# Homework Help: Moment of inertia lab graph reading assistance

1. Dec 2, 2013

### azhdrake

The problem
I am currently trying to right a lab report on a experiment trying to find a flywheels moment of inertia. In the lab we had a fly wheel connected to a mass on a string, and measured the position of a point on the wheel as the mass was accelerating downwards. The data was analysed with a program that generated a graph and has the line of fit that I came up with.
http://dl.dropboxusercontent.com/u/85778837/LabGraph.png [Broken]
X position vs time ---------------------------------------- -------------- Y position vs time
My problem is that I don't know how the equations relate to physical properties. For instance, the x position graph uses the equation "f(z) = -0.064 + 0.061 sin (-2.79 + 0.390088z2)" but I don't know how these numbers actually related to any physical properties.

2. Relevant equations
To find the moment of inertia I will use the equations mgh = Iω2/2 to find the moment of inertia.

3. The attempt at a solution
I really don't know where to start.

Last edited by a moderator: May 6, 2017
2. Dec 2, 2013

### collinsmark

Some of your fit equations seem a little weird to me, at least the ones in the picture.

If I'm imagining your setup correctly, then I would expect the x and y positions of a point on the wheel to both follow the general form,

$$x = C_1 + A \cos (\omega t + C_2)$$
$$y = C_3 + A \cos (\omega t + C_4)$$

where $C_1, C_2, C_3$ and $C_4$ are arbitrary constants. A is also a constant, and depends on the distance from the center of the wheel to the point being measured. Feel free to substitute sin for cos; it only changes what $C_2$ and $C_4$ are.

On the other hand, $\omega$, the wheel's angular velocity, is not constant, and is a function of time itself.

Assuming a uniform torque (which I think is reasonable here), $\omega = \alpha t$, where $\alpha$ is the wheel's angular acceleration and is a constant.

So what I suggest is to fix up the form of your fit equations, find a relationship between your variable z and time t, make your appropriate substitutions and solve for $\alpha$.

Not quite so fast Don't forget that as the attached mass falls downward, it gains kinetic energy too. You'll need to plug that in there as well. As part of the process, you'll likely need to know at what radius the string is attached to the wheel so you can determine a relationship between the mass's linear velocity and the wheel's angular velocity.

Edit: Oh, and azhdrake, welcome to Physics Forums!

Last edited by a moderator: May 6, 2017
3. Dec 3, 2013

### azhdrake

First of all, thank you so much for your help.

So what I am currently thinking is that the equations are in the form
$$x/y = C_1 + A \cos (\alpha t^2 + \omega t + C_2)$$
That would make it function similarly to linear kinetic equations, although I am not sure if $$\alpha t^2$$ should be kept as is, or have a constant attached like in the linear position vs time equation.

The variable z is equal to time already, so I am not sure what you are referring to when you say I should find a relationship between them.

Are you alluding to how $$m g h = I \omega^2 / 2 = m v^2 /2$$ or are you saying that energy is not conserved in this scenario? If energy isn't conserved, what is adding/taking energy out of the system?

4. Dec 3, 2013

### collinsmark

Okay, I see what you're saying, and that's not a bad idea. If the wheel has a non-zero, initial angular velocity $\omega_0$ at time t = 0, then you need to take that into account. Very good.

Also, I think I might have mislead you in my last post (sorry if I did). I forgot about a factor of 1/2.

The angular displacement in uniform angular acceleration is:
$$\theta = \frac{1}{2}\alpha t^2 + \omega_0 t + \theta_0$$

So you might wish to use instead,

$$x \ \mathrm{or} \ y = C_1 + A \cos \left( \frac{1}{2} \alpha t^2 + \omega_0 t + C_2 \right)$$

[Edit: and something that will come in handy later: the instantaneous angular velocity is $\omega = \alpha t + \omega_0$.]

Okay, that's fine. I wasn't sure. It wasn't obvious to me that z was already shown in units of seconds.

Energy is conserved. But you still need to modify your equation a little. The initial, gravitational potential energy is transformed into the rotational kinetic energy of the wheel plus the kinetic energy of the falling mass.

Last edited: Dec 3, 2013
5. Dec 3, 2013

### azhdrake

You didn't mislead me to much, I mostly just needed the nudge that the equation would be similar to the linear equations I am more familiar with.

Although now that I have an idea what I am looking at a new question arises. The angular velocities (0.391 for x and .03 for y) are off by an order of magnitude. Shouldn't they be about the same since the object is circular and rotating relativity smoothly? The origin was placed in a weird spot (the red lines in the graph are y=0) that meant that the positions numbers are going to be notably different for the x and y graphs, but I don't think that should effect the velocity much. Is there something that I am missing, or did my group and I just mess up fitting the equation?

Oh right! $$m g h = I \omega^2 / 2 + m v^2 /2$$ Thanks for pointing that out, that was silly of me for not noticing. That would be why you were telling me to find the relation between v and ω.

So if r is the radius of the spool that the string is rapped around, and d is the distance the string travels,
$$θr = d.$$
$$ω = θ/t$$
$$v = d/t = r * θ/t = rω$$

Is that right?

Last edited: Dec 3, 2013
6. Dec 3, 2013

### collinsmark

Yeah, that's what I meant when I said that your fit equations look a little weird to me. I would expect that the constant in front of the z2 under the sine/cosine to be the same in your x and y equations. Same thing for the constant in the z term.

Also, in your y fit equation, A seems to be function of z. I don't know what to make of that.

If it's not too late, you might consider tweaking your fit equations if you can.

If you can't, don't worry. It might not be a show stopper. The important thing you are solving for is $\alpha$, the angular acceleration. The initial angular velocity, $\omega_0$ looks to be fairly small, so it doesn't play as big of a role. The value of $\alpha$ is what dominates the behavior, and that is pretty similar in both of your fit equations.

Eventually, you will calculate out a $\omega$, at a time when things are speeding up, that you will use in your conservation of energy equation. But don't confuse that $\omega$ with the initial angular velocity, $\omega_0$.

Yes, that's right.

Be careful there. That equation is only valid for constant angular velocity which is not the case in this problem.

The $v = r \omega$ is correct.

Last edited: Dec 3, 2013
7. Dec 4, 2013

### Basic_Physics

You can get the speed of the point by differentiating your fitted equations with respect to time and then using pythagoras to get the resultant speed of the point. This speed can be used to determine the angular velocity of the point. The gradient of the angular speed vs time graph gives the angular acceleration. One can see that the recording was started while the point was already in motion since both graphs have a non zero gradient at t=0.

Last edited: Dec 4, 2013