinetic energy in the system for the final State now we were asked for a percentage here so just
getting a decimal approximation and multiplying by 100 we find that 57.1 percent of the final
kinetic energy is in that larger Mass 2N now we look at the kinetic energy in the smaller mass
and we can see that was a one-half MV squared and we plug in our V and we end up with a one-half m
times 4gd over seven one factor of two is going to cancel here and I end up with a 2 7
MGD and turning that into a percentage I get 28.6 percent finally we look at the kinetic energy in
the rotating disk in its final State I'll call that KD and that was given by this final term
in our original energy equation and I could write that as 1 4 M times V squared remember r squared
Omega squared can be replaced with v squared V squared again is 4G D over seven the fours cancel
and I end up with 1 7 MGD and this gives me 14.3 percent now we've done a little bit of
rounding here but if we add up all these percents we're happy to find that we end
up with a hundred percent of the energy and we're done if you enjoyed this video or at
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