How Does Friction Impact the Energy Calculations in a Rolling Ball Experiment?

In summary: So if you have a sphere with a radius of say, .5 ft and weigh 1 lb, the force on the sphere due to air drag would be .05 lbs.
  • #1
greep
2
0
My group was assigned to make a lab about potential/kinetic energy and present the lab to the class. The lab we designed was to roll a ball down an incline, which then made a loop upwards like a rollercoaster, and calculate how high we need to set the ball on the incline in order that the ball completely make the loop without dropping.

"without friction" we used the following formula, which worked perfectly for the ball that had little friction:

Kinetic Energy Final + Potential Energy Final + Rotational energy (of a sphere) = KE Initial + PE Initial

For Velocity final we used √(g*(radius of the circle)

The slick ball was off by 3 cm (62 cm theory, 65cm experiment), but the rougher ball was off by an entire 20 cm. So we decided to try and find the energy lost due to friction. Here's where things go downhill.

To find the coefficient of rolling friction we set up another experiment where we just used a stopwatch and see how fast each ball rolled down the incline part of the track, tilted to be almost flat so we could actually measure the time right. I'll use some hard numbers here since this might be where the problem is.

Avg time for "White ball": 2.4s
Avg time for "yellow ball": 2.9s
Horizontal length: 70cm
Vertical length: 2.5cm
sinθ: 2.05 degrees

To figure out the coefficient we just used ma = mg - μmgcosθ.

For "a" we used some kinematics. Vfinal^2 = Vinitial^2 +2ad. Vinitial is 0, leaving

a = Vfinal^2/2d

And Vfinal is just 2Vavg

So we got

μ(White ball) = .011
μ(Yellow ball) = .019

Which makes sense since rolling friction is pretty small. Unfortunately, I didn't really know what to do at this point. What I tried was alter the original equation like this:

KE(final) + PE(final) +RE(final) + Friction(incline) + Friction(semi-circle) = PE(inital) + KE(Final)

giving this: (H = diameter of circle, h = height of ball at start)

.5mv^2 + mgH + .5(Iω^2) + (μmghcosθ)/sinθ + .5μmg(Pi)(H/2) = mgh

I got the (μmghcosθ)/sinθ by μ(Normal force)(distance), where distance = h/sinθ

I got the .5μmg(Pi)(H/2) also by μ(Normal force)(distance). distance is (Pi)(H/2), and normal force I just made as a rough average of 0 at the top and μmg at the bottom, yielding .5μmg. Not perfect I know, but certainly not a big difference.

Solving for h... only changed the original 62cm by a few centimeters for each ball, nowhere near 20 cm for the yellow ball. The error must lie somewhere in the two frictions I used, if I'm even supposed to due frictional energy this way.

Anyways, sorry if this is a mess, but does anyone know where I went wrong? Was the frction incorrectly added to the energy equation? Wrong numbers? Other factors? If any more info is needed, I can change this. We can probably can get a decent grade without getting the friction since we're designing our own lab, but this is really bugging me :/
 
Last edited:
Physics news on Phys.org
  • #2
What is the weight of each ball? I'm wondering if you'll need to consider air drag on the ball. The drag due to the air is a function of the velocity squared. It is a separate issue from rolling friction.
 
  • #3
Well the white ball was 42.33 g and the yellow was 9.69 g, and the white ball had a much bigger radius. We sort of glossed over drag, so I'm not sure what information is needed. However, I thought friction was generally a lot stronger than air resistance anyways.
 
  • #4
Well, you can check it out. Look up the drag coefficient for a sphere. The force on the sphere due to air drag is:

F = rho*A*Cd*V^2/(2*g)

where
rho is air density in pounds/ft^3
A is area pi*d*d/4 in ft^2
Cd is drag coefficient - look it up for a sphere
V is velocity in ft/sec
g is gravitational acceleration in ft/sec^2

Units on F are pounds force.
 
Last edited:
  • #5


I can see that you have put a lot of thought and effort into your lab and calculations. Your understanding of potential and kinetic energy is evident in your use of the appropriate equations and your consideration of friction as a factor in the energy transfer.

It is possible that the discrepancies in your results could be due to some experimental error or inaccuracies in your measurements. It is also possible that there are other factors affecting the motion of the ball, such as air resistance or the shape of the ball itself.

In terms of your calculations, it seems that you have taken into account the frictional forces in the incline and the semi-circle, but it may be helpful to also consider the friction at the top of the loop. Additionally, the coefficient of rolling friction may vary for different surfaces and materials, so it could be worth exploring this further in your experiments.

Overall, I appreciate your attention to detail and your determination to find the source of the error. As scientists, we know that experimentation and analysis can lead to unexpected results, and it is important to continue exploring and questioning to improve our understanding. Good luck with your lab and keep up the scientific curiosity!
 

Related to How Does Friction Impact the Energy Calculations in a Rolling Ball Experiment?

1. What is friction?

Friction is a force that resists the motion or potential motion of two surfaces that are in contact with each other.

2. How is friction related to total energy?

Friction can convert kinetic energy (energy of motion) into thermal energy (heat), thus reducing the total energy of a system.

3. What factors affect the amount of friction between two surfaces?

The amount of friction depends on the type of surfaces, the force pressing the surfaces together, and the roughness of the surfaces.

4. Can friction ever be beneficial?

Yes, friction can be beneficial in certain situations, such as when it helps us grip objects or when it allows us to walk without slipping.

5. How can we reduce friction?

Friction can be reduced by using lubricants, such as oil or grease, between two surfaces, or by making the surfaces smoother.

Similar threads

  • Introductory Physics Homework Help
Replies
32
Views
2K
  • Introductory Physics Homework Help
2
Replies
60
Views
451
Replies
10
Views
512
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
14
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
909
  • Introductory Physics Homework Help
Replies
20
Views
2K
  • Introductory Physics Homework Help
Replies
8
Views
3K
Replies
39
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
1K
Back
Top