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Author Topic: Racing Balls  (Read 152447 times)
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Fu-Kwun Hwang
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on: January 29, 2004, 01:24:44 pm » posted from:,,Satellite Provider

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From : http://www.physics.umd.edu/lecdem/outreach/QOTW/arch1/q002.htm
Identical balls are launched at the same time with the same velocity from the left front end of the two-track gizmo photographed below. (Because this is a physics problem, there is no friction.) A race of the balls will then ensue. The ball on the flat track clearly proceeds across the track at a constant speed. The ball on the dipped track goes for a while at that same speed, goes faster while it is in the dipped part of the track, then returns to its original speed for the final segment of the track. Note that it also travels further.

What will happen?

   * (a) The ball on the straight track will reach the end first.
   * (b) The ball on the track with the dip will reach the end first.
   * (c) The race will end in a tie.
Click

The answer is (b); the ball on the dipped track gets to the end first and wins the race. The two balls go along together for the first part of the race. As the ball on the dipped track goes down, its horizontal velocity increases, so it gets ahead. When it returns to its original level, it slows down to its original horizontal speed, but in so doing it never goes slower than the ball on the flat track, so it never gets behind the other ball or even allows the flat track ball to catch up. The two balls then move along at the same speed with the dipped track ball remaining ahead of the straight track ball by a constant amount.

Click on the photograph of the apparatus above for a video showing this demonstration in action.

The same thing happens when two people are walking along a straight flat road. If one of the two runs for a short time (the dip) then slows down to the original walking speed, the runner will get ahead during the time he or she is running. After slowing back down to the same walking speed, the two will then move along at the same speed but the one who ran will remain a constant distance ahead of the person who walked the whole time.

A more mathematical way of "discovering" this result is to draw graphs of horizontal velocity versus time for each of the two balls. Then draw curves of the integral of the velocity curves for each, which are the horizontal position versus time for each of the balls. After the ball on the dipped track returns to its original level its horizontal position can be observed from the graph to be greater than that of the ball on the flat track for any time.
or play with the following simulation to find out answer.




When the two balls are launched from one end of the track with the same initial velocity, what will happen:
    1) the ball #1 on the straight track arrives at the other end first
    2) the ball #2 on the track with the dip arrives at the other end first
    3) the race is a tie - both balls reach the other end at the same time?
    This is a java version for one of our physics demostration. Think about it, select the answer to active the program.
      Did you get the correct answer?

    Press the start button to restart.




  1. Click the left button within the window will suspend the animation

    1. click it again to resume.
  2. Clcik the right mouse button and drag it up and down

    1. to change the shape of the lower track.

  3. Click more information checkbox to display more information

  4. Shape of the track ( from left to right)

    1. section 1 red curve: part of a circle (1/4)
        tangential component of the gravitation field is the source of the acceleration.

      section 2 blue curve: Trajectory for a projectile

        (with initial velocity when it entering this region)
        So, horizontal component of the velocity is a constant.

      section 3 red curve: part of a circle.
        The solpe for the curves match at the boundary.
        particle will accelerate (as section 1) but with different acceleration

      sectin 4 blue curve: Trajectory for projectile similar similar to section 2
      section 5 green line: horizontal line.
        The velocity is a constant in this region.

  5. The horizontal component of the velocities are also shown.



This applet was originated from one of the demonstrations developed by Prof. Berg (Dept. of physics at Maryland, College Park) many years ago.


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


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There are 11 translations,
Higher number at the end means more translation been done.
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Fu-Kwun Hwang
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Reply #1 on: January 24, 2005, 03:47:46 pm » posted from:Taipei,T'ai-pei,Taiwan

Velocity of ball 2 >= velocity of ball 1 !
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Reply #2 on: June 17, 2004, 11:40:07 am »

why (mechanical) energy conservation cannot be used?
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Reply #3 on: June 18, 2004, 11:06:44 am »

Energy is conservation in the above situation. Could you explain more why you think energy conservation can not be used?
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Reply #4 on: August 20, 2004, 09:47:50 pm »

Because of Friction I think...
(need source code of this program Sad )
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spoung45
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Reply #5 on: October 19, 2004, 12:06:01 pm »

Well its becuse of the acceleration of the ball down in the second dip that gives it more velocity. saying there is no friction involved ball 2 gains more velocity after the second dip.
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Fu-Kwun Hwang
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Reply #6 on: October 20, 2004, 07:07:15 am »

Please click the image to see the real demonstration (mpg)
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Fu-Kwun Hwang
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Reply #7 on: January 26, 2005, 10:58:07 am » posted from:Taipei,T'ai-pei,Taiwan

Let Vx is the horizontal component of the velocity

Vx2 >= Vx1 so ball 2 has longer horizontal displacement than ball 1.

