Good morning. We have already derived simple harmonic motion equations for a mass spring
system and for a simple pendulum. Today, we are going to derive the angular frequency and period
of a physical pendulum using calculus! Calculus!
Yep. ♪ Flipping Physics ♪
Bobby, what is the equation which is the condition for an object to be in simple harmonic motion?
An object is in simple harmonic motion when the second derivative of the position of the
object with respect to time equals the negative of the square of the object’s
angular frequency times position. Correct Bobby. Thanks. Now,
just to be absolutely clear, class, remind me, what is a simple pendulum?
A simple pendulum is an idealized, simple version of a pendulum. Hence “simple pendulum”.
Right, in a simple pendulum, the string has a small enough mass
such that the mass of the string is negligible. And the pendulum bob is a point mass, meaning
it has no shape and zero size. Also, there is no friction,
either with the medium the pendulum bob is moving through or between the string and
whatever it is attached to at the very top. And the string of a simple pendulum is
inextensible, meaning the length of the string cannot be increased.
The length of the string will not change. Very nice. Today we are going to look at a
physical pendulum. The only thing we are going to assume now about the pendulum in simple harmonic
motion is that there is no friction and actually, no string. In other words, a physical pendulum is
a rigid body, suspended at some point, and it is swinging back and forth in simple harmonic motion.
The rigid body could be any shape. Here I have animated a random blob
as a rigid object which is oscillating back and forth in simple harmonic motion.
(Lovely) Yep. You can see the point where the rigid
body is fixed in place is the axis of rotation. I have also identified the center of mass of
the rigid body so you can more easily see the object is oscillating in simple harmonic motion.
Why is the center of mass of the rigid blob not at the center of the blob?
Uh, the rigid body must not have a uniform mass density.
Oh, right. Thanks. Why is the rigid body only swinging back
and forth a little bit? We could see the simple harmonic motion better if the angle were larger.
The simple harmonic motion equations for pendulums only work for “small angles”.
Yeah. I forgot that. Thanks. Great. Now, let’s pause the oscillation
and exaggerate the angle so we can better see the geometry of the situation, however, please still
remember, as Bo pointed out, pendulums are only in simple harmonic motion at small angles. Again,
a physical pendulum is a rigid body suspended at a single pivot point, the axis of rotation,
which oscillates without friction. Therefore, there is a force of gravity acting on the
center of mass of the rigid body which causes a torque around the axis of rotation. Bobby,
please determine the net torque about the axis of rotation and see where things go from there.
Sure. The net torque on the rigid body about the axis of rotation equals, well the only
force causing a torque about the axis of rotation is the force of gravity acting down at the center
of mass of the rigid body. Net torque also equals well it equals rotational inertia times angular
acceleration, and it equals the derivative of angular momentum with respect to time.
Which equation should I use mr.p? (Let’s use rotational inertia
times angular acceleration.) Okay. The torque caused by the force of
gravity equals the magnitude of position vector r, or the distance from the axis of rotation to the
center of mass, times the force of gravity, times the sine of the angle between the direction of
r and the force of gravity. And that torque is negative because the direction of the torque is
always opposite the direction of the angular displacement from equilibrium regardless of
whether the center of mass of the rigid body is to the left or to the right of equilibrium.
Let’s replace distance r with capital L sub c m for the length from the axis of rotation to the
center of mass. And replace force of gravity with mass times gravitational field strength.
And we can leave sine theta there. But how is the angle between r,
or the position vector to the center of mass, and the force of gravity,
how is that angle theta? The r position vector is
directed from the axis of rotation along down along the string toward the center of mass
and the force of gravity is straight down. The angle between those two directions is the same
angle theta as shown in the illustration. Oh, yeah. Thanks!
Looking back at the equation condition for simple harmonic motion, we can see we need to replace
angular acceleration with the second derivative of angular position with respect to time.
We can rearrange that equation to get the second derivative of angular position with respect to
time equals the negative of the quantity mass times gravitational field strength
times the length to the center of mass all divided by rotational inertia
all times the sine of the angle theta. And here is where we use the small angle
approximation. For small angles, or angles less than roughly 15 degrees,
sine of theta is approximately equal to theta, when theta is in radians.
Exactly. And you can see we now have an equation which fits the condition for simple harmonic
motion. The second derivative of angular position as a function of time equals the negative of
angular frequency times angular position. That means the square of angular frequency
equals mass times gravitational field strength times length to the center of mass
all divided by rotational inertia. And we can take the square root of
that equation to solve for angular frequency. And previously we showed period equals two pi
divided by angular frequency. We can substitute in what we got for angular frequency.
We get that the period of a physical pendulum equals two pi times the square root of the
quantity rotational inertia divided by the quantity mass times gravitational field strength
times the length to the center of mass. Nice. Agreed. Very nice. Now, just for fun,
let’s see what happens if we use this equation for the period of a physical pendulum with a simple
pendulum. Billy, what happens when we do that? Well, if it were a simple pendulum instead of a
physical pendulum, then all the mass would be concentrated at the end of a string.
