Kinematics of a ball and track

[chopperwalker]

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

 

[Dale]

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)

 

[woodne]

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):

mgh\ =\ \frac{1}{2} mv^2\ +\ \frac{1}{2}I\omega^2

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

Solve that entire equation for v and you should get a more accurate approximation.

 

[rcgldr]

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.

 

[chopperwalker]

@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.