# How to find the moment of inertia K from rolling an object down a ramp

1. May 1, 2012

### ranger275

1. The problem statement, all variables and given/known data
I am teaching a class where we are trying to find the K of I=1/2Kmr^2 from rolling a very thin cylinder down a ramp.

2. Relevant equations
Linear Forces: mgsinθ-F(friction)=ma
angular forces: τ=Iα where τ=F(friction)r, I=Kmr^2 and a=αr
distance=1/2*a*t^2 since the initial velocity is 0

3. The attempt at a solution

If you solve for F(friction) for both linear and angular forces you get
Ia/r^2=mgsinθ-ma which gives
a=gsinθ/(1+I/mr^2)=gsinθ/(1+k)
If you solve for K you get K=gsinθ/a-1
Using the distance formula you get K=gsinθt^2/(2d)-1

We are running the lab using a video camera which allows us to time to the nearest .03 seconds and the distance measurements are within .005 meters but we are getting a K for a very thin cylinder manufactured on a 3-D printer of 1.2. I can't figure out how there is that much error in the lab so I assume I am leaving something out in my calculations. Any help would be appreciated.

2. May 1, 2012

### LawrenceC

I solved it by conservation of energy considerations and get:

K = (2*g*sin(theta)*t^2)/d - 1

so it looks like there is a factor of 4 difference in results.

I also did it your way using forces and torques and get the same result. Check your algebra.

3. May 1, 2012

### Staff: Mentor

The algebra in post #1 looks OK to me. I also did it via energy methods and got the same result as in post #1. (So one of us is making an algebra mistake! :uhh:)

4. May 1, 2012

### ranger275

When we do the lab with a solid disk we get a K of 1/2 like we are supposed to and when we use a metal ring we get a K of .91 which is believable because the thickness of the ring is about 5% of the radius. If my equations are a factor of four off then my lab results are even worse than I thought. Because the thin ring is so light we thought air resistance might be a problem so we lowered the ramp until it took around 5 seconds instead of around 2 seconds but we still got K's in the 1.2 range so it doesn't appear air resistance is the problem. It may be something wrong with our set-up which people on this forum couldn't see unless it is possible to put a video on here.

5. May 1, 2012

### Staff: Mentor

I also solved this problem, and got the same result as you and Doc Al did.

I would like to hear more about the experimental setup and the data processing. How did you determine K from the experimental data? I would have plotted d vs t on a log-log plot, and confirmed that I get a straight line, and that the slope of the line is 2. If the line is curved, or the slope is not 2, then this should be telling you something.

What was the length to diameter ratio of the cylinder? Did you try the test with a series of L/D ratios? Were the ends of cylinder open? If the L/D was an issue, the K should approach 1 as L/D becomes large.

The results should be independent of the radius of the cylinder and the density of the material. Have you tried the test with different radius cylinders, and with different materials?

How sure are you that you measured the angle correctly? Was the table properly leveled?

You should be able to get a very rough estimate of the air drag on the cylinder by using the results for fluid flow past a cylinder. This doesn't have to be exact. Just assume a constant drag coefficient based on the highest speed. Use this to get the order of magnitude of the drag force.

6. May 1, 2012

### ranger275

X R K X/R
Black Disk 0 76.5 0.50 0.00
TP Large 9.25 72 0.64 0.13 0.02
TP Small 8 55.5 0.65 0.14 0.02
Silver Ring 72 74 0.91 0.97
Masking Tape 38.5 51.8 0.91 0.74
Duct Tape 38.5 57.5 0.94 0.67
Small Masking 39.5 45.5 1.02 0.87
Small Duct Tape 38.5 41.5 1.18 0.93

(Sorry I lose the columns when I post it! So I attached a file.)
This is the data from our first runs. x is the inside radius (so the black disk is solid) and r is the outer radius. We measured used sin=opp/hyp so we didn't have to measure the angle. The ramp was a metal door from a bathroom stall. So the gsinθ/(2d) was fixed and measured to the nearest millimeter. We have had to change the set-up and remeasure a couple of times and the gsinθ/(2d) doesn't change enough to make much difference in the K. The table's angle didn't change once we set it up. We used the K=gsinθt^2/(2d)-1 formula derived in the first post to calculate K. We used a digital camera to measure the time to the nearest .03 seconds. When the masking tape didn't have a smaller K than the silver ring I thought it might be due to the cardboard and tape having different densities. I also thought that was why the small duct tape had a K of 1.2 (the amount of tape was about the same width as the cardboard core). But we used a 3-D printer to make a cylinder open at both ends which was paper thin and measured K's of around 1.2 for both a small cylinder and a larger one (the data wasn't entered in the above table). I don't see how we can be 20% off the predicted value as careful as we were with the measurements. I don't follow how to estimate the air drag, but there was no difference in the measured K when we lowered the table which more than doubled the time. If air resistance was an issue wouldn't slowing the speed of the cylinder change the measured K? Thanks for all the help!

#### Attached Files:

• ###### k question.doc
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Last edited: May 1, 2012
7. May 2, 2012

### LawrenceC

Whoops, sorry....

8. May 2, 2012

### Staff: Mentor

Thanks for the data. This is very helpful.

It is possible to derive an equation for the moment of inertia of an annular cross section. From this, you get K as a function of the ratio of the wall thickness to the outer radius:

K = 1 -T +T2/2

where T is the ratio of the wall thickness to the outer radius.

I plotted your data as K versus T on a scatter plot, and compared it (on the same plot) to a graph of the above theoretical equation. In 6 out of the 8 cases, your values of K were higher than the theoretical values (by about the same percentage as the 1.18 vs 1.0)

I wish that you had measured d as a function of t all along the incline. This would tell us much more.

Chet

Last edited: May 2, 2012
9. May 2, 2012

### ranger275

I still have some of the video. I will see if I can do that. Thanks!

10. May 2, 2012

### Staff: Mentor

If you have sufficient data, I suggest plotting d (ordinate) versus t2 (abscissa) on arithmetic paper. You should get a straight line through the origin with a constant slope equal to a/2. If the line is straight, you can do a least squares fit (using your graphics program capabilities) to get the slope. This should give you your best estimate of K. If the slope decreases with increasing t, you probably have air drag.

11. May 2, 2012

### ranger275

I will do the plot but for your info I've attached the data and put it below. The gsinθ/2d was .071 for this run. I'm going to bring a level in tomorrow to see if there is enough of a dip to cause a problem. Thanks for the help!

d t
5.2 1
22.5 2
55.4 3
46.8 4
149.0 5
201.0 5.77

#### Attached Files:

• ###### finding k d vs t.doc
File size:
29 KB
Views:
87
12. May 2, 2012

### Staff: Mentor

The data point at t = 4 should be 96.8.

I plotted this up the way we discussed and pretty much got a perfect straight line. At least for this run, it doesn't look like air drag was a significant factor.

Chet