in this video we introduce the cross product definition of torque
we derive a formula for the torque exerted by gravity on a rigid body and then we apply the
new formula to a short example so by now you're probably familiar with the definition of torque as
the perpendicular component of force multiplying the lever arm and this is often written as f
sine theta times r provided i have the angle labeled like i do in this diagram so there's
the perpendicular component of the force and i can see that that's given by an f sine theta and
we multiply by the length of the lever arm to find the magnitude of the torque exerted
with respect to that rotation axis and i'll note here that that torque exerted is clockwise
so as we get into more serious applications of torque it's advantageous to formalize the
definition of torque as a vector product called a cross product so instead of calling the torque
f perpendicular times r and then having to give its direction as clockwise or counterclockwise
we're going to give a vector definition of torque as r cross f where r is that vector that points
from the rotation axis to the point where the force is applied and then f is the force vector
applied at that point now the magnitude of a cross product is just given by the magnitude of the two
vectors that were involved in the cross product multiplied by the sine of the angle between them
so we still get the same magnitude of the torque out of this equation but the cross
product has this extra piece of information it actually has direction this is torque as a vector
and the direction is given by using what's called a right hand rule so to figure out the direction
of a cross product you need to visualize and i'm moving to the second diagram here
you need to visualize the tails of the two vectors located in the same place so i shifted f over and
attached the tail to the tail of r and there's a couple ways of doing this right hand rule my
favorite way is to put my fingers in the direction of the first vector so that's r sweep your fingers
out to the direction of the second vector through an angle less than 180 degrees so that means
this angle rather than going all the way through this angle that's bigger than 180 degrees
and if you do this sweeping motion your thumb is forced to be into the page and that's the
actual direction of the cross product and into the page is normally given by a circle with an x
in it and that's supposed to represent the tail feathers of the vector pointing into the page
there's a second popular way to visualize a right hand rule and that's to put your index finger in
the direction of the first vector so that's r put your middle finger in the direction of the second
vector that's f and then the orientation of your thumb will be the direction of the cross product
and again it turns out to be into the page so in the video clip here i'm showing the orientation of
my index finger middle finger and thumb and then i rotate into the correct position for the diagram
and my thumb is forced to point into the page now the direction of the torque vector is really
saying the same thing as before it's clockwise torque and you can always get the direction of
the torque out of the torque vector by putting your thumb in the direction of the torque vector
and then the fingers of your right hand are going to wrap around in the direction of the actual
twisting force so it's another sort of right hand rule put your thumb in the direction of the torque
vector your fingers are curling in the direction of the twist that's being caused by that torque
now that we have torque defined as a cross product we can approach this task of computing the torque
exerted by gravity on a rigid body and so in this diagram i have some rigid body that's been
broken into point masses m1 m2 m3 all the way to mn i've also labeled the center of mass of
this rigid body and in order to define torque you have to have a rotation axis in mind so if that's
at the origin given by this little white circle and all my position vectors are measured relative
to that rotation axis so to find the total torque exerted by gravity on this rigid body
we just look at the torque exerted on every single point mass that makes up the rigid body
and of course gravity is going to pull straight down on every one of these and every one of
those little weights is found by just taking the mass at that point and multiplying by g
and i'm going to put a vector hat on this to indicate that it's a vector force it points
straight down so m1g is the force of gravity on m1 and i'll do the rest and now i can write down the
total torque exerted by gravity by expressing all these torque contributions using the cross product
so the torque exerted on m1 is given by r cross f again r1 is a vector pointing
from the rotation axis to the location where that force is applied so i have r1 cross m1g
present r2 cross m2g and so on now in each of these terms the mi is just a scalar quantity
and i can attach that scalar quantity to either one of the vectors in the cross
product or i could even factor it out in front of each cross product what i'm going
to do in this case is move each of these m's next to the corresponding position vector
now each of these is crossed into a g vector which really means a magnitude of g and pointing
straight down and as it turns out and we're not going to prove this in this video the cross
product is distributive and i notice that every single one of these is crossed into the exact
same vector that's a g pointing straight down so the sum of those cross products could be rewritten
as the sum of all the m i times ri vectors crossed into this vector g pointing straight down
and if i switch to summation notation here this may become more familiar
that expression in parentheses is the sum of all the m i r i's
crossing that total vector into the vector g pointing straight down well that sum is familiar
it's related to the center of mass formula that's why i have a center of mass in my diagram
and here's just a quick reminder of that and i'll post a link to the video where it was derived
the center of mass position vector is 1 over the total mass of the object times the sum
of all the mis times their position vector ri so that sum in parentheses is actually the total mass
multiplying the center of mass position vector and to wrap up our derivation here again we're going
to take that scalar big m and we're going to move it onto the second vector instead of the first
and i end up with the total torque exerted by gravity is the center of mass position vector
crossed into mg where m is the total mass of the rigid body so that's an r cross f
where the force is the total mass of the body multiplied by g in other words the total weight of
the body and the position vector is the position of the center of mass so in other words the way
things turned out here if i just draw the total weight vector in there located at center of mass
the way things turned out is the total torque exerted by gravity can be found
by pretending all the mass of this rigid body is located at the center of mass
you find the weight at that point and then you use r cross f to find the torque and i should
put a vector hat on that torque and we've got our formula so let's apply our new formula to
a simple example in this example i have a rod of mass m it has a length of l it's hinged at one end
and i'm given the moment of inertia of a rod rotating about one end that's one-third
ml squared and what i want here is the initial angular acceleration of the rod so the way our
torque formula turned out for the torque exerted by gravity all i have to do is pretend that all
the mass has been concentrated at the center of mass of the object and then i compute the
torque at that location so i have the force of gravity exerted at the center of mass location
and that's using the total mass of the rod so that's capital m times g and then just by
symmetry here the center of mass of this uniform rod is located at a distance of l over 2 from the
rotation axis and i immediately have my torque and my weight vector is perpendicular to the lever arm
already so i'm just going to write down the torque quickly here as an mg multiplied by l over 2.
that's the perpendicular component of the force multiplied by the length of the lever arm
and of course this is acting in the clockwise direction and then i use the rotational equivalent
of newton's second law so this tau equals i alpha and i'll plug in the torque that's mgl over 2
and i'll plug in the moment of inertia that's 1 3 ml squared and alpha is the only unknown there
we're going to cancel an m and we'll cancel on l i multiply both sides by 3 and i get a 3 g
in the numerator on the left hand side divide by l and i get a 2l in the denominator and i
arrive at a formula for alpha it's 3g over 2l and the last thing i should probably do here
is just check the units on this expression to make sure that it makes sense so the units of
g that's meters per second squared the units of l that's meters i end up with total units here
one over seconds squared but radians are unitless and i can pop those into the numerator and i end
up with units of radians per second squared which are the proper units for angular acceleration
and we're done with the problem if you enjoyed this video or at least found it useful
check out another one by clicking one of the links on the left or click the zax lab logo on
the right to explore dozens of physics and math playlists as always you can leave your questions
comments and requests in the comments section below and i'll get back to you within 24 hours
thanks for watching zach's lab and best of luck on your math and physics journey