in this video we quickly derive a formula for the gravitational potential energy of a rigid
body and then we apply the formula to an example so we start by looking at a rigid body and any
rigid body can be broken into point masses as we've shown in the diagram so that's m1 m2
m3 all the way to sum m n where n is the total number of particles that we broke it into in
addition we've shown the center of mass position vector pointing from the origin to the center
of mass of the rigid body and the formula for the center of mass position vector is given by
1 over the total mass of the object times the sum of each mass multiplying each position vector for
those masses so the sum of mi times ri and i'll post a link to the video where i derive that so
that's a vector equation and any vector equation can be broken down into its components so i could
write down the x coordinate of the center of mass as 1 over the total mass times the sum of
m i times the x coordinate of the ith mass and the y center of mass coordinate is similar
one over the total mass times the sum of each mi multiplied by its y coordinate now that second one
is the one that we want because we're looking for gravitational potential energy and that depends on
the y-coordinate not the x-coordinate so to find the total gravitational potential energy of this
rigid body we just write down the potential energy of every single point mass that it's made of
and each one of those point masses contributes an mgy so i'm going to sum the m i g y i terms
and i'm adding together the potential energy of every single point mass the rigid body is made of
next i can factor out the g because that's a constant common to every term in that sum and
i end up with g times the sum of m i times its y coordinate y i now i recognize that expression
from the center of mass y coordinate the sum of m i times y i is equal to m
multiplying the y center of mass coordinate and to write things in a little more familiar order
i have capital m g y center of mass and we've got our formula for the gravitational potential
energy of a rigid body so it's important to see the physical interpretation here we're taking the
total mass of the object multiplying by g and then multiplying by the y center of mass coordinate so
what we're doing is pretending the entire object has its mass concentrated at the center of mass
and then we're finding the gravitational potential energy of that point mass so let's apply our new
formula for gravitational potential energy to a quick example in this example we have a rod
of mass m it has a length of l so we're being very general here the rod is hinged at one end
and released from rest in the horizontal position and i want to compute the angular velocity of the
rod when it passes through the lowest point and that's when the rod is in the vertical position
in the second picture and we get a reminder here that the moment of inertia of a rod rotated about
one end is one-third ml squared so this is a pure rotation problem and we can figure out the final
angular velocity by using energy conservation and all i have to do is locate the center of mass in
my initial configuration and i'm going to go ahead and call the y coordinate there zero we're going
to put the origin at the hinge point and that center of mass is located a distance of l over 2
from the hinge but again the y coordinate of the center of mass and its initial position
is 0. then we follow that center of mass until we hit the vertical position and now my center
of mass has a negative y coordinate i have a reduction in gravitational potential energy
and that's going to be turned into the rotational kinetic energy of the rod so we write down the
formula for energy conservation i'll start right from the beginning e initial is equal to e final
and in my initial state i have zero gravitational potential energy simply because i put the origin
of coordinates at that height and i have zero kinetic energy so i'm just going to put a 0 on
the left hand side and in my final state if i look at my gravitational potential energy i have the
entire mass of the rod times g times the center of mass y coordinate which is negative l over 2
and i have kinetic energy in the rod and that's going to be a one half i omega squared
where i is given by this formula 1 3 ml squared so i'll go ahead and put that
gravitational potential energy term over on the left and i have an mgl over two
and then on the right i have one half times the moment of inertia which is one third ml squared
multiplying omega squared and omega is what we're after here we can cancel a factor of m out of this
and we can cancel a factor of l out of this and 2 times 3 that's 6. i'm going to multiply both sides
by 6 and i get a 6g over here and i'm going to divide by l on both sides so 2l in the denominator
and again that's all equal to omega squared so we solve for omega and i'm going to go ahead and
cancel a factor of 2 here giving me a 3g over l on that left hand side when i take the square root of
that i get that omega is the square root of 3g over l and that's just a quick check on our work
i want to look at the units of omega so inside that square root i have units of acceleration in
the numerator that's meters per second squared and i have units of length in the denominator
that's meters the meters are going to cancel and i get the square root of one over seconds squared
which gives me one over seconds and radians are unitless so i can just pop the radians into the
numerator and i get units of radians per second as i should and we're done if you enjoyed this video
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