Author Topic: Racing Balls  (Read 176032 times)

Fu-Kwun Hwang

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Re: Racing Balls
« Reply #15 on: September 20, 2008, 10:56:23 pm »
The condition is the ball has to be always on the track.
If the slope of the track is to deep, the ball will fall out.

lookang

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Re: Racing Balls
« Reply #16 on: September 22, 2008, 09:03:14 am »
I wonder if both ball 2 experiment is conducted at the same time, however one is at ground level, the other is at 10km above ground level, will they arrive at the same time?

Based on the concept of Newton’s law of universal gravitation,
http://en.wikipedia.org/wiki/Newton%27s_law_of_universal_gravitation

g = G m1.M2 / r^2
where:

  * F is the magnitude of the gravitational force between the two point masses,
  * G is the gravitational constant, G is approximately equal to 6.67 × 10^?11 N m^2 kg^-2
  * m1 is the mass of the first point mass,
  * m2 is the mass of the second point mass,
  * r is the distance between the two point masses.

Let's assume m1 = mass of ball say = 1 kg for easy substitution and calculation of g.

mass of Earth = 5.9742 × 10^24 kilograms = M_2
http://www.google.com.sg/search?q=mass+of+earth&ie=utf-8&oe=utf-8&aq=t&rls=org.mozilla:en-US:official&client=firefox-a

radius of Earth = 6 378.1 kilometers
http://www.google.com.sg/search?hl=en&client=firefox-a&rls=org.mozilla:en-US:official&hs=vAF&pwst=1&sa=X&oi=spell&resnum=0&ct=result&cd=1&q=radius+of+earth&spell=1

therefore using, g = G m1.M2 / r^2

g = (6.67 × 10^?11)(1)(5.9742× 10^24) / (6378 x 10^3)^2 = 9.80 m/s^2

at a point where R' = R + 10 km =

g' = (6.67 × 10^?11)(1)(5.9742× 10^24) / ([6378+10] x 10^3)^2 = 9.77 m/s^2

so assuming the ball is on a slope tilt of angle teta,

a = g.sin (teta) = 9.80.sin (teta)

a' g'.sin(teta) = 9.77. sin (teta) where teta is say = 30 degrees

assuming motion is under constant acceleration,

equation of motion says, s = u.t + 1/2.a.t^2 and s' =u'.t + 1/2.a'.t'^2

subs in s = s' = say 1 m of simple substitution  , u = u' =0
simplified........


solving which gives t =  Math.sqrt[(1)(2)/9.81.sin(30^o)] =  0.639 s  & t' = 0.640 s approximately.

in conclusion to answer your question, the answer should be roughly the same time unless you can conduct the experiment at a height above Earth where the g' is very different of the sea-level g .:)

understand?
« Last Edit: September 22, 2008, 09:09:36 am by lookang »

lookang

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Re: Racing Balls
« Reply #17 on: April 13, 2009, 09:17:56 am »
<applet code="racingBall.class" width=600 height=350 codebase="http://www.phy.ntnu.edu.tw/java/racingBall/"><param name="Reset" value="Reset"><param name="Start" value="Start"><param name="info" value="more information"><param name="MSG" value="Which ball will reach the other end first or they will arrive at the same time?"><param name="ANS" value="Select an answer to start"><param name="ANS1" value="1) ball 1 arrive earlier"><param name="ANS2" value="2) ball 2 arrive earlier"><param name="ANS3" value="3) Arrive at the same time"><param name="Path" value="complete Path"></applet>
trying to figure out the embed code for the applets without the need to host it

in a different forum http://iresearch.edumall.sg/cos/o.x?ptid=80&c=/iresearch/forum&func=showthread&t=223

done!

lookang

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Re: Racing Balls
« Reply #18 on: May 07, 2009, 10:29:30 pm »
wonder if this video is good?
unfortunately it require login

http://iresearch.edumall.sg/cos/o.x?ptid=80&c=/iresearch/forum&func=showthread&t=223





« Last Edit: May 07, 2009, 10:39:30 pm by lookang »

juliaamit

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Re: Racing Balls
« Reply #19 on: May 16, 2009, 11:25:48 pm »
Dear sir,

Thank you very much for Designing such a great site.

