in this video we're solving a rigid body dynamics problem where the rotation axis
is actually moving so the setup here is that we have a string wound around the outside of
a disc we attach the string to a fixed surface above the disc and then we release it from rest
and this disc is going to fall down and it's unwinding at the same time so it's spinning
faster and faster as it falls and our basic toolkit for solving general dynamics problems
is first of all that the sum of all the external forces on a rigid body is going to be equal to
the mass times the center of mass acceleration and i'll post a link to where this was derived
and then second the sum of all the torques with respect to the center of mass
is equal to the moment of inertia with respect to the center of mass times the angular acceleration
with respect to the center of mass so the whole challenge of the problem is to set
up these two equations in terms of all the parameters that we're given for the problem
this should result in a system of two equations and two unknowns and we should be able to solve
for what we want so in part a we're trying to find the acceleration of the disk and the tension in
the string and just as a qualitative check on our work that acceleration should be less than free
fall because i have a tension pulling upward on the disk so it's going to be something less than g
and the tension should be something less than the weight of the disc otherwise the disc wouldn't be
accelerating downward so to get into the solution we're going to start by looking at all the forces
that are acting on the disc and i have the force of gravity and that's big m g and we're going to
locate that on the center of mass and then i have the tension and that's coming off the edge of the
disc so tangential to the disc and that points straight up so when i apply my first equation
the sum of all the external forces equals mass times center of mass acceleration
i'm going to call downward positive here so that a is positive and i get mg minus t is equal to the
total mass times the center of mass acceleration then i get to my torque equation and the force
of gravity is not exerting any torque with respect to the center of mass here because for the purpose
of computing torque we can locate the total force of gravity at the center of mass well that's the
rotation axis under consideration so the torque exerted by gravity vanishes the only torque that
we have in the problem is the torque exerted by the tension and that's acting through a lever arm
of magnitude r the radius of the disk and the tension is already at right angles to that
radius so the torque equation gets really simple it's just tension multiplied by r so that torque
is going to be equal to the moment of inertia with respect to the center of mass and that's one half
mr squared for a disc multiplied by alpha with respect to the center of mass in other words
the angular acceleration around the center of mass and so one of the challenges of the problem is we
have to relate the angular acceleration around the center of mass to the linear acceleration of the
center of mass and the key to this problem is to realize that we're just looking at another case of
rolling without slipping so if i turn my head sideways this is totally equivalent to seeing a
disc rolling along the ground this way where it's not slipping in other words the point of contact
right here is always stationary because the disc is contacting a stationary string
there's no slipping between the string and the disc so with that contact point along that
vertical line always being stationary this disc is really rolling without slipping on its way down
and i'll post a link to the video where i derive this but there's a nice relationship between
center of mass velocity and angular velocity when we're rolling without slipping and it
turns out to just be that the speed of the center of mass is equal to r times omega where omega is
the angular velocity of the disk well we can take the time derivative of this and get the center of
mass acceleration is equal to r times the angular acceleration around that center of mass and this
gives us a substitution allowing us to get rid of the angular acceleration in our second equation
it's just the center of mass acceleration divided by the radius of the disk so my second equation
becomes t times r the torque is equal to one half m r squared times center of mass acceleration
divided by r so now one of these r's is going to cancel on the right hand side of that second
equation and i can cancel an r from the left side of the equation with the remaining r on
the right side of the equation and now our system of equations is getting really simple again the
force equation was mg minus t equals m a center of mass and then my torque equation has reduced to t
equals one half m a center of mass and this equation is set up to solve by elimination
very simply i just add both equations and the t's are going to cancel and this gives me mg
on the left hand side and then i have an m a plus a one half m a that's three halves
m a center of mass and the m's are going to cancel and i arrive at an expression for the center of
mass acceleration that's 2g over 3 or 2 3 g and just to check with the comment that i made at
the beginning here the center of mass acceleration should definitely be less than free fall because
i have a force pulling up on this disc and two-thirds g is definitely less than g itself
so then the second part is to find the tension in the string
and that pops out of our second equation really easily the tension in the string
is just one half times the mass times the center of mass acceleration which we now have
that's 2g over 3. i can cancel the twos there and i find that the tension is equal to 1 3 mg in
other words 1 3 the weight and again this squares with the comment that i made at the beginning of
the problem the tension better be less than the weight otherwise the net force wouldn't point
downward and the thing wouldn't accelerate so then in part b we get into a kinematics
problem i want to use the acceleration that i just computed to figure out the speed of the
disc after it falls through a distance of d and this is just an acceleration from rest through
a distance of d so i'm going to write down the kinematics equation v squared is v naught squared
plus 2a times d again i'm treating a as positive downward is the positive direction here and i'm
starting at a location of x equals zero finishing at a location of x equals d and this started from
rest so v naught vanishes and i can plug in my acceleration that was two g over three
and i take the square root of both sides and i find that v is the square root of four g d
divided by three now the reason i asked part b is because it opens up the possibility of checking
our answer using part c so in part c we're going to take a different approach to the entire problem
and we're going to use energy conservation to predict the speed after the disk has fallen
by a distance of d so the initial height of my center of mass i'm going to call that y
equals d and then the disc will fall all the way down say to here i'll call that y equals zero and
i'm trying to get the speed of the disc by the time it's fallen through that total distance
so i apply energy conservation to this and in the initial state everything is stationary so
there's no kinetic energy it's all just the potential energy of this rigid body
and that's given by pretending all the mass is located at the center of mass and i'll post a
link to the video where i derive that fact so we pretend the disc is a point mass located at
the center of mass and the potential energy then is going to be m g times d in my final state my
center of mass is dropped to a height of zero so there is no potential energy and all i have
is kinetic energy but it's two different types of kinetic energy i have the spinning of the disc and
the translation of the disc so the translational kinetic energy piece is going to be a one-half m
times the center of mass speed squared and then the rotational piece is going to be one half i
with respect to the center of mass so the moment of inertia with respect to the center of mass
times the angular speed with respect to the center of mass squared and again this thing
is rolling without slipping i know it's totally equivalent to something rolling along a level
surface because i see that contact point with the string is always stationary and that allows
us to make a substitution for omega i go back to my rolling without slipping condition and omega
is going to be the center of mass velocity divided by r so in the next step i'll go ahead and plug in
the rolling without slipping condition and i'm going to plug in the moment of inertia of the disc
that's one half mr squared and omega is v center of mass over r and i still have to square that
so when i square that the r squareds are going to cancel so those are gone i also see that the
mass cancels out of both sides of the equation so those are gone and on the left hand side i
end up with just the g d and on the right hand side i have a one half v center of mass squared
plus a one fourth v center of mass squared and one half plus one fourth that's three fourths
and now i multiply both sides by four thirds and take the square root to get v center of mass and
i arrive at the same formula four g d over 3 square rooted for the center of mass velocity
after it falls through a distance of d now solving the problem in two different ways gives me a high
degree of confidence that i've got everything right and we're done if you enjoyed this video
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