Kinematics of a ball and track

In summary, the conversation discusses designing a track with a high-to-low transition and finding the exit velocity for a ball rolling down the track. The formula used is y=5.5cos(xpi/18)+5.5 and the method used is to use conservation of energy to approximate the velocity. However, to get a more accurate approximation, one should also account for the rolling of the ball.
  • #1
Hello,

New here. I have designed a segment of a track that I am building that I need to pull some data out of. It is a high-to-low transition (down hill) that takes the form of cosine over the domain [0,pi]. Thea graph of the transition's profile is y=5.5cos(xpi/18)+5.5 It is 11 inches high and 18 inches long.

I want to find out its exit velocity (x=18) for a ~100g ball that starts form rest at the top. The graph of the formula I used above is position-vs-position. I know that velocity is the graph of position-vs-time's first derivative.

How do I get from the position-vs-position graph/formula to the position-vs-time graph/formula? I think I could do it if the slope was constant, but I'm not sure about the curve and I don't want to average the slope. Shouldn't it be the same graph, just compressed along the x-axis?

Thanks,
Chopper
 
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  • #2
To a first approximation just use conservation of energy. You have PE = mgh and KE = 1/2 mv² so by setting them equal you get

mgh = 1/2 mv²
v = sqrt(2gh)
 
  • #3
Though to get a more accurate approximation you should account for the rolling of the ball (assuming there is enough friction to prevent it from slipping):

[tex]mgh\ =\ \frac{1}{2} mv^2\ +\ \frac{1}{2}I\omega^2[/tex]

Where I is the moment of inertia for a sphere ([tex]\frac{2}{5}MR^2[/tex]) and [tex]\omega[/tex] is [tex]\frac{v^2}{R^2}[/tex].

Solve that entire equation for v and you should get a more accurate approximation.
 
  • #4
Note that w^2 is not v^2 / R^2 if the ball running on a pair of rails as opposed to rolling along a surface. w^2 = v^w / r^2, where r is the perpendicular component of distance from center of ball to the surface of the ball where it meets the supporting track.
 
  • #5
@everyone: Thank you for the input. This will be very helpful. I had considered Inertia and that the point of contact of the ball/track would have an impact, but this is very helpful.
 

1. What is kinematics?

Kinematics is the branch of physics that studies the motion of objects without considering the forces that cause the motion.

2. What is a ball and track?

A ball and track, also known as a ball ramp or ball run, is a simple physical toy or educational tool that consists of a track or ramp with a ball or marble rolling down it.

3. How does a ball and track demonstrate kinematics?

A ball and track demonstrates kinematics by showing the motion of the ball as it moves along the track. This includes concepts such as acceleration, velocity, and displacement.

4. How does the height of the track affect the ball's motion?

The height of the track affects the ball's motion by determining the potential energy of the ball. The higher the track, the more potential energy the ball has, which translates to a higher velocity and longer distance traveled.

5. What are the main factors that affect the ball's motion on a track?

The main factors that affect the ball's motion on a track are the height of the track, the angle of the track, and the presence of any obstacles or friction. These factors can affect the ball's acceleration, velocity, and trajectory.

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