# How do you calculate theoretical ratios of angular velocity final and angular velocity initial? [closed]

I'm asked to calculate the "Theoretical Ratio". ($\omega_f/\omega_i$)

I have Inertia of the Disk = $I_{DISK} = .0094115713$,

I have Hoop Inner Radius = $I_R = 55.725$

So I am supposed to use the formula $\frac{I_D}{I_D + I_R} = \omega_f/\omega_i$

But I have 6 rows of data.. So would the Theoretical Ratio just be what ever I calculated for each row? Would the theoretical ratio always yield the same calculation? I'm confused what they are asking me to calculate here, I've also googled "Theoretical Ratio" and had no luck.

Also, I am told to find the Average "Measured Ratio of $\omega_f/\omega_i$, so would I just add up all of my data in my $\omega_f/\omega_i$ column and then divide by how ever many rows I have?

Data:

## closed as off-topic by ACuriousMind♦, user36790, Kyle Kanos, Bill N, GertNov 20 '15 at 15:27

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – ACuriousMind, Community, Kyle Kanos, Bill N, Gert
If this question can be reworded to fit the rules in the help center, please edit the question.

You don't explain what you are doing in the experiment, but I would guess it is supposed to demonstrate conservation of angular momentum. If we calculate the angular momentum $L$ using:

$$L = I\omega$$

then unless some external force is applied $L$ will be a constant.

When you start the experiment the disk has moment of inertia $I_\text{Disk}$ and is rotating at a speed $\omega_i$. So the initial angular momentum is:

$$L_i = I_\text{Disk}\omega_i \tag{1}$$

When you put the hoop onto the the disk the combined angular momentum of the disk and hoop is:

$$I_{\text{total}} = I_\text{Disk} + I_\text{Hoop}$$

where $I_\text{Hoop}$ is the moment of inertia of the hoop. If the final angular velocity is $\omega_f$ then the final angular momentum is:

$$L_f = (I_\text{Disk} + I_\text{Hoop})\omega_f \tag{2}$$

If angular momentum is conserved that means $L_i = L_f$, and that means the right hand sides of equations (1) and (2) must be equal:

$$I_\text{Disk}\omega_i = (I_\text{Disk} + I_\text{Hoop})\omega_f$$

and rearranging this equation gives:

$$\frac{\omega_f}{\omega_i} = \frac{I_\text{Disk}}{I_\text{Disk} + I_\text{Hoop}} \tag{3}$$

Now, in your experiment you are changing $\omega_i$, and therefore $\omega_f$, but you are keeping $I_\text{Disk}$ and $I_\text{Hoop}$ constant. That means the ratio $\omega_f/\omega_i$ given by equation (3) should be constant i.e. it will have the same value for each of your six experiments and it should be the same as the value you calculated in the Measured ratio column. That means the values in the Measured ratio column should all be the same as well, and actually they're all pretty close - the differences are presumably due to experimental error.

The column title Theoretical ratio is possibly a bit misleading. I would have called it Calculated ratio meaning it's what you calculate the ratio should be.

Incidentally, on your sheet you have $I_\text{Hoop} = 2204.900958$ (without any units!). I don't know where you got this from. The moment of inertia of an annulus is:

$$I_\text{hoop} = \tfrac{1}{2}M\left(r_\text{Outer}^2 + r_\text{Inner}^2\right)$$

And for your hoop I make this $I_\text{hoop} \approx 0.00508$ kgm$^2$.

• Also, I am told to find the Average "Measured Ratio of ωf/ωi, so would I just add up all of my data in my ωf/ωi column and then divide by how ever many rows I have? – Shammy Nov 21 '15 at 21:13
• @Shammy: yes, what you describe is a form of average called the arithmetic mean, and that's how most of us experimental physicists would calculate an average. You might also be interested in calculating the standard deviation, which will give you an idea of the ecxperimental error. – John Rennie Nov 22 '15 at 5:57