Did Galileo's Experiments Prove Aristotle Wrong?

[Viraam]

Homework Statement

Homework Equations

$F = \frac{m(v-u)}{t}$

The Attempt at a Solution

The mass of cricket ball is greater than mass of plastic ball.
Change of velocity is same for both. Now since the rate of change of momentum of cricket ball is greater the force needed to bring it to stop is greater. Since $F \alpha \frac{1}{t}$ the time taken by the cricket ball to stop is small. Therefore the plastic ball will cover greater distance.

Is my answer correct. Some sites claim that the cricket ball will cover larger distance. Is that true, Please answer ASAP [/B]

 

[andrewkirk]

Since $F \propto \frac{1}{t}$ the time taken by the cricket ball to stop is small. Therefore the plastic ball will cover greater distance.

That proportionality is only correct where the other variables remain constant. But $m$ differs between the two cases, so the proportionality does not hold.
The problem is not solvable without more information about what you are supposed to take into account, in particular air resistance and rolling resistance, the size of the two balls, whether the plastic ball is solid, and so on.
In general one would expect the cricket ball to roll further than a hollow plastic ball because, if they are the same size, they'll encounter the same air resistance but the cricket ball will have more initial momentum and KE. But without that extra info, the problem is not solvable.

 

[Viraam]

That proportionality is only correct where the other variables remain constant. But $m$ differs between the two cases, so the proportionality does not hold.
The problem is not solvable without more information about what you are supposed to take into account, in particular air resistance and rolling resistance, the size of the two balls, whether the plastic ball is solid, and so on.
In general one would expect the cricket ball to roll further than a hollow plastic ball because, if they are the same size, they'll encounter the same air resistance but the cricket ball will have more initial momentum and KE. But without that extra info, the problem is not solvable.

Why is it that the cricket ball will go farther distance than the plastic ball. Please assume that the plastic ball is hollow and lighter than the cricket ball. Please explain in terms of $F=\frac{m(v-u)}{t}$ The $v-u$ is equal for both the balls only mass $m$ is differing. Therefore we can say that a larger force will be needed to stop the cricket ball because of which it travels a larger distance. Is this correct?

 

[NickAtNight]

I am leaning towards the plastic ball going farther.

Why is it that the cricket ball will go farther distance than the plastic ball.

Why are we making this assumption? Is this an assumption provided by the provider of the problem, or one you are making?

A cricket ball is well defined by rules. The plastic ball is open to your imagination the way this problem is stated..

A cricket ball is a hard, solid ball used to play cricket. A cricket ball consists of cork covered by leather, and manufacture is regulated by cricket law at first class level...The next major change in 1809 saw further standardisation of the weight of the ball from between 5 and 6 ounces (142 to 170 g) to between 5.5 and 5.75 ounces (156 to 163 g),...

Please assume that the plastic ball is hollow and lighter than the cricket ball.

Why did you pick this equation? What Force(s) work on the balls to slow them down and cause them to stop?

Please explain in terms of $F=\frac{m(v-u)$

I do not think so.

Is this correct?

 

[NickAtNight]

Some sites claim that the cricket ball will cover larger distance

Interesting web page on the physics of a cricket ball. (not relevant to the problem at hand.

 

[andrewkirk]

in terms of $F=\frac{m(v-u)}{t}$ The $v-u$ is equal for both the balls only mass $m$ is differing. Therefore we can say that a larger force will be needed to stop the cricket ball because of which it travels a larger distance. Is this correct?

That equation is not correct for this case because it assumes that force is constant over time. In this case it will not be constant because air resistance increases with velocity. A correct version of the equation is $v-u=\int_{t_1}^{t_2} F(t)dt$.

We can approximately ignore that problem if we replace F in your equation by the average force over time. Then it will be true that, to stop the cricket ball in the same time as a much lighter plastic ball will require a greater average force. Conversely, because the forces will not be very different between the two (Air resistance is roughly similar. The cricket ball will suffer greater rolling resistance than the plastic ball because of its greater weight but if it's rolling on a smooth surface then I think that will be much less than the air resistance), the cricket ball will take a longer time to stop, and hence will travel further.

 

[NickAtNight]

Homework Equations

[/B]

How are you doing. Did you figure out the correct equation yet?

Hint: read this page

 

[Viraam]

How are you doing. Did you figure out the correct equation yet?