The rotational inertia of a point mass about an axis of rotation equals the mass of the pendulum
bob times the square of r, the distance from the axis of rotation to the center of mass of the
point mass. We can substitute that equation for rotational inertia into the period of a
physical pendulum equation. Mass cancels out of the equation. r, the distance from the axis of
rotation to the center of mass of the pendulum bob is capital L, the length of the pendulum.
Actually, the length to the center of mass of the pendulum is also capital L. One of
the pendulum lengths cancels out and we are left with the period of the simple pendulum
equals two pi times the square root of the quantity length of the simple pendulum
over gravitational field strength. And we know that is the equation we derived previously for the
period of a simple pendulum. That is pretty cool! I completely agree Billy. Thanks. Now, realize all
the motion equations we solved for last time which describe the simple harmonic motion of a simple
pendulum are also valid for a physical pendulum, it is just that the angular frequency has changed.
Again, we have the same equations for the angular position, angular velocity, and
angular acceleration of the pendulum. For a simple pendulum angular frequency equals the square root
of gravitational field strength over pendulum length, which is different than the angular
frequency of a physical pendulum we just solved for. But, again, while the angular frequencies are
different for a simple pendulum and a physical pendulum, the basic equations of motion for simple
harmonic motion are the same. {Hear Bo} Yes Bo? (Mr.p) Can we show that the physics works now?
Yeah! (Yeah, let’s do that.)
Actually, something is bothering me.
May I ask a question first? (Sure Billy. But,
I feel like you just asked a question.) Right. Yeah. Uh, Way back in the problem when
Bobby asked you if we should use net torque equals rotational inertia times angular acceleration
or the derivative of angular momentum with respect to time, you chose rotational inertia
times angular acceleration. Why did you do that? I mean, if I were trying to derive this on my own,
I would not know which one to choose. Okay. Sure. The truth is you can solve this either
way. If we use net torque equals the derivative of angular momentum with respect to time,
we can substitute rotational inertia times angular velocity in for the angular momentum of the rigid
body. The rotational inertia of the body does not change as a function of time because it is a rigid
body and does not change shape, so we can take rotational inertia out from the derivative. And
the derivative of angular velocity with respect to time equals the second derivative of angular
position with respect to time. That gets us to the same equation we had before in the derivation;
the right hand side of the equation again equals rotational inertia times the second derivative of
angular position as a function of time. I think the way we did it before was a little bit easier,
and it also avoids the issue of having capital L stand for both angular momentum and length,
which is not good. But the physics works either way you choose to do the derivation.
Yeah. That makes sense. Thanks. Alright, now. To show that the physics works,
I have picked out an object we have already derived the rotational inertia of.
We are going to use a uniform, long, thin, metal rod which you can see I have set up such that it
can rotate about one end. And, we can set that physical pendulum into simple harmonic motion.
To determine the period of the physical pendulum, we need the time for one cycle. Rather than
measuring just one cycle, we are going to measure how long it takes for the pendulum to go through
10 cycles. And the time for 10 cycles works out to be 15.63 seconds. That means the observed period
of this physical pendulum equals 1.563 seconds per cycle. Given that the mass of this uniform, long,
thin rod is 0.667 kilograms and its length is 0.909 meters, Bo, please use all this information
to determine the observed and accepted values for the rotational inertia of this physical pendulum.
Sure. We already derived that the rotational inertia of a uniform, long, thin rod,
about one end equals one third times the mass of the rod times the length of the rod squared. That
means the rotational inertia of the rod equals one third times 0.667 times 0.909 squared or 0.18371
or 0.184 kilograms times meters squared with 3 sig figs. Oh, hold up. We need to rearrange the
period equation to solve for rotational inertia. Square both sides to get rid of the square root
and divide both sides by four pi squared. And rearrange the equation for rotational inertia.
We get rotational inertia equals period squared times mass times gravitational field strength
times length to the center of mass all divided by the quantity four pi squared.
We can now substitute in all the numbers. We get 1.563 squared times 0.667 times 9.81 times uh oh,
half the length of the rod because the rod has uniform density and its center of mass
is in the middle of the rod, so, 0.909 divided by 2. And then all divided by four pi squared.
That works out to be 0.18403 or 0.184 kilograms times meters squared.
(Very nice Bo. And what is the percent difference between the observed and the accepted values?)
Okay. Relative error equals observed value minus accepted value all divided by accepted value times
100 or woah both the observed and accepted value equal 0.184 kilogram meters squared
so we get zero percent relative error. Awesome! Yeah! Zero percent error!!
Bo and The physics works, the physics works, uh huh, uh huh.
Uh, wait a second. Bo and What? Yeah?
We should never round in the middle of a problem. We need to use the unrounded
numbers for the observed and accepted values for the period of the physical pendulum.
Oh. Right. Yeah. Okay. So, it works out to be
0.17360 or 0.174 percent relative error with 3 sig figs. That’s still pretty amazing though…
The physics works, the physics works, uh huh, uh huh, the physics works. The physics works,
the physics works, uh huh, uh huh, the physics works.
Thank you very much for learning with me today, I enjoy learning with you.