in this challenge problem we're computing the work done by gravity as this simple pendulum bob
descends through a height of h now if you know about conservation of energy already
then you know there's a workaround to this that makes it so we don't have to use any calculus
and that's the whole point of energy methods is that they simplify your life a lot
but for now we're trying to do a direct calculation of the work so i'm going to
start to get force vectors in here and say okay gravity is doing what gravity always does
it's pulling straight down with a magnitude of mg so there's my force vector at this moment in time
and then i can look at a small increment of time going by
and i end up moving along this tiny little section of arc
that i'll call dr and then the work increment here
i may as well write that down and then we have to go back and do some geometry but
the work increment during that small that small little chunk of time is going to be given by mg
dotted
into dr and our goal is to integrate that
add up all the dws and find the total work done by gravity on the way down this path
quick side note when i set up the work increment i totally ignored the tension force
just intuitively but that should be explained why am i able to just ignore the tension force and
the answer is that the tension is always pointing exactly perpendicular to the direction of motion
because the direction of motion is tangential and the tension force is pulling radially and
this is actually not just a special case but it's a general truth about forces of constraints
so forces of constraints
that means a force that is used purely to keep an object on a particular path
so the force of constraint that we use most often is a normal force that's what
it's doing it's keeping an object in contact with the surface and normal forces are also
perpendicular to displacement so they do no work so tangent does no work here
and that's why we're allowed to ignore it so first off if i'm going to get this dot product
i need to find the angle right here i'll call it phi
and i'm going to take advantage of the fact that this angle
is theta the same as this angle of incline of the pendulum
so it looks like theta and phi must add up to 90 degrees so i could say that phi
is 90 minus theta and we'll just save that over here for when it comes in handy
all right then i have um another piece that i need to figure out i need to express h
in terms of l and theta my ultimate goal here is to be able to write down
an integral that's entirely phrased in terms of theta so i'm working on it
so this height right here that's also l this by the way is a really common geometry step
that has to be taken in calculus based physics and this little chunk right here
well that's a leg of a right triangle with a hypotenuse of l happening over here so
that's going to be a l cosine theta it's a cosine because it's the adjacent side
and that means i just figured out
that h plus l cosine theta is equal to l in other words h is l
minus l cosine theta which i might want to clean up a little bit
and then i'll put that in a box l times the quantity 1 minus cosine theta
so let's see um maybe i'll get back to working on dw
that's going to be mg dotted into a little dr which gives me mg times the magnitude of dr
times the cosine of the angle between them which would be the cosine
of 90 minus theta i see one more little piece of prep work that has to be done i need to get the
magnitude of dr and this is a little section of arc on a circle and we have a formula
i'm going to call that little angle d theta we
have a formula for writing down the arc length in terms of the radius
um and the angle subtended it's just s equals r theta or arc length is equal to radius times
the angle so i have just a little side note here i'll just put it over here in green that little
dr's magnitude is going to be l times d theta i just use the formula s equals r theta there
let's get that into our dw so i have mg times l
um and i'll put the d theta at the end and then the cosine of 90 minus theta there's an identity
for that that's the sine of theta and then a little d theta at the end this stuff out in front
is just constants so that's not a problem and then this stuff is all phrased in terms of theta
so it's starting to look like we can set up the integral
my final value of theta is going to be zero when i'm down here my initial value of theta
i may want to express that in terms of other quantities so let's write down the work integral
i'll just leave it like this for now because we have to work on these limits of integration
mgl
sine theta d theta all right
so my initial angle i suppose that i could write it in terms of other quantities hmm
maybe what i should do is take this expression and solve for theta that's my initial angle
i didn't need to solve for that height at all and a lot of problems you do so
i just kind of did it reflexively um so if i go back to this top formula
i get h minus l is negative l cosine theta if i then divide by negative l it reverses
the order on the left hand side because of the minus sign and i end up with a 1
minus h over l equals cosine theta which means my initial value of theta
is going to be the angle whose cosine is 1 minus h over l
let's plug that into our work integral and i have an initial theta of inverse cosine
one minus h over l and a final theta of zero and an integrand of mgl sine theta d theta
but there's one more trick lurking in this integral before we can just put on our calculus
hats and evaluate it and the issue is that d theta is actually negative here so if i look at
what's happening in this process theta is getting smaller that means my little d theta counts as a
negative number and that causes a sine error in the integral so the work done by gravity even
this little work contribution done by gravity it must be a positive number because i can see
that the angle between the force of gravity and the little displacement is less than 90 degrees
but d theta is a negative number so how are we going to fix it we just throw in a minus sign
to correct the problem now i can finally get to integrating this thing and i'm just going to pull
the constants out in front and then i have to guess the anti-derivative oops i forgot a theta
in there i have to guess the anti-derivative of the sine function and that's just negative cosine
and those minus signs will cancel out and then i need to evaluate this
from the angle whose cosine is 1 minus h over l
all the way to 0. and when i plug in 0 the cosine of 0 is 1 so my first term is going to be mgl and
then i'll have minus the thing that i get when i plug in the lower limit so that's going to be mgl
cosine of the angle whose cosine oh those undo each other
so the cosine of the angle whose cosine is this thing is just that thing
and i distribute this
and something nice happens to the first term i have an mgl but when i distribute
the minus sign to it i end up with mgl minus and mgl so those are gone
and then i have only one survivor i have negative mgl multiplied by negative h over l
and the l's cancel and i find out that the work done by gravity
is just mgh
now we're going to develop a theory of energy that allows us to bypass all this integration stuff and
with a bit of experience you could just look at the problem right from the beginning and say oh it
descended through a height of h i know the work of gravity is going to be mgh regardless of the path
you