Galileo’s Method Non-Uniform Motion

RACING BALLS

• Which ball reaches the end of the track first? – Ball on straight track – Ball on bent track – Neither, it’s a tie!

Non – Uniform Motion Examination of Leslie's swim Timers stationed at approx. 4.6 m intervals along a pool who record the time it takes Leslie to pass them.

Overall average speed. Use the equation V = distance traveled / elapsed time V = (45.7 - 0) / (53 - 0) = .862 m / sec Where did Leslie have her greatest speed? Unless you can do a great number of calculations very quickly, the answer may take some time to figure out.

Split times First half speed V = (21.5 - 0)/(22 - 0) = 0.977 m/sec Second half speed V = (45.7 - 21.5)/(53 - 22) = .781 m/sec Fortunately there is an easier method!

Position (m) 0.0 4.6 9.1 13.7 18.3 21.5 27.4 32.0 36.7 41.2 45.7

Time (s) 0.0 2.5 5.5 11.0 16.0 22.0 26.5 32.0 39.5 47.5 53.0

Graph Interpretation The part of the graph with the greatest slope represents the part of Leslie's swim where her speed was the greatest.

Position vs Time 50.0 45.0 40.0 35.0

Notice that the graph starts off very steep then trails off then has another steep part before finally trailing off again. Can you explain the second steep part of the graph?

) m (30.0 n25.0 o i t i s20.0 o P 15.0 10.0 5.0 0.0 0.0

10.0

20.0

30.0

40.0

50.0

60.0

Time (s)

We can see the turn and estimate the length of the pool by looking at the graph. We might not be able to spot this in just the data table alone.

Instantaneous Speed By split breaking swim into Her speeds her are the slopes of Notice that Leslie's the lines that connect end smaller and smaller overall speed is the the time points of herof swim with the point slope the line intervals we get a that better and d = 22.9 m andthe t = end 22.0 sec. connects better representation of her points of her swim. actual swim. This line is anot verysense These lines give better indicative of her for what her swim was really like. swim.

Position vs Time 50.0 50.0 45.0 40.0

)35.0 ) ) m m ( (30.0 n n o o25.0 i t i s20.0 o P 15.0 10.0 10.0 5.0 5.0 0.0 0.0 0.0 0.0

10.0 10.0

20.0 20.0

30.0 30.0

40.0 40.0

Time (s)

If the time interval is infinitesimally small we refer to the speed computed as the instantaneous speed; and is represented by the equation:

50.0 50.0

60.0 60.0

Example In this equation Vinst refers to the speed at a particular instant of time and lim t->0 means to limit the time interval to a value very close to zero - but not equal to zero. Position vs Time

Example - find Leslie's speed at t = 10 sec. 50.0

On the graph I would select two point that are very close to the 10 sec mark (perhaps at t = 9.5 and t = 10.5)

Next I need to estimate the distance coordinates for these times. (Perhaps d = 11 m with t = 9.5 and d = 13 with t = 10.5) Now use the two point slope formula that were are accustomed to using, i.e. d/t = v. Hence v(at 10 sec) = (13 - 11)/(10.5 - 9.5) = 2 m/sec.

45.0 40.0

)35.0 m (30.0 n o25.0 i t i s20.0 o P 15.0 10.0 5.0 0.0 0.0

10.0

20.0

30.0

40.0

50.0

60.0

Time (s)

The procedure can be repeated for as many instances as we need or desire - the more often we do this the better will be our understanding of the entire swim This concept, of instantaneous speed, can be extended and developed into the fundamental theorem of calculus. Watch for it in your math class! Your math class will have the time to develop quicker and easier techniques for finding the limit. However, the technique that I've outlined above will suffice for this course.

Galileo’s Observation The Greeks view of motion - Accepted unchallenged for over 2000 years Based upon common sense. Four terrestrial elements and their placement could account for all natural motion Violent motion needed a force to cause it. (Since the earth itself was so large, and already in its natural position, they could imagine no force large enough to move it. Hence the earth in their mind had to be at rest)

Hidden in their logic was the idea that an object composed of twice as much "earth" as another object would seek its natural position twice as fast as another object that might be partly composed of "air". This could never have been tested.

It All Falls Apart Galileo noticed this discrepancy while watching a chandelier swinging in a breeze. The chandelier was made of many small pieces of glass attached to many larger pieces of glass. If Aristotle was right, the chandelier should have pulled itself apart on the first swing. (The larger pieces would have traveled faster than the smaller ones.)

Galileo timed the swinging of various parts (with his pulse) and found them to be the same. He wrote that he became so excited with his discovery that he had to quit timing due to the quickening of his pulse.

