# How does the diluted gravity experiment make sense?

We know that if a ball rolls down a ramp, then the component of its weight acting down in the direction of the ramp is $$g\sin\theta$$. Therefore, we can substitute that into $$s=ut+at^2/2$$ to get $$s=(g\sin\theta)t^2/2$$, where $$s$$ denotes the height of the ramp.

But if you put in $$s/\text{(length of ramp)}$$ for $$\sin\theta$$, the $$s$$ cancels out and the equation breaks.

• Just curious, I've never heard term "diluted gravity", where did you find that? Commented Oct 11, 2018 at 23:49
• @Charlie It's a famous (I think) experiment done by Galileo. Google "galileo diluted gravity" or something like that.
– Ivan
Commented Oct 11, 2018 at 23:54
• Oh I guess I knew it by another name since I'm not a native English speaker. Commented Oct 12, 2018 at 0:23

where s becomes the vertical height of the ramp

This is incorrect. The displacement $$s$$ is the distance covered, i.e. the length of the ramp.

The confusion arises because you're mixing variables. Let $$h$$ be the height of your ramp and $$L$$ the length, with $$\theta$$ being the angle with respect to the ground. Therefore, this means that:

$$Sin\theta=\frac{h}{L}$$

As the ball rolls down it traverses the hypotenuse, so your equation, assuming no initial velocity, should instead read:

$$s=\frac{1}{2}gSin\theta t^2$$

Where $$s$$ denotes the displacement component parallel to the hypothenuse (where the ball rolls). In other words, you used $$s$$ both as the distance traversed and the height of the ramp, so got that inconsistency. Now if you substitute the definition of sinus:

$$s=\frac{1}{2}g(\frac{h}{L})t^2$$

Which doesn't lead to any problems. So be careful with naming your variables.

As an exercise, you may want to see what happens when $$s=L$$. What would this mean?