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Pinewood Derby Design Idea

  1. Feb 12, 2015 #1
    Hello everyone! This is my first post, so excuse my improper etiquette (if any). Anyways, the Pinewood Derby, just in case you guys are unfamiliar with it, is an event that is held as a competition where small, homemade cars, race. The track is contains a ramp, which then leads onto a flat surface.
    Although I have not personally competed in this competition, I do have a design proposal for it. Without further or do, here it is:


    So, the basic idea is that the car will have a container on it that will house a relatively heavy object that can move freely. As the car goes down the ramp, the center of gravity, represented by the ball, will move forward, making the acceleration of the car increase. Then, as the car becomes parallel to the ground, the ball (center of gravity) will move to the back, which will in return will reduce the acceleration. The point of the moving center of gravity is that it will provide more acceleration for the car going down the ramp, as opposed to a regular car with a permanent center of gravity.

    To culminate, please express your ideas and comments on my proposal! Tell me my mistakes, and please give me an explanation behind any hypothesis you make! Thank you!

    (NOTE: Disregard the aerodynamic and frictional factors on my design)
  2. jcsd
  3. Feb 12, 2015 #2
    Can you please explain the physics involved in your proposal?

  4. Feb 12, 2015 #3
    @tduell Hey! Thanks for the reply! Well, I'm merely just an amateur with Newtonian and Classical Physics, but I will try to give an explanation using my current knowledge. When the car is set at the top of the ramp, the center of mass, represented by the ball in the diagram, will be at the front of the car. With the center of mass at the front of the model, the gravity will "pull" it down with more force. Thanks to the triangular shape of the ramp, the force will be transferred to the diagonal direction, which in turn will provide more acceleration in that direction as opposed to a car model with the center of mass at the center itself.

    Well, that's as thorough of a physical explanation that I can give. I know I am subject to mistake, so please, help me! Thanks!

  5. Feb 12, 2015 #4
    I think you are mistaken. I reckon the "pull" due to gravity is the M*g*sin(theta) (theta is angle to horizontal) and M=the total mass of the car. I reckon it doesn't make any difference where your centre of mass is located.
    That's how I see it, and hopefully others will correct me if I'm wrong.

  6. Feb 12, 2015 #5
    @tduell Ah. I see. Thank you! I will research more about this topic. Honestly, this idea was developed with my minuscule knowledge on Newtonian Physics and my intuition. So, thanks for correcting me!

  7. Feb 12, 2015 #6
    @tduell Also, your reply was in regards to when the car was going down the ramp, correct? What about when the car is parallel to the ground? Will the center of mass matter?

  8. Feb 12, 2015 #7


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    Staff: Mentor

    This is correct.

    No, except the ball will not move toward the back of the car as you have shown when on level ground. Since the car is decelerating on the flat part, the ball will stay pinned in the front of the car.
  9. Feb 12, 2015 #8
    @berkeman Ah, I see. Thanks a lot!

  10. Feb 12, 2015 #9


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    Science Advisor

    You want as much weight as possible to start as far rearward as possible. That gives you an edge because the rear wheels are on the steep starting ramp longer. Any initial speed advantage pays off for the remainder of the course.
  11. Feb 13, 2015 #10
    @jbriggs444 Ah, I see. I guess I misunderstood the concept. Thyanks a lot for the info!

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