Uniform Acceleration Galileo could not explain the motion of a freely falling object. The motion occurred too fast for the technology of his day to measure. He assumed that an object that falls gains its' speed in the most simple manner possible, i.e. Equal increments of speed are gained in equal increments of time - Uniform Acceleration

This can be represented by an equation to compute the average acceleration:

a

v

t

(v f

vi )

(t f

t i )

Note that this equation and the definition of uniform acceleration closely parallel the equation for uniform speed and its definition.

Thus the slope of a speed versus time graph is the acceleration

The Three Motions

There is a similarity and a possible connection between the types of motion that is suggested by the graphs.

Speed vs time graphs Let’s examine the speed vs time graph for an object traveling at constant speed of 3 m/s for 8 s. In 2 seconds the object travels: d = v*t = 6 m The area of the rectangle under the graph at 2 seconds is A = l*w = 3*2 = 6 the same as the distance traveled The distance that it travels in 8 seconds is d = t*v = 24 m which is equal to the area under the graph at 8 seconds

Advantage to Area under the graph

A3 = ½ *4*4 =8m

A2 = 4*2 =8m

A1 = 4*6 = 24 m

An object is traveling at 2 m/s and then increases its’ speed by one m/s each second for 4 seconds until it is traveling at 6 m/s. It then travels at 6 m/s for the next 4 seconds. How far has it traveled?

d

= AT = A1+A2+A3 =24 + 8 + 8 d = 40m

Merton Mean Speed Rule An object whose speed in increasing 2 m/s every second travels for 8 seconds. How far did it go? Area under the graph is A = ½ 8*16 = 64 m NOTICE, that’s the same as if it traveled at the average speed for the entire time vavet

Area under the graph is A = 8*8 = 64 m

vave

vi

v f

2

vi

v f

2

t or,

How far? vave vave

vave

12

vi

v f

2 6 18 2 d t

d

8

d 96

12

DON’T USE The Merton Mean Speed Rule only works when the change in speed is constant and only during the time period when the speed is changing at this constant rate, We’re looking for expressions that work all of the time. The MMSR is too particular to be of much use and I don’t encourage you to use it. We will make one use of this once in the near future and then never return to it. You’re welcome to use it but please be aware of the restrictions of when it will work for you and when it will not.

His Approach Galileo had to test his hypothesis that objects gain equal amounts of speed in equal time periods. The technology of his day would not allow him to measure the very small time intervals associated with falling objects Also, to test the equation directly would entail measuring two instantaneous speeds. To this day we cannot measure even one instantaneous speed we can only calculate it.

Galileo would need to derive an indirect test - i.e. he would have to test a consequence of his hypothesis instead of the hypothesis itself. (This is quite common in science)

Algebra Galileo knew that if an object uniformly increased its' speed then the average speed would be midway between the initial and final speeds i.e vave = (vf + v i)/2 This is known as the Merton Theorem or Merton Rule

From the work that had been completed on uniform motion: vave = (df - di)/(tf - t i) His hypothesis:

aave = (vf - vi)/(tf - t i)

The Ramp Experiment The ramp experiment: 1. A ball is rolled from rest down a ramp. 2. Distance is measured from the point of release. 3. Time is measured from the instant of release. vi = 0 ti = 0 di= 0

vave = (vf ) /2 vave = (df )/(tf ) aave = (vf )/(tf ) Combining top two eq gives:

vf = 2 df /(tf )

Substituting this into 3rd equation yields: df =

1 / at 2 2 f

This equation can be tested since it involves only measuring distances and times. If a is uniform (or constant) as Galileo hypothesized, what would a graph of d versus t look like? Answer - parabola (let y = d, x = t, k = 1 / 2a) Then equation looks like y = kx 2

Some algebra vave vave a

2vi

v f

2

d f

d i

v f

t

v f

vi

2

vi

v f

d f

d i

(vi

at ) vi

t

2

d f

d i

t

vi at

t

at

2

vi

d f

d i

t

or...

d f

vi

1

2 at

d f

d i

t

d i vi t

or...

1

2

vi t 1 2 at

2

2

at

(Also works all of the time)

d f

d i

The Kinematic Expressions Basic Definitions (work all of the time) vave a

d f

d i

t

v f

vi

vave

t

Derived Expressions (work all of the time)

d f v f

d i vi t 2

Special Rule (Does NOT work all of the time)

2

1

2 a t 2

vi 2ad

v f

vi

2

Only works when the acceleration is constant or the object travels at each of the two speeds for the same amount of time.