Hint: read this page

Actually, this question is appearing in our Chapter called Force and motion where the following equations are given:
$F=ma \\ F= \frac{m(v-u)}{t} \\ v = u+at \\ v^2 - u^2 = 2as \\ s = ut + \frac{1}{2} at^2 \\ \text{Conservation of Linear Momentum}$

 

[Viraam]

That equation is not correct for this case because it assumes that force is constant over time. In this case it will not be constant because air resistance increases with velocity. A correct version of the equation is $v-u=\int_{t_1}^{t_2} F(t)dt$.

We can approximately ignore that problem if we replace F in your equation by the average force over time. Then it will be true that, to stop the cricket ball in the same time as a much lighter plastic ball will require a greater average force. Conversely, because the forces will not be very different between the two (Air resistance is roughly similar. The cricket ball will suffer greater rolling resistance than the plastic ball because of its greater weight but if it's rolling on a smooth surface then I think that will be much less than the air resistance), the cricket ball will take a longer time to stop, and hence will travel further.

Actually, this question is appearing in our Chapter called Force and motion where the following equations are given:
$F=ma \\ F= \frac{m(v-u)}{t} \\ v = u+at \\ v^2 - u^2 = 2as \\ s = ut + \frac{1}{2} at^2 \\ \text{Conservation of Linear Momentum}$
It would be great if you could give me an answer with regard to this

 

[andrewkirk]

Actually, this question is appearing in our Chapter called Force and motion where the following equations are given:
$F=ma \\ F= \frac{m(v-u)}{t} \\ v = u+at \\ v^2 - u^2 = 2as \\ s = ut + \frac{1}{2} at^2 \\ \text{Conservation of Linear Momentum}$
It would be great if you could give me an answer with regard to this

All of those equations except the first are for a force that is constant over time, which will not be the case for this question. So they are not applicable.
Given the lack of information provided in the question, the best answer you can give will be based on the broad facts outlined in my second para in post 2. The cricket ball will win if the plastic ball is the same size as the cricket ball and the floor is polished wood. Why? Because air resistance is the dominant force and it's approx the same for both balls. So, given the mass of the cricket ball is greater, its stopping time will be longer and so its distance will be further.

 

[NickAtNight]

Actually, this question is appearing in our Chapter called Force and motion where the following equations are given:

We are not allowed to just give the answer. :(

1) You know that both balls have an initial velocity of $v_i$
2) Look more closely at the left side of the equation. The F is actually the sum of all the Forces acting on the object. $\sum F$
3) The initial force that started the motion (your hand) is gone.
4) there are two Forces (mentioned by the other poster) working pro slow the ball.
5) What are the formulas for those two forces?
6) What properties affect those two forces.
Hint: the other poster mentioned size of the ball. Skydivers change the rate that they fall at by changing their shape to increase/decrease their size and increase/decrease the wind resistance.
Hint: the web page I posted previously gives a formula for the other force.
7) it is a logic problem, so the answer is the possible answers depending on the circumstances. Not saying that well. But something like A goes further if x and y are more important. And B goes further if z and a are dominant.

 

[CWatters]

I think some of you are over thinking the problem.

Cricket balls are usually heavier than plastic balls of the same size.
Both start with the same velocity so one has more initial KE and momentum.
In the absence of info it would be reasonable to assume both encounter the same resistance (whatever the source might be).

 

[Viraam]

We are not allowed to just give the answer. :(

1) You know that both balls have an initial velocity of $v_i$
2) Look more closely at the left side of the equation. The F is actually the sum of all the Forces acting on the object. $\sum F$
3) The initial force that started the motion (your hand) is gone.
4) there are two Forces (mentioned by the other poster) working pro slow the ball.
5) What are the formulas for those two forces?
6) What properties affect those two forces.
Hint: the other poster mentioned size of the ball. Skydivers change the rate that they fall at by changing their shape to increase/decrease their size and increase/decrease the wind resistance.
Hint: the web page I posted previously gives a formula for the other force.
7) it is a logic problem, so the answer is the possible answers depending on the circumstances. Not saying that well. But something like A goes further if x and y are more important. And B goes further if z and a are dominant.

4) Air Resistance and Rolling Friction
5) Not Sure. The formula for rolling friction in the site is $F_R = μ_RWcos(a)$
6) Friction: Irregularities between the surfaces. Air Resistance: Shape of the body.

