Mr.p: Good morning.
Today we're going to derive the rotational inertia of a long thin rod of uniform density
about three different axes of rotation and then demonstrate that our solution is correct.
"Flipping Physics" Intro
Bobby, what is the equation for the rotational inertia of a rigid object with shape?
Bobby: The rotational inertia of a rigid object with shape equals the integral of r squared with respect to mass.
Where r is the distance each infinitesimally small piece of the object is from the axis of rotation.
Mr.p: Billy, please solve for the rotational inertia of a long, thin rod of uniform density
about an axis perpendicular to its length and through its center of mass.
Billy: The rotational inertia about the y axis equals the integral of r squared with respect to mass.
That's weird. What does r have to do with mass?
Bo: Try starting with lambda, the linear mass density. It is constant for a uniform rod.
Bobby: Yeah, linear mass density equals the total mass of the rod divided by the total length of the rod,
but it also equals the mass of the
infinitesimally small piece of the rod, which is dm,
divided by the length of that infinitesimally small piece dm in the x direction, which is dx.
Billy: Okay, that all makes sense, but I still do not get how we integrate r with respect to mass.
The goal here is to substitute in for dm with an expression which includes dr, or actually dx
because "r", in this case, is in the x direction.
If we multiply both sides by dx, we have dm equals linear mass density times dx
and the linear mass density equals the total mass divided by the length of the rod.
So we can substitute that in for lambda. Billy, see if you can solve it from here.
Billy: Okay. Substitute mass over length times dx in for dm in the rotational inertia equation.
The total mass of the rod and the length of the rod are not dependent on position,
so mass over length can come out from underneath the integral.
and r is just a generic letter for direction, so let's substitute in x instead.
Which means we can now put in the limits for the integral,
x varies from negative one half L on the left side of the rod to positive one half L on the right side of the rod,
so those are our limits. The integral of x squared with respect to x is x cubed over 3.
We can substitute in our limits and cube both positive and negative L over 2.
We now have L cubed over 24 plus L cubed over 24 in the brackets, which is 2 L cubed over 24.
L cubed over L is just L squared.
So, the rotational inertia of a uniform rigid rod about its center of mass is
one-twelfth times the mass of the rod times the length of the rod squared.
Mr.p: Nice job, Billy.
Bobby, please determine the rotational inertia of a uniform rigid rod about one end.
So it's the same object, I've just moved the axis of rotation from the center of mass to one end of the rod.
Bobby: Okay, let's see. It's the same object, but we've moved the axis of rotation.
So it's actually the same integral, but with different limits.
We can start from the middle of the previous example with the rotational inertia about the y axis
equals the total mass of the rod divided by the length of the rod times the integral of x squared with respect to x.
And the limits are now...
Well, x initial is at the axis of rotation, so x initial is zero
and x final is at the other end of the rod, so x final is the total length of the rod, L.
Again, the integral of x squared with respect to x is x cubed over 3 from 0 to L.
Substituting in our limits gives us, mass over length times length cubed over 3 minus 0 cubed over 3.
One L cancels out, and the rotational inertia of a uniform rigid rod about one end equals
one third times the mass of the rod times length of the rod squared.
Mr.p: Absolutely, Bobby, thank you. Now, I said we are going to be able to test our solution.
To do so, we can use the rotational inertia demonstrator from Arbor Scientific.
Previously, we measured the rotational inertia of the central pulley system
to be 0.00067 kg times meters squared.
We can add four spokes to the rotational inertia demonstrator
and each spoke is a long, thin uniform rod.
Therefore, the total rotational inertia of the rotational inertia demonstrator
equals the rotational inertia of the pulley plus four times the rotational inertia of one long, thin, uniform spoke.
We just need the rotational inertia of one of the spokes.
Notice this is similar to what we have already done,
only the axis of rotation is no longer on the long thin uniform rod,
it is now located to the left of the rod a certain distance.
The rod screws into the pulley a small distance
and the inner end of the rod ends up being 3.05 cm from the axis of rotation of the pulley.
The length of the rod is 30.5 cm.
That means the initial position of the rod as a fraction of the length of the rod is
3.05 divided by 30.5 times L or 0.1 L.
And the final position of the rod as a fraction of the length of the rod is
3.05 plus 30.5 all divided by 30.5 all times L, or 1.1 L
In other words, the initial position is 0.1 L and the final position is 1.1 L.
So we go back to our integral and solve it using our new limits. Bo, please do that.
Bo: Sure. The rotational inertia of a single spoke
equals the mass of the spoke divided by the length of the spoke
times the integral from 0.1L to 1.1L of x squared with respect to x.
The integral of x squared with respect to x is x cubed over 3.
Substitute in our limits. Factor out 1/3 and cube 1.1L and 0.1L.
1.331L cubed minus 0.001L cubed equals 1.33L cubed.
And divide 1.33L cubed by 3L to get the rotational inertia of one spoke equals, with 3 significant digits,
0.443 times the mass of the spoke times the length of the spoke squared.
Mr.p: Thanks Bo.
Before we substitute in numbers, take a moment to appreciate
that we showed that the rotational inertia of a
long, thin rod increases as the axis of rotation gets farther from the center of mass of the rod.
And it should, because, as the axis of rotation gets farther from the center of mass of the rod,
more of the mass of the rod is farther from the axis of rotation.
In other words, the average "r" value increases.
That means the farther the axis of rotation is from the center of mass of the rod
the larger the resistance of the rod to angular acceleration.
All right. Let's plug in numbers and check our answer.
The mass of one spoke is 0.0742 kilograms and the length of one spoke is 0.305 meters.
Plugging those numbers into our equation gives us
a total rotational inertia for the rotational inertia demonstrator with 4 spokes
of 0.0129 kilograms times meters squared with three significant digits.
Previously we solved for the rotational inertia of an object
in terms of the pulley radius, angular acceleration,
and the force of tension acting on the pulley. So, I'm not going to repeat that here.
However, you are more than welcome to check out that video.
I'm simply going to use the equations from that solution with new numbers we collect right now.
It took the pully 133 frames or 2.216 repeating seconds starting from rest
to go through 2 revolutions or 4 pi rad.
That means the angular acceleration of the pulley was 5.11492 radians per second.
The radius of the pulley was 0.0286 meters
and the average measured force of tension while the system was accelerating was 2.452 newtons.
This gives us a measured rotational inertia of the rotational inertia demonstrator to be,
with three significant digits, 0.0137 kilograms times meters squared.
Using the relative error equation with the measured rotational inertia as our observed value,
and the predicted rotational inertia as our accepted value,
we get a difference of roughly 6 percent. I would consider that to be pretty darn close.
What do y'all think? BBB: Looks good to me. Seems close. Sure.
Mr.p: That's great. Thank you very much for learning with me today, I enjoyed learning with you.