Calculating Flywheel Inertia Using Conservation of Energy Equations?

[forces9912]

Homework Statement

Problem in the images

Relevant Equations

mgh = 1/2 m v^2 + 1/2 I w^2 - Work
s = u t + 1/2 a t^2
a = v/t
I = 1/2 m R^2

I am stuck on what to do to calculate the inertia of a flywheel using the method described.
I am supposed to use conservation of energy equations to calculate the inertia.
I have a picture of the experiment and also the measurements I have taken.

It seems each method I try I get a different number for Inertia.
Any help on how to tackle this problem would be greatly appreciated.

 

[erobz]

The instructions seem to outline an approach. You must make a reasonable attempt at solving the questions before receiving help.

Also, for posting your equations please see the latex guide.

https://www.physicsforums.com/help/latexhelp/

 

[forces9912]

Sure, I will show you what I have worked out so far. I am stuck because the I value I worked out of 1.0273 kg/m^2 is very different to when I use I = 1/2 m R^2, where I got 1/2 * 4.01 * 0.08025^2 = 0.0129 kg/m2

 

[erobz]

For one thing, It should be + W_loss on the RHS, not minus. The initial potential energy goes into three reservoirs. Your equation says you start off with potential energy and the heat that is then converted to kinetic energy. Do you see why that doesn’t make sense?

Also…again please write math equations using latex. The link is in the last post, or there is another on the lower left corner of the reply box. Your other pick is illegible( I’m on my phone-too small), cleanly write equations using the formatting available on the site, so errors can be directly quoted and addressed.

 

[haruspex]

Did it really take over 10 seconds to traverse the 1m? That implies less than a tenth of the energy went into linear KE, which is hard to believe.

 

[forces9912]

Did it really take over 10 seconds to traverse the 1m? That implies less than a tenth of the energy went into linear KE, which is hard to believe.

Yes it did. It was a very gentle slope and also it was the shaft of the flywheel that was in contact with the surface. So most of the energy would have gone into rotating the central cylinder of the flywheel, if that makes sense.

 

[erobz]

@forces9912 Do you see why $W_{loss}$ has the wrong sign in your Work-Energy equation?

 

[erobz]

You also appear to be using the wrong radius $r$ in your calculations.

 

[haruspex]

You also appear to be using the wrong radius $r$ in your calculations.

Quite so. @forces9912, which radius should you use to find the rotation rate from the linear speed?

 

[Lnewqban]

Welcome, @forces9912 !

Perhaps irrelevant, but I believe that the complete shape of the flywheel (one disc and two journals) should be considered.
The above calculation of the moment of inertia is actually for a simple disc of wider dimension, resulting in a higher value for I.

Please, see:
http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html#mix

 

[forces9912]

Quite so. @forces9912, which radius should you use to find the rotation rate from the linear speed?

Right, I should be using the shaft radius of 0.01005m because that is the part of the flywheel that is in contact with the surface.

Welcome, @forces9912 !

Perhaps irrelevant, but I believe that the complete shape of the flywheel (one disc and two journals) should be considered.
The above calculation of the moment of inertia is actually for a simple disc of wider dimension, resulting in a higher value for I.

Please, see:
http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html#mix
View attachment 333973
View attachment 333974

Amazing graphic of the problem, thank you for that. I will have a read of that

@forces9912 Do you see why $W_{loss}$ has the wrong sign in your Work-Energy equation?

Yes I can see that now, thank you for making it clear
Potential Energy should = KE + KE rot + Energy loss.

 

[forces9912]

These are my new calculations for $I$

$0 = E-E_0 = (PE + K_{trans} + K_{rot}) - (PE_0 + K_{0trans} + K_{0rot})$

$0 = \frac {1}{2}Mv^2 + \frac {1}{2}Iw^2 - Mgh$

$0 = \frac {1}{2}Mv^2 + \frac {1}{2}I\left(\frac{v}{r}\right)^2 - Mgh$

rearranged for $I$

$I = \frac{Mgh - \frac{1}{2}Mv^2}{\frac{1}{2}\left(\frac{v}{r}\right)^2} = \frac{Mr^2\left(2gh-v^2\right)}{v^2}$

$v = \frac {2d}{t}$

$I = \frac{Mr^2\left(2gh-\left(\frac{2d}{t}\right)^2\right)}{\left(\frac{2d}{t}\right)^2}$

$I = \frac{Mr^2\left(ght^2-2d^2\right)}{2d^2}$

$I = Mr^2 \left( \frac {ght-2d^2} {d^2}\right)$

$M=4.01$
$r=0.01005$
$h=0.04$
$d=1$
$t=13.28$

$I = 4.01*0.01005^2 \left( \frac {9.81*0.04*13.28-2*1^2} {1^2}\right)$

$I = 0.00065$

 

[erobz]

These are my new calculations for $I$

$0 = E-E_0 = (PE + K_{trans} + K_{rot}) - (PE_0 + K_{0trans} + K_{0rot})$

$0 = \frac {1}{2}Mv^2 + \frac {1}{2}Iw^2 - Mgh$

Where did the frictional work term go? You should be solving for $I$ and $E_{loss}$ simultaneously using the (both) experimental results.

 

[forces9912]

Where did the frictional work term go? You should be solving for $I$ and $E_{loss}$ simultaneously using the (both) experimental results.

Right, so then my formula will end up like

$0 = (PE + K_{trans} + K_{rot} + E_{loss}) - (PE_0 + K_{0trans} + K_{0rot} + E_{0loss})$

$0 = \frac {1}{2}Mv^2 + \frac {1}{2}Iw^2 + E_{loss} - Mgh$

$E_{loss} = Mgh - \frac {1}{2}Mv^2 - \frac {1}{2}I\left(\frac{v}{r}\right)^2$

and then I put in the data from each experiment and set it to equal each other, and finally solve for $I$?

such as

$Mg(h_1) - \frac {1}{2}M(v_1)^2 - \frac {1}{2}I\left(\frac{(v_1)}{r}\right)^2 = Mg(h_2) - \frac {1}{2}M(v_2)^2 - \frac {1}{2}I\left(\frac{(v_2)}{r}\right)^2$

 

[erobz]

Correct, then sub that result into either equation to get $E_{loss}$.