RACING BALLS

• Which ball reaches the end of the track first? – Ball on straight track – Ball on bent track – Neither, it’s a tie!

Non – Uniform Motion Examination of Leslie's swim Timers stationed at approx. 4.6 m intervals along a pool who record the time it takes Leslie to pass them.

Overall average speed. Use the equation V = distance traveled / elapsed time V = (45.7 - 0) / (53 - 0) = .862 m / sec Where did Leslie have her greatest speed? Unless you can do a great number of calculations very quickly, the answer may take some time to figure out.

Split times First half speed V = (21.5 - 0)/(22 - 0) = 0.977 m/sec Second half speed V = (45.7 - 21.5)/(53 - 22) = .781 m/sec Fortunately there is an easier method!

Position (m) 0.0 4.6 9.1 13.7 18.3 21.5 27.4 32.0 36.7 41.2 45.7

Time (s) 0.0 2.5 5.5 11.0 16.0 22.0 26.5 32.0 39.5 47.5 53.0

Graph Interpretation The part of the graph with the greatest slope represents the part of Leslie's swim where her speed was the greatest.

Position vs Time 50.0 45.0 40.0 35.0

Notice that the graph starts off very steep then trails off then has another steep part before finally trailing off again. Can you explain the second steep part of the graph?

) m (30.0 n25.0 o i t i s20.0 o P 15.0 10.0 5.0 0.0 0.0

10.0

20.0

30.0

40.0

50.0

60.0

Time (s)

We can see the turn and estimate the length of the pool by looking at the graph. We might not be able to spot this in just the data table alone.

Instantaneous Speed By split breaking swim into Her speeds her are the slopes of Notice that Leslie's the lines that connect end smaller and smaller overall speed is the the time points of herof swim with the point slope the line intervals we get a that better and d = 22.9 m andthe t = end 22.0 sec. connects better representation of her points of her swim. actual swim. This line is anot verysense These lines give better indicative of her for what her swim was really like. swim.

Position vs Time 50.0 50.0 45.0 40.0

)35.0 ) ) m m ( (30.0 n n o o25.0 i t i s20.0 o P 15.0 10.0 10.0 5.0 5.0 0.0 0.0 0.0 0.0

10.0 10.0

20.0 20.0

30.0 30.0

40.0 40.0

Time (s)

If the time interval is infinitesimally small we refer to the speed computed as the instantaneous speed; and is represented by the equation:

50.0 50.0

60.0 60.0

Example In this equation Vinst refers to the speed at a particular instant of time and lim t->0 means to limit the time interval to a value very close to zero - but not equal to zero. Position vs Time

Example - find Leslie's speed at t = 10 sec. 50.0

On the graph I would select two point that are very close to the 10 sec mark (perhaps at t = 9.5 and t = 10.5)

Next I need to estimate the distance coordinates for these times. (Perhaps d = 11 m with t = 9.5 and d = 13 with t = 10.5) Now use the two point slope formula that were are accustomed to using, i.e. d/t = v. Hence v(at 10 sec) = (13 - 11)/(10.5 - 9.5) = 2 m/sec.

45.0 40.0

)35.0 m (30.0 n o25.0 i t i s20.0 o P 15.0 10.0 5.0 0.0 0.0

10.0

20.0

30.0

40.0

50.0

60.0

Time (s)

The procedure can be repeated for as many instances as we need or desire - the more often we do this the better will be our understanding of the entire swim This concept, of instantaneous speed, can be extended and developed into the fundamental theorem of calculus. Watch for it in your math class! Your math class will have the time to develop quicker and easier techniques for finding the limit. However, the technique that I've outlined above will suffice for this course.

Galileo’s Observation The Greeks view of motion - Accepted unchallenged for over 2000 years Based upon common sense. Four terrestrial elements and their placement could account for all natural motion Violent motion needed a force to cause it. (Since the earth itself was so large, and already in its natural position, they could imagine no force large enough to move it. Hence the earth in their mind had to be at rest)

Hidden in their logic was the idea that an object composed of twice as much "earth" as another object would seek its natural position twice as fast as another object that might be partly composed of "air". This could never have been tested.

It All Falls Apart Galileo noticed this discrepancy while watching a chandelier swinging in a breeze. The chandelier was made of many small pieces of glass attached to many larger pieces of glass. If Aristotle was right, the chandelier should have pulled itself apart on the first swing. (The larger pieces would have traveled faster than the smaller ones.)

Galileo timed the swinging of various parts (with his pulse) and found them to be the same. He wrote that he became so excited with his discovery that he had to quit timing due to the quickening of his pulse.

