in this video we're looking at an atwood variant where the pulley is heavy
so i have a mass sliding down a frictionless incline it's connected to a string wrapped around
a heavy pulley with a massive am and a radius of r and as that mass slides down the incline it's
unwinding the pulley but remember the pulley has some inertia now so that slows things down we're
reminded in the problem that the formula for the moment of inertia of a disc is one-half mr squared
and we're kind of guided through how to find the solution to this problem in part a we're
told to set up two equations relating tension and acceleration by applying newton's second
law that's f equals m a to the block and then the purely rotating thing the pulley we're going
to apply tau equals i alpha to that that's the rotational equivalent of newton's second law
we're also reminded that the relationship between linear and angular acceleration
for a rotating object is a equals r alpha and ultimately our goal here is to find the
acceleration of this block as it slides down the ramp so the first step is to get all the relevant
forces into the diagram we'll go ahead and start with the tension acting at the ends of the string
and so there's the tension acting at the edge of the disc
here's the tension pulling up the ramp on the mass then the mass is also going to experience the
force of gravity pulling straight down and we want to decompose this into the standard components
and this is our standard diagram where we have gravity pulling straight down
and then the angle between the force of gravity and the perpendicular is the same as the angle of
incline for the ramp so if our force of gravity is mg this perpendicular piece is mg cosine theta
and the parallel piece which is the one we're concerned with is mg sine theta
now that perpendicular piece that determines the normal force but that would only matter if we had
friction in this problem all right one last thing i want to do before i start writing down equations
is to note that our disk is accelerating in the angular sense with an acceleration of alpha
and that the linear acceleration on the edge of the disk which is the same as the acceleration
of the string is given by this relationship a equals r alpha now let's apply newton's
second law to the sliding block in the parallel direction i'm going to call the actual direction
of acceleration positive so we apply newton's second law very responsibly starting from f net
equals m a and then our positive force here down the ramp is mg sine theta the tension pulls up the
ramp which makes it negative and there's our net force and that's equal to the mass of the block
times its acceleration so there's equation number one now we analyze the disk and we
look at the net torque on it and that's equal to i times alpha it's angular acceleration
and there's only one force exerting a torque on this disc it's that string pulling tangentially
and it has a radius of r from the center of rotation because it's coming off tangentially that
automatically makes it perpendicular which makes torque very simple to compute and my net torque is
just t times r and that's equal to i times alpha this is a disk so i is one half m r squared and
then finally instead of writing alpha i'm going to use this relationship a equals r alpha and solve
for alpha that's a over r one of those r's cancels on the right hand side between the numerator and
denominator the other one cancels by dividing by r on both sides of the equation so the r's
are all gone and we can clean up our system of equations that's mg sine theta minus t equals ma
and t equals one half m a now in part b we want to compute the acceleration of the block in terms of
g and theta so this system of equations is already set up to use elimination we just simply add them
and the negative t and the positive t cancel out so one of our variables is gone
on the left hand side we get mg sine theta and on the right hand side we get m a plus another one
half m a which is three halves m a now the m's are going to cancel and we solve for a
and obtain an acceleration of two thirds g sine theta now when i gave this problem on an exam
i needed a two part question at that point in the test so i didn't ask kind of the obvious follow-up
question here what's the tension in the string so i want to show that real quick all we have to do
is plug in the a we just got into the second equation and i get t equals one-half m a and we've
got the tension so this is kind of our bonus for this problem t is one half m a and i just figured
out a is two thirds g sine theta the twos cancel and i find a tension in the string of one-third mg
i can tell that's a force so that's good times sine theta if you enjoyed this video or at least
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