An experiment question: Rotational Motion

[Warlord]

Homework Statement

I am doing the rotational motion experiment with a rotary disk that has three spindle sizes. A string goes around the spindle and is connected to a hanger. I calculated my ratio as a percentage of total final KE to decrease in potential energy and it comes out to be around 189 to 200%. I know 100% means ME is fully conserved but it's way greater than 100%. What does this mean and was ME conserved?

Homework Equations

$\frac{K_{final}}{mg\Delta h} · 100\%$

 

[Simon Bridge]

Welcome to PF;
Total energy is always conserved. IF you ever see more energy out than you put in, you did something wrong.
There are not enough details to work out what is happening - how are you working out KE(final) for example?
Is the setup like a yo-yo where you fix the end of the string and the disk rolls down it?

 

[Warlord]

Ah yes. There's the rotational motion apparatus with three spindles of different radiuses. The wire is wound on at say the smallest size spindle and it is let go. It stop when the hanger which is connected to the wire hits the ground. I calculated my KE(final) by adding KE(rot) = $\frac{1}{2}I\omega^2$ and KE(tr) = $\frac{1}{2}m_hv_f^2$ where $v_f$ is the final velocity and $m_h$ is the mass of hanger.
My trial 1 data goes like this if it helps: r(m) =0.0299m, $m_h$=.07005kg, $\alpha$=1.154 $rad/s^2$, a= .03450 $m/s^2$, $\omega_f$=10.454 rad/s, $v_f$=.3126 m/s. This gave me $K_{final}$=.9719J and a ratio of 189.53% which is huge so can you tell me what can cause this? Thanks

 

[Simon Bridge]

I still don't have a clear picture of what the experiment is - now it sounds like you drop a mass and the mass turns the wheel - so why would the wheel translate?

Your trial 1 data includes angular and linear acceleration - how did you measure that?
You appear to have independent measurements of angular and linear velocities ...
Accelerations do not appear in your equations so how did you use the acceleration data?

There is still not enough detail - pretend I am on the other side of the World and have no idea about what your experiment looks like.

 

[Warlord]

It's exactly that. The rotational motion apparatus is connected to the computer which measures angular acceleration and the final angular velocity. I used
v = r$\omega$ to calculate linear final velocity and a = r $\alpha$ to calculate linear acceleration. I used the linear acceleration to calculate Torque,$\tau$ = $rm_h(g-a)$ which is then used to calculate inertia, I = $\frac{\tau}{\alpha}$ which is used in the formula for rotational KE, KE(rot) = $\frac{1}{2}I\omega^2$. That's all there is to it.

 

[Simon Bridge]

It's exactly that. The rotational motion apparatus is connected to the computer which measures angular acceleration and the final angular velocity.

"rotational motion apparatus"?
What does that mean? If I don't know the experiment setup I cannot help you: there is not enough information to know what the experiment was doing.

I used $v = r\omega$ to calculate linear final velocity

OK - what is "r" in this equation ... radius of something yeah: but of what?

and $a = r \alpha$ to calculate linear acceleration. I used the linear acceleration to calculate Torque, $\tau=rm_h(g-a)$ ...

... what is $m_h$ the mass of?
... redo that derivation for me please.

... which is then used to calculate inertia, $I = \frac{\tau}{\alpha}$ which is used in the formula for rotational KE, $KE_{(rot)} = \frac{1}{2}I\omega^2$.

Pretend you are describing the experiment setup to someone who has never been in your class and has never seen the equipment you are using.
Being able to communicate this sort of stuff clearly is an important skill in science.
The way you are writing about it leads me to think you may not understand the experiment and what it is doing.

Here's an example of the sort of description I am looking for:
A wheel mass M radius R with a hub radius r. There is a string wound around the hub, with the free end fixed to something. The wheel is dropped so it rolls down the string a vertical distance h. Since gravity is the only energy source for the motion, we can write: $$Mv^2 + I\omega^2 = 2MgL$$ ... where v is the final speed of the center of mass of the wheel, I is it's total moment of inertia, and $\omega$ is it's final angular velocity.
In real life we expect the LHS < RHS in that equation.

The experiment uses a "magic box" to directly determine the final angular velocity and acceleration without influencing the motion.
You measure masses using a sensitive balance, and lengths with a good ruler.

If you define the LHS as the total KE, then the ratio you are looking for (which amounts to v^2/gL) should be less than 1. Always.
You can use the theoretical relations for angular velocity and moment of inertia to get an idea of what sort of figures to expect.

A ratio bigger than 1 means you have made a mistake someplace (because it implies the law of conservation of energy is violated) ... maybe there is another energy source you have not accounted for, or maybe you have put the wrong numbers in someplace.
Have you trued comparing your work with your classmates?