Designing a Ramp for Consistent Ball Rolling: Exploring Formulas and Variables

[Crystal Faye]

I have to make a ramp, eek!

Here is my assignment. I would appreciate help with choosing what formula to use. I am so lost.

Design a Ramp:​

The ramp can be made of any material and be any width, length, and/or height. The only rule is that there must be a ball of any material that will consistently roll from a noted point A to a noted point B on this ramp.

The ball much be rolling from point A to point B in 1.8 consistently over a one meter stretch. [ It is also an option to create a scenario in which the ball with take .9 seconds to travel over a .5 m ]

 

[andrevdh]

The linear acceleration of an object rolling down an incline, without slipping, with an angle of $\theta$ with respect to the horizontal is given by

$$a = \frac{g\sin(\theta)}{1 + A}$$

where

$$A = \frac{I_{com}}{MR^2}$$

where $I_{com}$ is the rotational inertia of the rolling object around its centre of mass.

 

[Crystal Faye]

Thanks for the help :)

What ball and ramp material do you recommend?

And also, if the ball must travel 1 meter in 1.8 seconds [or .5 meters in .9 seconds], should we get the ball to accelerate BEFORE the "measured meter".

We are only in our first few weeks of physics.

 

[andrevdh]

The ball need to be of a uniform material (not hollow) because you need to evaluate $I_{com}$. It should be rather hard, but it must not slip on the incline while going down.

 

[Crystal Faye]

Do you think that a marble will work? We are considering creating the ramp (the surface) out of lamenated card stock.

We also have bouncy balls at our disposal :)

I am really happy that you are helping, yay!

 

[andrevdh]

I would rather go for a ball that has less chance of slipping, like a superball. If the ball is of such dimensions and mass that the angle becomes too steep the chances of slipping becomes great. So calculate $\theta$ for the particular ball (working backwards from the given time to the acceleration to the angle) and then you will have an idea if the angle is such that the ball will probably slip.

 

[Crystal Faye]

ok! I am going to try and draw a diagram...

 

[andrevdh]

I am not sure why you need a diagram. But I wo'nt so no to nice pictures.

 

[Crystal Faye]

Are we identifying theta correctly?

 

[andrevdh]

That is the danger one faces when one have'nt derived a formula oneself: misinterpreting the meaning of it. $\theta$ is the angle that the ramp makes with the horizontal. So it is the bottom right hand angle of the triangle in your drawing. You can see its meaning by multiplying the formula with the mass of the ball on the left and right. The numerator then becomes the component of the weight of the ball parallel to the ramp.

 

[andrevdh]

The $g$ in the formula is the value of the gravitational acceleration. Its value changes a bit with lattitude. Where are you on this planet?

 

[Crystal Faye]

I live in America, in the south...near the Gulf of Mexico.

 

[Crystal Faye]

Oh, I understand what you meant by sine now. I had looked it up on the internet and I was misled.

 

[andrevdh]

I am not at work now so I don't have the information at hand to look what your g will be at appoximately 30 degrees latitude. Will post it on Monday. When is the project due and how are you going to check the time?

I guess that is now about one o'clock in the afternoon there by you.

 

[Crystal Faye]

Yes, it is 1:54 as I am writing this.

My project is due on Friday.

What do you mean by how will I check the time?

 

[andrevdh]

That it comes to about 1.8 seconds. It is too fast to measure accurately with a stopwatch.

 

[Crystal Faye]

Whoa. It is part of my project to make the ball go 1 meter in 1.8 seconds.

Are you saying that my teacher is asking us to be too accurate?

 

[Crystal Faye]

oh. I think I understand what you are saying. We may possibly measure the ball's time with an electrical timer.

 

[andrevdh]

Sorry, I am a scientist and are use to strive for perfection. If it takes 1.7 or 1.9 seconds I would consider it not good enough. So I was wondering how your teacher was going to evaluate your design, maybe by using photogates? Just by changing the angle of the ramp you could get to a better result.

Do the rest of the class have different projects? Do you work in groups? I myself work with students in the Physics pratical lab all day so I find this project rather interesting.

 

[Crystal Faye]

My teacher told us that it had to be exactly 1.8 seconds. I am really hoping that he does not rely solely on a timer. I can see how that would cause problems.

I find this project rather interesting as well; however it is also extremely challenging for me. I am a senior in high school and so far we have only learned the scientific method and a little about velocity. We are pretty much left to figure this out on our own.

I am in a group with three other people, and the entire class was assigned the same project.

 

[andrevdh]

I was toying with the idea of using some sort of lever system to start and stop a stopwatch so that you could record the time more accurately. Envisage a holder for the stopwatch (fixed to the incline) at the top so that when the ball is released it starts the stopwatch (normally you can hold the button down and it only starts when the button is released). The button rest against the side of the holder and a pin runs through the side wall of the holder. The ball opersates this pin at the release point as I explained. At the bottom is a buffer at the 1 meter mark that connects via a stiff piece of wire to the rear of the stopwatch. This wire runs back up the bottom of the incline and pushes on the other side of the stopwatch to stop it. Sounds good in theory but in practice... Maybe you can design something similar along these lines to operate the stopwatch mechanically.

If you feel even more adventurous you can open up the stopwatch and check whether it is possible to connect two wires to the two contacts that the button short out (crocodile clips, paper clips). This will enable you to duplicate a button press by shorting the ends of the two wires.

Anyway, if you need more help feel free to ask.

 

[Crystal Faye]

I love your idea, it sounds so fun!

