Mr.p: Good morning.
Today, we are going to measure the rotational inertia, or moment of inertia, of a bicycle wheel.
"Flipping Physics" Intro
To measure the rotational inertia of a bicycle wheel,
I have attached a known hanging mass to a string and wrapped the string around the bike wheel.
When I release the hanging mass, the torque it applies to the wheel through the force of tension
angularly accelerates the wheel and we can solve for the wheel's rotational inertia.
Please realize we are going to assume the axle of the bicycle wheel is frictionless.
Bobby, please draw a free body diagram of all the forces acting on the wheel and hanging mass
while the hanging mass is causing the wheel to angularly accelerate.
Bobby: Well, there is a force of gravity acting down on the center of mass of the wheel.
There must also be a force acting upward on the axle to hold the wheel up. Let's call that a force normal.
There's also a force of gravity which acts on the hanging mass.
Actually, there have to be forces of tension acting on the ends of the string, too.
So there is a force of tension acting upward on the hanging mass and
one acting downward on the wheel right where the string is last touching the wheel.
Those two forces of tension are on either end of the same string so they have the same magnitude.
Mr.p: Very nice Bobby.
Now we are going to use the rotational form of Newton's second law to sum the torques.
Bo, what do we need to identify when we use the rotational form of Newton's second law?
Bo: We need to identify what object, or objects, we are summing the torques on,
the axis of rotation, and the direction of a positive torque.
Mr.p: Great, Bo. Could you please do that?
Bo: Okay. Let's sum the torques on just the wheel with the axis of rotation at the axle of the wheel.
The wheel will spin in the counterclockwise direction, or out of the board.
So let's make that the positive torque direction.
Mr.p: Billy, please sum the torques.
Billy: Well, because
torque equals the r vector times force times the sine of the angle between the r vector and the force vector,
neither the force normal nor the force of gravity acting on the wheel cause a torque on the wheel
because each force acts on the axle of the wheel right where the axis of rotation is.
Therefore their r value is zero, hence, no torque.
So, the only torque acting on the wheel is caused by the force of tension acting on the wheel.
And it is positive because it will cause the wheel to rotate in the positive direction.
Net torque also equals rotational inertia times angular acceleration, or fishy thing!
Substitute in the equation for torque I gave you earlier,
and the r vector has a magnitude equal to the radius of the wheel,
the force of tension is just the force of tension,
but the angle between the force of tension and the r vector is 90 degrees.
The sine of 90 degrees is one. Therefore the rotational inertia of the bicycle wheel equals
the radius of the wheel times the force of tension all divided by the angular acceleration of the wheel.
Mr.p: Absolutely. Okay, we can measure the radius and angular acceleration of the wheel,
however, what about the force of tension in the string?
Bobby, do you have any ideas about how we could determine that?
Bobby: We could sum the forces on the hanging mass, right?
Mr.p: Yes, we could Bobby. Could you please do so?
Bobby: The net force on the hanging mass in the y- direction equals
the force of tension minus the force of gravity. The force gravity is negative because it's down.
Which then equals the hanging mass times the acceleration in the y-direction.
We can substitute in mass times acceleration due to gravity for the force of gravity.
And then solve for the force of tension. Then we can factor out the hanging mass to get the force of tension
equals the hanging mass times the acceleration in the y-direction plus the acceleration due to gravity.
Why hasn't he started writing on the board yet?
Bo: Yeah, this means something.
Billy: We must have done something incorrect.
Bo: It's the direction. Look at the direction of positive torque. It's counterclockwise,
which means the direction of positive force on the left side of the wheel is actually down.
Bobby: Right. That means the force of gravity is positive and the force of tension is negative.
When we make that change, the force of tension equals the hanging mass times the quantity
acceleration due to gravity minus the acceleration in the y-direction.
Mr.p: Please, always pay careful attention to direction
and therefore what should be positive and what should be negative.
I have substituted that equation for the force of tension
back into the equation we had previously solved for for rotational inertia.
Notice now, we have both a linear acceleration and an angular acceleration in the same equation.
Does anybody see a way to relate the linear acceleration of the hanging mass
to the angular acceleration of the wheel?
Bobby: The string is connected to the hanging mass and the outside edge of the wheel.
So as the hanging mass goes down, the edge of the wheel travels the same linear distance or arc length.
Billy: That means the linear velocity and linear acceleration of the hanging mass have the same
respective magnitudes as the tangent velocity of the tangential acceleration
of the outside edge of the bicycle wheel.
And tangential acceleration equals radius times angular acceleration, or in this case,
the radius of the wheel times the angular acceleration of the wheel.
Bo: And we can substitute that back into our equation for rotational inertia.
Mr.p: Absolutely!
Now, notice that everything in the equation for the net force on the hanging mass is constant.
This means the force of tension acting on the wheel is constant.
In addition, the "r" vector of the hanging mass and the angle in the torque equation are both constant.
This means the net torque acting on the wheel is constant.
The rotational inertia of the wheel is also constant.
Therefore, according to the rotational form of Newton's second law of motion,
the angular acceleration of the wheel is constant
and we can use the uniformly angularly accelerated motion equations.
BBB: U fishy M!
Mrp: Correct, the uniformly angularly accelerated motion or U fishy M equations.
We can use the U fishy M equation,
angular displacement equals angular velocity initial times change in time
plus one half times angular acceleration times change in time squared.
The initial angular velocity of the wheel is zero,
so we can solve for angular acceleration. (Billy: "fishy thing!")
Which equals two times change in angular position all divided by change in time squared.
We could then plug that back into our equation for rotational inertia, however I think we will leave it as
two equations instead because the rotational inertia equation would be rather cumbersome
if we substituted in our equation for angular acceleration.
So, given that we have two equations, one for rotational inertia and one for angular acceleration,
Bo, which variables do we need to know the value of
in order to determine the rotational inertia of the bicycle wheel?
Bo: Well, for the rotational inertia equation,
we need to know the external radius of the wheel, the mass of the hanging mass,
and the acceleration due to gravity.
We also need the angular acceleration of the wheel,
but to find that we need an angular displacement and change in time of the wheel.
Mr.p: The radius of the wheel is 0.332 meters, the hanging mass is 0.205 kilograms.
Here on planet Earth, the acceleration due to gravity is 9.81 meters per second squared.
As the wheel goes through a change in angular position of 185 degrees, the change in time is 1.28 seconds.
We need the angular displacement in radians,
so I have multiplied by two pi radians over 360 degrees to get 3.228859 radians
for the change in angular position of the wheel.
When we substitute our values for angular displacement and change in time
into the equation for angular acceleration, we get 3.941478 radians per second squared.
Which we can then substitute into our equation for rotational inertia.
Substituting in our values for radius, hanging mass, acceleration due to gravity, and angular acceleration,
we get 0.146800 or 0.147 kilogram meters squared with three significant digits
for the rotational inertia of this bicycle wheel. Billy: Mr. P?
Mr.p: Yes, Billy?
Billy: I have no frame of reference to even have an iota of a clue as to have 0.147 kilogram meters squared
makes even a scintilla of sense for the rotational inertia of a bicycle wheel.
Bobby: Yeah, I do not even know a modicum of rotational inertias to compare that to.
Bo: Perhaps you can help us remedy this dearth of real rotational inertia information we seem to have?
Billy: Please?
Mr.p: Ummm. Oh, there's the bell. We will have to address those issues next time.
Thank you very much for learning with me today, I enjoyed learning with you.
Billy: This does not really make sense to me.
Bo: That figures.
Bobby: When was the last time we heard the bell?