 

[NickAtNight]

4) Air Resistance and Rolling Friction
5) Not Sure. The formula for rolling friction in the site is $F_R = μ_RWcos(a)$
6) Friction: Irregularities between the surfaces. Air Resistance: Shape of the body.

Let's assume air resistance is negligible.

The only force to stop the balls is now rolling friction.
... Since we are on a flat surface, we can drop the cos(a). So $F_r = μ * W$

The rolling resistance can be expressed as. Fr = c W (1)

where
Fr = rolling resistance or rolling friction (N, lbf)
c = rolling resistance coefficient - dimensionless (coefficient of rolling friction - CRF)
W = m g = normal force - weight - of the body (N, lbf)
m = mass of body (kg, lb)
g = accelaration of gravity (9.81 m/s2, 32.174 ft/s2)
The rolling resistance can alternatively be expressed as
Fr = cl W / r (2)

where
cl = rolling resistance coefficient with dimension length (coefficient of rolling friction) (mm, in)
r = radius of wheel (mm, in)

Rolling Friction Coefficients
Some typical rolling coefficients:

Rolling Resistance Coefficient

So since the balls have the same initial velocity and F=m*a. Then F/m=a. So does the mass term on both sides of the equation cancel leaving us with just the rolling friction coefficient.

So if you have a plastic ball, of same mass and surface area as the cricket ball, then the friction coefficient is the determining factor for which ball stops later.

Now there are all kinds of plastics out there. So depending on how this ball is made and its resulting rolling friction coefficient, sets the answer.

A nice smooth, hard plastic ball would probably have a lower friction coefficient and roll a longer distance than a cricket ball.
... One of those super elastic rubber plastic bouncing balls, softer, but with energy recovery from deformation, and smooth, would also probably go further.
... But a squishy plastic, with a deliberately made rough surface - as long as it is rougher than the cricket balls surface, would go less distance.

 

[NickAtNight]

4) Air Resistance and Rolling Friction
5) Not Sure. The formula for rolling friction in the site is $F_R = μ_RWcos(a)$
6) Friction: Irregularities between the surfaces. Air Resistance: Shape of the body.

Now add back in air resistance...

Cross sectional area is important. As is the skin friction. What can you say about this force?

https://en.m.wikipedia.org/wiki/Air_resistance

Drag depends on the properties of the fluid and on the size, shape, and speed of the object. One way to express this is by means of the drag equation:

where

is the drag force,

is the density of the fluid,[11]

is the speed of the object relative to the fluid,

is the cross sectional area, and

is the drag coefficient – a dimensionless number.
The drag coefficient depends on the shape of the object and on the Reynolds number:

[\quote]

 

[andrewkirk]

In the absence of info it would be reasonable to assume both encounter the same resistance (whatever the source might be).

I do not think it would be reasonable. As I understand it, rolling resistance is proportional to weight, so the retarding force of rolling resistance would be greater for the cricket ball. So, ignoring rotational effects, it seems plausible that, in a vacuum, both balls may take roughly the same distance to stop. In fact it even seems plausible that the plastic ball would take longer to stop if hollow (as plastic balls typically are), because its ratio of moment of inertia to mass will be higher than for the solid cricket ball.

It appears to be only the air resistance that provides the strong differential effect that makes the cricket ball travel further.

 

[NickAtNight]

Interesting. Learning a lot about cricket balls.

What will the prominent stitching do to the cricket balls friction?

The Leather. The outer casing of a cricket ball is leather, which will be dyed to produce the desired colour, most commonly red or white. The leather outer is composed of four pieces of leather, two on each side of the ball. The side pieces are stitched together internally, whilst the two sides are then stitched together externally around the ‘equator’ of the ball, which will form the ball’s prominent seam.

Are we talking anew cricket ball or a used one. (Ignore me !)

A cricket ball which has been used for 80 overs will have significantly different characteristics from a ball which has never been bowled. Multiple overs of the ball being bowled into the pitch and hitting the bat will soften the ball. A softer ball will ‘stay in the pitch’ for longer when it bounces (i.e. the ball is in contact with the ground for a longer time than a harder ball would be when bowled at the same speed).

I was looking for the density of the cricket ball. No luck so far. Quotes are from http://www.decathlon.co.uk/blog/cricket/everything-you-need-to-know-about-cricket-balls/ [/quote][/quote]

 

[NickAtNight]

?..the plastic ball would take longer to stop if hollow (as plastic balls typically are)...