I would be very much obliged if u have time to clear my doubts.

1) U have written that only tangential component of "g" will be considered at the circular path. But do i have to consider centripetal force ?

2)How did u get the time to reach at the end of 1st. circle without knowing the actual radius of the first quarter circle( common to both ball)?
kindly send me the calculation of time to reach at the end of first circle.I mean just before the first ball starts horizontal journey.If u give the total time calculation, then it be very nice.

Regards,
Amit

Fu-Kwun Hwang

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Re: Racing Balls
« Reply #20 on: May 17, 2009, 10:13:05 am »
 When I did the calculation in the simulation, I already assume the particle will move alone the circular path. The tangential component of "g" is the only source of acceleration along the circular path. Actually, it was transform to angular acceleration.
The equation of motion is d2?/dt2= -g*sin?/L , where g*sin? is the tangential component of "g".
The centripetal force is provided from normal force to keep it move in circular path (to restrict it's path or change moving direction only, the centripetal force did not change the magnitude of the velocity).

If you need to know the detail of the motion before it reach the end of the first circular, it is exactly the same as a pendulum. Please check out Force analysis of a pendulum and Pendulum.

It is assumed the same time step dt=0.05s in the above simulation, and the program calculate new position from velocity of the particle(also update particle velocity from the external force at each time step).
We did not calculate the time it will reach the end of first circle in advance, what we did was checking at each time step if the particle pass the end point.

May I know why you think you need to know  how to calculate the time for the particle to reach at the end of first circle? May be there are better way to solve your problem.

sithy

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Re: Racing Balls
« Reply #21 on: June 03, 2009, 12:29:34 pm »
Hi

Thanks for this simulation. I just downloaded the folder. The downloaded folder does not have a .xml file. I would like to open it and see the source if you will allow. Thanks.

Sithy.

Fu-Kwun Hwang

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Re: Racing Balls
« Reply #22 on: June 03, 2009, 02:19:01 pm »
All the simulations under category: JDK1.0.2 simulations (1996-2001), were not created with EJS.
So there is no EJS source (xml files) exists.
The above applet is one of those. All of them were created with JDK1.0.2 more than 10 years ago.

cmscritic

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Re: Racing Balls
« Reply #23 on: September 10, 2009, 10:51:23 pm »
I've always found it fascinating how the initial reaction is to assume the shortest path leads to the quickest finish but when incorporating the velocity and speed increase that is achieved from the dip, the 2nd ball of course arrives first.

Very cool demo.

janelavis

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Re: Racing Balls
« Reply #24 on: September 22, 2009, 02:48:46 pm »
I'm getting excited. Gonna try and get the hitch on the car this weekend align it and fix the oil leak so I can be ready for this! Been looking to get back to Grattan for a while now.

lookang

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Re: Racing Balls
« Reply #25 on: April 30, 2010, 09:08:23 am »
FlexTrackRaces.mov
A video suitable for video analysis found here http://www.cabrillo.edu/~dbrown/tracker/mechanics_videos.zip



It will be cool to have an Ejs version of the racing balls! For your consideration  ;D
« Last Edit: April 30, 2010, 09:12:23 am by lookang »

lookang

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Re: Racing Balls
« Reply #26 on: March 02, 2012, 09:57:44 pm »
enjoy!

http://www.youtube.com/watch?v=bpEPW6aowAA

http://www.youtube.com/watch?v=bpEPW6aowAA


A demonstration to show that acceleration enables an object to reach the end of a slope sooner.Collaboration with NUS Physics Associate Professor Sow Chorng Haur.
Directed, Scripted and Acted by lookang
Narration by: foong yin
permission granted to upload on social platform by ETD :) as part of a research project to understand how to improve delivery of educational resources to schools in Singapore!
« Last Edit: March 02, 2012, 10:05:34 pm by lookang »

wyatt

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Re: Racing Balls
« Reply #27 on: June 30, 2012, 12:19:13 am »
good video

papagall

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Re: Racing Balls
« Reply #28 on: April 05, 2013, 04:01:31 pm »
It is possible to modify the conditions in order horitzontal ball wins? For exemple with more inclination?
Thanks