How Can You Design a Ramp for Consistent Ball Rolling?

[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.

 

[Crystal Faye]

Thanks. I looked in my book for the meaning of the equation. The closest that I have came is centripical accerlation. It does not look like the same thing. I can not figure out what the "u" in the equation means.

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[andrevdh]

s is the distance the object travels (in a straight line) [we want the ball to travel 1 meter down the ramp in this case] during the time t [1.8 seconds is the time it must take the ball to travel this distance of 1 meter down the ramp].

u is the object's speed when the timing is started (t = 0 seconds)
[in this case the ball is started out of rest so u = 0 m/s].

and a is the constant acceleration of the ball down the ramp that we can calculate with the given information.

The density of the ball seems to be about $980\ kg/m^3$ which means it should float in water with a little bit of head above the water. Please check that this is the case because I would have thought that it should be a bit heavier.

 

[andrevdh]

Answers to some questions that might have bothered you:

http://www.Newton.dep.anl.gov/askasci/phy00/phy00124.htm"

http://www.physics.uoguelph.ca/tutorials/torque/Q.torque.inertia.html"

http://www.physics.ucla.edu/demoweb/demomanual/mechanics/kinematics/acceleration_down_an_inclined_plane.html"

 

[Crystal Faye]

I really want to thank you. Our ramp came out very well. We presented it today and I made a 98 percent on it. I know that a 98 percent is not perfect, but I am pleased.

We named ourselves "Team Andrevdh" haha.