Let's go with a professional plastic ball... Hmmm, what would that be. How about a bowling ball ! https://en.m.wikipedia.org/wiki/Bowling_ball

Is a cricket ball more or less dense than a plastic bowling ball?

The USBC and FIQ specifies that bowling balls may only be made from uniform, solid materials with a density less than or equal to 3.80 g/mL. The weight of the ball must not exceed 16.00 pounds (7.26 kg), with no lower bound for weight. The hardness of the ball must be at least 72, as measured by a Type D Shore durometer at room temperature (68-78 degrees Fahrenheit). A ball may have a circumference between 26.704 inches (67.83 cm) and 27.002 inches (68.59 cm), and a diameter in the range of 8.500 inches (21.59 cm) to 8.595 inches (21.83 cm).[

In a test of a cricket ball versus a 16lb plastic bowling ball, the bowling ball would probably win hands down. :)

Our logic problem here did not limit us on mass or volume of the plastic ball - so we cheat and use a massive bowling ball.

Control for volume and make a mini urethane ball, and it probably still wins.

All kinds of cool information on the tricks professional bowlers got up to to enhance their score... https://en.m.wikipedia.org/wiki/Bowling_ball

Note: I have an old friend who's kid was the third youngest to score a perfect 300 in bowling. That is about 3x better than my typical score. They told me that bowling alleys oil their lanes differently for different games. One popular pattern is a Christmas tree shape. The location of the oil sets where location where the balls skid and where they roll. Once the ball stops skidding and starts to roll, then the spin the bowler put on the ball comes into play causing the ball to hook.

 

[NickAtNight]

Actually, this question is appearing in our Chapter called Force and motion where the following equations are given:
$F=ma \\ F= \frac{m(v-u)}{t} \\ v = u+at \\ v^2 - u^2 = 2as \\ s = ut + \frac{1}{2} at^2 \\ \text{Conservation of Linear Momentum}$

So did they explain where these equations come from? I suppose they just presented them.

So you have been using: $$\sum_{net}F = F_1 + F_2 + ... + F_i = m \times a$$

Now acceleration (a) is the rate of of change in velocity with respect to time.: $$ a = \frac {\delta V} {\delta Time} = \frac {V_2 - V_1} {T_2 - T_1}$$

An object with zero initial velocity is accelerated at 10 m/s^2 for 10 seconds. What is the final velocity?
...$$ a = \frac {\delta V} {\delta Time} = 10 m/s^2= \frac {V_2 - 0} {10 - 0}$$
Rearrange to solve for V_2
...$$ V_2 = 10 m / s^2 * 10 s= 100 m/s$$

graphical solution:
On graph paper, Draw a graph of gravitational acceleration (use 10 m/s^2 for simpler math) versus time for time from zero to ten seconds. It is a nice flat line. The change in velocity is simply the shaded area under the curve ! So count off the number of boxes for each second.

In the first second, we have 10 boxes, so we are at 10 m/s after the first second.
The second second adds ten more boxes, so we are now at 20 m/s.
The third second adds ten more, so 30 m/s
4,5,6,7,8,9. Are 40, 50, 60, 70 , 80, and 90 m/s respectively.
At 10 seconds, we will have 100 squares, and thus be at 100 m/s.

On graph paper, draw the velocity as a function of time for ten seconds. Notice that it is a nice straight line with a slope of 10 m/s.

Integration:
When we integrate, we are just adding up the area under the curve!

Position:
Now velocity (v) is just the rate of change in position with respect to time. $$ v = \frac {\delta S} {\delta Time} = \frac {S_2 - S_1} {T_2 - T_1}$$

If the initial position was zero, what is the final position? We would need to use the average velocity (50 m/s is 1/2 for 100 and 0)
$$ S_2 = 50 m/s * 10 seconds + o m = 500 m$$

Graphical solution:
Count up the squares underneath the curve (straight line)

So for 0 seconds, 0 m
1 second... 0+ 10 = 10 m
2 seconds. ... 10 + 20 = 30 m
3 s...... 30 + 30 = 60 m
...
10 s ....,,,,, 400 + 100 = 500 m

Plot your result position versus time on your graph paper. You should have a nice curve.

Note: if we go the other way, position to velocity to acceleration, we reverse the process. We 'differentiate'. No you are finding the slope of the line at each instant in time.
The slope of a squared line (position) is a straight line.
The slope of a straight line is a constant.