The above condition is true if the slope is not too large so that ball 2 is always stay on track.
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Fu-Kwun Hwang
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Reply #8 on: February 12, 2005, 01:54:33 pm »

Horizontal velocity of ball 2 is either the same or greater than Horizontal velocity of ball 1 during the who trip.
So the horizontal displacement of ball 2 is larger than the horizontal displacement of ball 1. :-)
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Reply #9 on: October 13, 2007, 05:38:54 pm »

Horizontal velocity of ball 2 is either the same or greater than Horizontal velocity of ball 1 during the who trip.
So the horizontal displacement of ball 2 is larger than the horizontal displacement of ball 1. :-)
  I can't download this file
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Fu-Kwun Hwang
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Reply #10 on: October 13, 2007, 10:20:15 pm »

I just click the "get file for offline use" button, and I received the file in my email account.
I think there is a problem in your browser. Please change to another computer, may be it will work!
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lookang
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Reply #11 on: April 16, 2008, 03:23:26 pm » posted from:SINGAPORE,SINGAPORE,SINGAPORE

http://quark.edumall.sg/quark/slot/mlib/dc07/2d94749fd_6407.mp4

http://quark.edumall.sg/cos/o.x?c=/quark/mlib&uid=106&ptid=33&func=prop2&id=6407
to rate the video

For a video i made for the same concept!





« Last Edit: September 25, 2008, 08:24:24 am by lookang » Logged
lookang
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Reply #12 on: June 27, 2008, 01:18:38 pm » posted from:SINGAPORE,SINGAPORE,SINGAPORE

1. can check why i click download, it becomes check my email ?

2. how to get rid of this message?
Emails send your account at [XXXXXXXXXX@yahoo.com.sg] were returned at least 30 times, please check/modify your email address/email quota setting!

 Grin
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Fu-Kwun Hwang
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Reply #13 on: June 27, 2008, 02:02:30 pm » posted from:Taipei,T\'ai-pei,Taiwan

I added code to check bounced email because a lot of user enter incorrect email which cause email sent out by our system were returned.
The records shown that at least 30 emails sent to you were returned by your mail server (for some unknown reason).
I just reset your record manually. It should work fine now.
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Reply #14 on: September 20, 2008, 02:56:49 pm »

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?
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Fu-Kwun Hwang
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Reply #15 on: September 20, 2008, 09:56:23 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.
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lookang
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Reply #16 on: September 22, 2008, 08:03:14 am » posted from:SINGAPORE,SINGAPORE,SINGAPORE

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 .Smiley

understand?
« Last Edit: September 22, 2008, 08:09:36 am by lookang » Logged
lookang
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Reply #17 on: April 13, 2009, 08:17:56 am » posted from:SINGAPORE,SINGAPORE,SINGAPORE

Code:
<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!
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lookang
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Reply #18 on: May 07, 2009, 09:29:30 pm » posted from:Singapore,,Singapore

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, 09:39:30 pm by lookang » Logged
juliaamit
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Reply #19 on: May 16, 2009, 10:25:48 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
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Fu-Kwun Hwang
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Reply #20 on: May 17, 2009, 09:13:05 am » posted from:Taipei,T\'ai-pei,Taiwan

 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 d2θ/dt2= -g*sinθ/L , where g*sinθ 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.
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sithy
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Reply #21 on: June 03, 2009, 11:29:34 am » posted from:Lansdale,Pennsylvania,United States

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.
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Reply #22 on: June 03, 2009, 01:19:01 pm » posted from:Taipei,T\'ai-pei,Taiwan

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.
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Reply #23 on: September 10, 2009, 09:51:23 pm » posted from:Edmonton,Alberta,Canada

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.
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Reply #24 on: September 22, 2009, 01:48:46 pm » posted from:Phagwara,Punjab,India

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.
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lookang
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Reply #25 on: April 30, 2010, 08:08:23 am » posted from:SINGAPORE,SINGAPORE,SINGAPORE

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  Grin


*** There are 1 more attached files. You need to login to acces it!
« Last Edit: April 30, 2010, 08:12:23 am by lookang » Logged
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Reply #26 on: March 02, 2012, 08:57:44 pm » posted from:SINGAPORE,SINGAPORE,SINGAPORE

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 Smiley as part of a research project to understand how to improve delivery of educational resources to schools in Singapore!
« Last Edit: March 02, 2012, 09:05:34 pm by lookang » Logged
wyatt
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Reply #27 on: June 29, 2012, 11:19:13 pm » posted from:Yogyakarta,Yogyakarta,Indonesia

good video
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Reply #28 on: April 05, 2013, 03:01:31 pm » posted from:Madrid,Madrid,Spain

It is possible to modify the conditions in order horitzontal ball wins? For exemple with more inclination?
Thanks
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