Why does the hoop have lower moment of inertia than the cylinder?

[cal35182]

Homework Statement

Hopefully you can see the equations I’ve attached...In my lab this week, we are rolling objects down a ramp. That gives us a time, so we are able to find the object’s acceleration. An object shaped like a hoop—a plastic lid—traveled faster than other objects, like cylinders—a AA battery. Ok, so everything would be fine, but I’m told that Moment of Inertia for a hoop should be higher: closer to I = MR^2, while other shapes have lower values, I = 0.5MR^2. That coefficient is what we are solving for, after we cancel out MR^2. If you look at the attached equations, and solve for I, youll get:

I + 1 = t^2 g sin θ / a(x)

So ***HOW DOES a higher value for a(x), acceleration, give me a larger value for the I coefficient?*** Again, faster objects like a hoop, should be closer to 1. But doesn’t having acceleration in the denominator mean these will be lower values? THANKS!

Relevant Equations

I = X * MR^2

a(x) = g sin θ / (1+I/MR^2)

Like I said, objects with the higher acceleration are giving me the lowest values. For a hoop, I got I=0.1*MR^2
For a cylinder, I got I=0.7*MR^2

this seems backwards, no?

 

[BvU]

Hello @cal35182 , !

Important lesson to learn here: In a lab you should evaluate immediately and not afterwards.
So that you at least have a chance to investigate why things go wrong -- as they will, according to Murphy's law

Your observation

An object shaped like a hoop—a plastic lid (*) —traveled faster than other objects, like cylinders—a AA battery.

Is rather contradictory to e.g. what we see here or here

With this lab you are in luck --- you can redo the experiment at home

(*) Is a plastic lid shaped like a hoop ?
are you sure a battery's mass is evenly distributed ?

 

[cal35182]

Allow me to simplify my question:

Regadless or what the shapes are, regardless of any expected values,
Don't objects traveling down the ramp with higher Acceleration...If you plug that in, to solve for moment of Inertia,doesn't higher acceleration give you a LOWER number in front of MR^2?

Let's say we are going down a ramp that 20 degrees. So g *sin θ =9.8 * sin (20°) = 3.352

So I will give you some different accelerations.

A(x) = 2 m/s/s, compared to A(x) = 3 m/s/s

...Remember, we are solving 1 + I/MR^2 = g sin θ/Ax
We are going to simplify that to 1 + X, IF I=X*MR^2

Okay

So anyway,

lol...

1+ X = 3.352 [g * sin θ]/ 3 [Ax]
X = 0.117

1+ X = 3.352 / 2
X = 0.676See? That's my only question--Is Higher Acceleration supposed to give a LOWER coefficient for Moment of Inertia
Because my teacher says otherwise

Thank U

 

[haruspex]

Is Higher Acceleration supposed to give a LOWER coefficient for Moment of Inertia

Yes.

If you look at the attached equations, and solve for I, youll get:

I + 1 = t^2 g sin θ / a(x)

No you don't. For a start, that equation is horribly inconsistent dimensionally. On the left you are adding a constant term to a term of dimension ML², and on the right you have T².
Try again.