 

[NickAtNight]

\begin{equation}
V (t) = \int_{0s}^{10s} a\, \delta t= [a \times t] + V_i = [10 \frac{m}{s^2} \times 10 s] + 0 = 100 \frac{m}{s}
\end{equation}

\begin{equation}
S (t) = \int_{0s}^{10s} (at + c)\, \delta t = [\frac{1}{2} a \times t^2] + [V_i \times t] + S_i = [\frac{1}{2} * 10 \frac{m}{s^2} \times 100s^2] + [0 * 10 s] + 0 = 500 m
\end{equation}

 

[CWatters]

Viraam.. Can I ask if you have covered air resistance/drag and rolling resistance on your course yet?

 

[Viraam]

Viraam.. Can I ask if you have covered air resistance/drag and rolling resistance on your course yet?

Yes, we do know about these terms in the most simplest basis. Just definitions and the basic concept behind them. No formula, detailed explanation etc.

 

[NickAtNight]

Yes, we do know about these terms in the most simplest basis. Just definitions and the basic concept behind them. No formula, detailed explanation etc.

Galileo performed a classic experiment with rolling balls of different masses to determine gravitational acceleration.

Perhaps your answer is 'they travel the same distance'...

Galileo set out his ideas about falling bodies, and about projectiles in general, in a book called “Two New Sciences”. The two were the science of motion, which became the foundation-stone of physics, and the science of materials and construction, an important contribution to engineering.

The ideas are presented in lively fashion as a dialogue involving three characters, Salviati, Sagredo and Simplicio. The official Church point of view, that is, Aristotelianism, is put forward by the character called Simplicio, and usually demolished by the others. Galileo’s defense when accused of heresy in a similar book was that he was just setting out all points of view, but this is somewhat disingenuous---Simplicio is almost invariably portrayed as simpleminded.[\quote]

SALV. The argument is, as you see, ad hominem, that is, it is directed against those who thought the vacuum a prerequisite for motion. Now if I admit the argument to be conclusive and concede also that motion cannot take place in a vacuum, the assumption of a vacuum considered absolutely and not with reference to motion, is not thereby invalidated. But to tell you what the ancients might possibly have replied and in order to better understand just how conclusive Aristotle's demonstration is, we may, in my opinion, deny both of his assumptions. And as to the first, I greatly doubt that Aristotle ever tested by experiment whether it be true that two stones, one weighing ten times as much as the other, if allowed to fall, at the same instant, from a height of, say, 100 cubits, would so differ in speed that when the heavier had reached the ground, the other would not have fallen more than 10 cubits.

SIMP. His language would seem to indicate that he had tried the experiment, because he says: We see the heavier; now the word see shows that he had made the experiment.

SAGR. But I, Simplicio, who have made the test can assure you that a cannon ball weighing one or two hundred pounds, or even more, will not reach the ground by as much as a span ahead of a musket ball weighing only half a pound, provided both are dropped from a height of 200 cubits.

. From http://galileoandeinstein.physics.virginia.edu/tns61.pdf

And from the unreliable wiki. https://en.m.wikipedia.org/wiki/History_of_experiments

One prominent example is the "ball and ramp experiment."[2] In this experiment Galileo used an inclined plane and several steel balls of different weights. With this design, Galileo was able to slow down the falling motion and record, with reasonable accuracy, the times at which a steel ball passed certain markings on a beam.[3]Galileo disproved Aristotle's assertion that weight affects the speed of an object's fall. According to Aristotle's Theory of Falling Bodies, the heavier steel ball would reach the ground before the lighter steel ball. Galileo's hypothesis was that the two balls would reach the ground at the same time.

Other than Galileo, not many people of his day were able to accurately measure short time periods, such as the fall time of an object. Galileo accurately measured these short periods of time by creating a pulsilogon. This was a machine created to measure time using a pendulum.[4] The pendulum was synchronized to the human pulse. He used this to measure the time at which the weighted balls passed marks that he had made on the inclined plane. He measured to find that balls of different weights reached the bottom of the inclined plane at the same time and that the distance traveled was proportional to the square of the elapsed time.[5] Later scientists summarized Galileo's results as The Equation of Falling Bodies.[6][7]Distance d traveled by an object falling for time t where g is gravitational acceleration (~ 9.8 m/s2):
[\quote]