I will let you know once I actually figure out the equation [yes, I am still working on it,haha] Then I can move on to the timing aspect.

 

[andrevdh]

So more ideas for your project.

The stopwatch is operated by a lever at the starting position.

The idea behind the threaded bolt is to adjust the angle to obtain the correct time.

What problems dou you have with calculating the angle of the incline?

 

[Crystal Faye]

I like the bolt idea, my little brother had mentioned something like that to me. I believe that he is going to help me in the actual construction of the ramp, along with my other group members.

Where did you find that image, or did you draw it up?

 

[andrevdh]

I drew it with CorelDraw.

If you operate the stopwatch via some interaction with the ball just check that it does not deliver a substancial force to the ball. Although I think that the rules of the design did not specify that the ball should start out of rest!

You should also check that the surface that the ramp is resting on is level. Maybe by pouring some warm water with a bit of detergent on it. If this will get you into trouble you should maybe put some plastic on it first.

Show us your acceleration calculation thus far so that we can help you. The best appoach is to use the calculated angle just as an initial setting. It is best to then set final adjustment via a calibration curve.

 

[Crystal Faye]

ok, calculations are coming soon. I am at school right now, but by the time I go to my physics class we should have it figured out. I will post them when I got home.

Don't think we are stupid [ hahahaha] but how exactly will the detergent in the water help?

 

[Crystal Faye]

Generally, will my "g" in the equation be 9.81 m/s^2 or will it be 6.6742... like I had labeled in my picture.

 

[Crystal Faye]

Another question... When I am calculating rotation interia does kinetic energy come into play? I have looked it up and I keep seeing this equation :

 

[andrevdh]

g is your local gravitational acceleration. At your lattitude it should be about 9.79 m/s^2.

The rotational inertia is a quantity that describes how difficult it is to rotate an object about a certain axis. It appears in the rotational kinematic equations just like mass do in the linear kinematic equations and therefore play a similar role. For a solid uniform ball that is rotating about its centre of mass (an axis through its centre) it comes to

$$I_{com} = 0.4 MR^2$$

The detergent will break the surface tension of the water and will make it flow easier across the surface. If the surface is slanted the flow will then be more apparent.

 

[Crystal Faye]

Here are my calculations so far. I am sorry if they seem messy, they are handwritten. [ yes, I know that I have unit problems and sig. fig. problems.. sorry]
PAGE ONE:
http://i108.photobucket.com/albums/n18/inferius_fire/pg1ramp.jpg


PAGE TWO:
http://i108.photobucket.com/albums/n18/inferius_fire/pg2ramp.jpg

 

[andrevdh]

In your calculation for the acceleration of the ball down the ramp your end velocity is a sort of average velocity since the ball are accelerating down the ramp. The initial velocity is assumed to be zero, but then in your calculation you've got

$$v_i = \frac{?}{.9s}$$

?

Another way of calculating the acceleration of the ball down the ramp is via the known distance that the ball will travel and the given time. Also assume that the ball will start out of rest.

Are you contemplating in using a ping-pong ball? The mass is awfully small. So I assume that the ball is hollow. The formula I gave you is for a solid, not hollow ball. So you need to consider another type of ball. I feel that the chances of you current ball slipping is great. Rather go for a hard rubber type ball - a super ball with more mass.

Please write with a very soft pencil, maybe 2B. It is difficult to see your work as it is.

 

[Crystal Faye]

We are using a super ball. It is possible that we took the measurement wrong, or that I wrote down the wrong number. I will double check at school today.

For initial velocity, I wrote 0 over .9 seconds to prove why it was zero. Is my acceleration wrong?

Is my X value really wrong?

 

[andrevdh]

As I say. Please bear with me I am a Physicist. Let's try and get things right.

Rather do the acceleration calculation as follows. We assume that the given information is correct for the 1 meter distance, that is that the conditions must be such that the ball need to cover the distance in 1.8 s. The acceleration can then be calculated with

$$s = 0.5at^2$$

The assumtion that the ball need to cover 0.5 meter in half the time is a bit cock-eyed. Some one's physics is a bit off, unless I misunderstand the problem. So calculate the acceleration with the info for the 1 meter distance and then recalculate the time for the half meter distance since it seems that you are going to measure the time over half a meter.

The acceleration should come out a bit different from your $$0.7 m/s^2$$.

Your mass is definitely wrong for a four centimeter super ball. I would estimate that it should be in the order of 100 - 200 grams.

Would'nt it be easier to use a longer ramp with 2 lines on it half a meter apart? Maybe test the concept with a piece of scrap to weigh the pros and cons. The starting position could then be a buffer nailed to the ramp. This way the ball always start at exactly the same point.

 

[Crystal Faye]

You were correct. The mass of the ball was actually 50.6 grams. I am not sure how we could have made such a simple mistake. Thanks for pointing that out.

So, to calculate acceleration.. Should I merely just insert 1 meter where it was previously .5 meters and 1.8 seconds where I had put .9 seconds?

Or do I use the new equation that you had given me? Can you tell me more about the new equation.

 

[andrevdh]

The constant acceleration equation

$$s=ut+\frac{1}{2}at^2$$

should be in your handbook. Use it and the distance 1.0 meter and time of 1.8 s to cover this distance. Then use this equation again with the calculated acceleration and the "new" distance of 0.50 meters to calculate the time to cover for this "new" distance.

I am sure your teacher knows his Physics very well. But anyone makes mistakes from time to time and I think this might be one of them.

Just follow my lead and all will become clear. I am not that good at explaining things, or do not have the patience for it.