Uniform Acceleration Galileo could not explain the motion of a freely falling object. The motion occurred too fast for the technology of his day to measure. He assumed that an object that falls gains its' speed in the most simple manner possible, i.e. Equal increments of speed are gained in equal increments of time - Uniform Acceleration

This can be represented by an equation to compute the average acceleration:

a

v

t

(v f

vi )

(t f

t i )

Note that this equation and the definition of uniform acceleration closely parallel the equation for uniform speed and its definition.

Thus the slope of a speed versus time graph is the acceleration

The Three Motions

There is a similarity and a possible connection between the types of motion that is suggested by the graphs.

Speed vs time graphs Let’s examine the speed vs time graph for an object traveling at constant speed of 3 m/s for 8 s. In 2 seconds the object travels: d = v*t = 6 m The area of the rectangle under the graph at 2 seconds is A = l*w = 3*2 = 6 the same as the distance traveled The distance that it travels in 8 seconds is d = t*v = 24 m which is equal to the area under the graph at 8 seconds

Advantage to Area under the graph

A3 = ½ *4*4 =8m

A2 = 4*2 =8m

A1 = 4*6 = 24 m

An object is traveling at 2 m/s and then increases its’ speed by one m/s each second for 4 seconds until it is traveling at 6 m/s. It then travels at 6 m/s for the next 4 seconds. How far has it traveled?

d

= AT = A1+A2+A3 =24 + 8 + 8 d = 40m

Merton Mean Speed Rule An object whose speed in increasing 2 m/s every second travels for 8 seconds. How far did it go? Area under the graph is A = ½ 8*16 = 64 m NOTICE, that’s the same as if it traveled at the average speed for the entire time vavet

Area under the graph is A = 8*8 = 64 m

vave

vi

v f

2

vi

v f

2

t or,

How far? vave vave

vave

12

vi

v f

2 6 18 2 d t

d

8

d 96

12

DON’T USE The Merton Mean Speed Rule only works when the change in speed is constant and only during the time period when the speed is changing at this constant rate, We’re looking for expressions that work all of the time. The MMSR is too particular to be of much use and I don’t encourage you to use it. We will make one use of this once in the near future and then never return to it. You’re welcome to use it but please be aware of the restrictions of when it will work for you and when it will not.

His Approach Galileo had to test his hypothesis that objects gain equal amounts of speed in equal time periods. The technology of his day would not allow him to measure the very small time intervals associated with falling objects Also, to test the equation directly would entail measuring two instantaneous speeds. To this day we cannot measure even one instantaneous speed we can only calculate it.

Galileo would need to derive an indirect test - i.e. he would have to test a consequence of his hypothesis instead of the hypothesis itself. (This is quite common in science)

Algebra Galileo knew that if an object uniformly increased its' speed then the average speed would be midway between the initial and final speeds i.e vave = (vf + v i)/2 This is known as the Merton Theorem or Merton Rule

From the work that had been completed on uniform motion: vave = (df - di)/(tf - t i) His hypothesis:

aave = (vf - vi)/(tf - t i)

The Ramp Experiment The ramp experiment: 1. A ball is rolled from rest down a ramp. 2. Distance is measured from the point of release. 3. Time is measured from the instant of release. vi = 0 ti = 0 di= 0

vave = (vf ) /2 vave = (df )/(tf ) aave = (vf )/(tf ) Combining top two eq gives:

vf = 2 df /(tf )

Substituting this into 3rd equation yields: df =

1 / at 2 2 f

This equation can be tested since it involves only measuring distances and times. If a is uniform (or constant) as Galileo hypothesized, what would a graph of d versus t look like? Answer - parabola (let y = d, x = t, k = 1 / 2a) Then equation looks like y = kx 2

Some algebra vave vave a

2vi

v f

2

d f

d i

v f

t

v f

vi

2

vi

v f

d f

d i

(vi

at ) vi

t

2

d f

d i

t

vi at

t

at

2

vi

d f

d i

t

or...

d f

vi

1

2 at

d f

d i

t

d i vi t

or...

1

2

vi t 1 2 at

2

2

at

(Also works all of the time)

d f

d i

The Kinematic Expressions Basic Definitions (work all of the time) vave a

d f

d i

t

v f

vi

vave

t

Derived Expressions (work all of the time)

d f v f

d i vi t 2

Special Rule (Does NOT work all of the time)

2

1

2 a t 2

vi 2ad

v f

vi

2

Only works when the acceleration is constant or the object travels at each of the two speeds for the same amount of time.