Galileo, inclined plane and geometry.

[talisman2212]

Homework Statement

Galileo in seeking to discover the laws governing the motion of bodies under the action of their weight, conducted a series of experiments on inclined planes. Choosing as unit of length the distance traveled by the ball the first unit of time, measuring at subsequent time points the distance traveled and repeating the experimental procedure with different mass ball finds the same numerical result: the distance [itex]s[/itex] traveled by any ball in the inclined plane under the influence of the weight, is square proportional to the time elapsed: [itex]s(t)=kt^2[/itex].
Gradually increasing the slope of the plane, finds that the value of [itex]k[/itex] increases to a maximum value which takes at the vertical gradient, i.e. freefall. Can you determine the relation between [itex]k[/itex] and the slope of the inclined plane?

Homework Equations

[itex]s(t)=kt^2[/itex]

The Attempt at a Solution

I tried this one: let the inclined plane have angle [itex]\theta[/itex], then [itex] sin \theta= \frac{AB}{AO}[/itex], we use now the fact that the distance traveled by the ball is [itex]s(t)=kt^2[/itex] so we can find the distances [itex]AO[/itex] and [itex]AB[/itex], that is [itex] sin \theta= \frac{AB}{AO} = \frac{gt^2}{kt^2} = \frac{g}{k}[/itex], but while [itex]\theta[/itex] increases, [itex]k[/itex] decreases, where is the wrong?

I tried also this: the first unit of time the ball travels distance: [itex]s(1)=k[/itex],
the second unit of time distance: [itex]s(2)=4k[/itex]
and the third unit of time distance: [itex]s(3)=9k[/itex].
I tried to use the Thales' theorem (Intercept theorem), but I don't know how to move on.

 

[Basic_Physics]

k should be related to the component of the weight along the incline.

 

[talisman2212]

k should be related to the component of the weight along the incline.

That's wrong. [itex]k[/itex] is related to the slope of the inclined plane and is independent from the weight. If we use Newton's law we have [itex]mk=mgsin \theta \Rightarrow k=gsin \theta[/itex] I must prove this (without using Newton's law).

 

[haruspex]

That's wrong. [itex]k[/itex] is related to the slope of the inclined plane and is independent from the weight.

Basic_Physics means that it's related to the fraction of the weight that acts along the slope. You have k sin(θ) = g, which makes k > g. Clearly that's wrong. What should the equation say?

 

[talisman2212]

The equation should say k=sin(θ)g, I am writing above what i have tried. Can someone help me to prove it?

 

[Basic_Physics]

For Galileo's choice of units of distance k will always be one:
" Choosing as unit of length the distance traveled by the ball the first unit of time..."
A graph of s vs t² will be directly proportional with gradient k, but the graph goes through the point 1,1. This means its gradient will be k=1.

 

[Basic_Physics]

sin(θ) ≤ 1.0 so you are fine. With a little bit of reasoning one can came to the right conclusion then. When the plane is flat k should be 0 (no acceleration). When the plane is vertical k should be equal to g.

 

[aralbrec]

This ball is rolling not slipping

 

[talisman2212]

I must determine the relation between k and the slope of the inclined plane. I am writing at the top of the topic what I have tried, I found g=ksin(θ) this is wrong. I am trying to use the trigonometry of the problem (no sees that??). Can someome really help me to find the relation between k and the slope of the inclined plane? (without using Newton's law, but only the problem's data)

 

[Basic_Physics]

One meaning that can be given to k is that it is the distance traveled along the incline when t = 1. Galileo did these "diluted acceleration" experiments to determine the relation beween distance covered and time for uniform accelerated motion. The component of gravity along the incline is
g[itex]\;[/itex]sin([itex]\theta[/itex])
This means that the speed obtained after 1 time unit will be
v[itex]_{1}=g\;sin(\theta)[/itex]
The average speed is then
v[itex]_{avg}=\frac{g\;sin(\theta)}{2}[/itex]
if it starts out of rest. The distance covered along the incline for 1 time unit will then be such that
[itex]\frac{g\;sin(\theta)}{2}\;t=k\;t^{2}[/itex]
so that
[itex]\frac{g\;sin(\theta)}{2}\;=\;k[/itex]

 

[aralbrec]

I tried this one: let the inclined plane have angle [itex]\theta[/itex], then [itex] sin \theta= \frac{AB}{AO}[/itex], we use now the fact that the distance traveled by the ball is [itex]s(t)=kt^2[/itex] so we can find the distances [itex]AO[/itex] and [itex]AB[/itex], that is [itex] sin \theta= \frac{AB}{AO} = \frac{gt^2}{kt^2} = \frac{g}{k}[/itex], but while [itex]\theta[/itex] increases, [itex]k[/itex] decreases, where is the wrong?

If the vertical acceleration is 'a' then:

[itex] sin \theta= \frac{AB}{AO} = \frac{0.5at^2}{kt^2} = \frac{a}{2k}[/itex]

When the plane is vertical (ie freefall), [itex]sin \theta = 1[/itex] and [itex]k = \frac{a}{2}[/itex].

From Newton's Law, at freefall, [itex]s(t) = \frac{1}{2}gt^2[/itex]. From Galileo, [itex]s(t) = kt^2 = \frac{1}{2}at^2[/itex] so that the vertical acceleration a = g.

Is this the answer being sought? The mistake was using [itex]gt^2[/itex] as the vertical distance AB since the vertical acceleration is only equal to g in freefall.

 

[talisman2212]

You haven't understand the problem at all, anyway I found the solution.