Example videos

• https://youtu.be/4pkAFmGZ6_M
• https://youtu.be/nr64gAbiZQw

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• In your computer, open a new text file and write the experimental data
• three columns: msec, position, and replica
• one row for each time the object crosses a mark

An ideal falling ball

Simulated data

What is the good formula this time?

We can use a logarithmic scale to see if we get a straight line

• If we get straight line in semi-log then the formula is \[y=A\cdot B^x\]
• If we get straight line in log-log then the formula is \[y=A\cdot x^B\]

Semi-Logarithmic scale

Logarithmic scale

This is not any exponential

There are many possible functions that connect \(x\) and \(y\)

We will try a polynomial formula

A polynomial is a formula like \[y=A + B\cdot x + C\cdot x^2 +\cdots+ D\cdot x^n\]

The highest exponent is the degree of the polynomial

We will try a 2-degree formula \[\text{height}=A+B\cdot\text{time}+C\cdot\text{time}^2\]

We really have to add a new column

You may be tempted to use

model <- lm(height ~ time^2, data=ball)

But that does not work.

Exponents and multiplications in formulas have a different meaning

(that we will see later)

A new column for polynomial model

We need to add an extra column, with the square of the time

ball %>% mutate(time_sq = time^2) -> ball
head(ball)

  height speed     time   time_sq
1  2.000 -0.20 0.000638 4.070e-07
2  1.978 -0.69 0.049256 2.426e-03
3  1.931 -1.18 0.100938 1.019e-02
4  1.860 -1.67 0.148492 2.205e-02
5  1.764 -2.16 0.200394 4.016e-02
6  1.644 -2.65 0.251069 6.304e-02

Linear model for polynomial

Now we use two independent variables in the model. This is different from the plot() function

model <- lm(height ~ time + time_sq, data=ball)
model

Call:
lm(formula = height ~ time + time_sq, data = ball)

Coefficients:
(Intercept)         time      time_sq  
      2.000       -0.205       -4.873  

Interpretation

We used the formula \[\text{height}=A+B\cdot\text{time}+C\cdot\text{time}^2\] The linear model gave us \[\begin{aligned} A & =2\\ B & =-0.205\\ C & =-4.9 \end{aligned}\]

Predicted v/s original

Interpretation

What is the meaning of the coefficients? \[\begin{aligned} \text{height} & =A+B\cdot\text{time}+C\cdot\text{time}^2 \\ A & =2\\ B & =-0.205\\ C & =-4.9 \end{aligned}\]

First view

plot(position~msec,
     data=fall)

Let’s keep only replica A

For this class, we will use only the first replica

fall %>% filter(replica=="A") -> fall
print(fall)

# A tibble: 9 x 3
   msec position replica
  <dbl>    <dbl> <chr>  
1  2355       10 A      
2  2457       20 A      
3  2533       30 A      
4  2558       40 A      
5  2639       50 A      
6  2690       60 A      
7  2742       70 A      
8  2742       80 A      
9  2797       90 A      

Time should start at zero

• We started the timer before dropping the ball
• The first data is when the ball crosses 10
• We can choose this as our time reference
• The ball is already moving
  • Initial speed is not zero
• The column name is msec. It means milliseconds

Normalizing time data

Let’s add a new column, called time, with the correct value of seconds

fall %>% mutate(time = msec/1000) -> fall
print(fall)

# A tibble: 9 x 4
   msec position replica  time
  <dbl>    <dbl> <chr>   <dbl>
1  2355       10 A        2.36
2  2457       20 A        2.46
3  2533       30 A        2.53
4  2558       40 A        2.56
5  2639       50 A        2.64
6  2690       60 A        2.69
7  2742       70 A        2.74
8  2742       80 A        2.74
9  2797       90 A        2.80

Normalizing time data

Now we change time to start at 0

fall %>% mutate(time = time - min(time)) -> fall
print(fall)

# A tibble: 9 x 4
   msec position replica  time
  <dbl>    <dbl> <chr>   <dbl>
1  2355       10 A       0    
2  2457       20 A       0.102
3  2533       30 A       0.178
4  2558       40 A       0.203
5  2639       50 A       0.284
6  2690       60 A       0.335
7  2742       70 A       0.387
8  2742       80 A       0.387
9  2797       90 A       0.442

Normalizing position data

We should also transform position to meters. The first data should be at 0

Each black/white strip was a little more than 19 centimeters

fall %>% mutate(height = (position/10 - 1) * 0.192) -> fall
print(fall)

# A tibble: 9 x 5
   msec position replica  time height
  <dbl>    <dbl> <chr>   <dbl>  <dbl>
1  2355       10 A       0      0    
2  2457       20 A       0.102  0.192
3  2533       30 A       0.178  0.384
4  2558       40 A       0.203  0.576
5  2639       50 A       0.284  0.768
6  2690       60 A       0.335  0.96 
7  2742       70 A       0.387  1.15 
8  2742       80 A       0.387  1.34 
9  2797       90 A       0.442  1.54 

Normalized data

plot(height~time, data=fall)

We will use a polinomial model

This is a parabola, which can be modeled with a second degree polynomial \[y=A + B\cdot x + C\cdot x^2\] When we use a polynomial model, we need to add new columns to our data frame

fall %>% mutate(time_sq = time^2) -> fall

This is used only for the model

Fitting the model

model <- lm(height ~ time + time_sq, 
            data=fall)
model

Call:
lm(formula = height ~ time + time_sq, data = fall)

Coefficients:
(Intercept)         time      time_sq  
    0.00131      1.55751      4.26664  

(Intercept) should be 0

Since we choose that at time=0 we have height=0, we know that (Intercept) is 0

In other words, our model here is \[y= B\cdot x + C\cdot x^2\] The old \(A\) has a fixed value of 0

How do we do that in R?

Fixing the intercept to 0

new_model_A <- lm(height ~ time + time_sq + 0, 
            data=fall)
new_model_A

Call:
lm(formula = height ~ time + time_sq + 0, data = fall)

Coefficients:
   time  time_sq  
   1.57     4.25  

Meaning of the coefficients

If you remember your physics lessons, you can recognize that in the formula \[y= B\cdot x + C\cdot x^2\] the value \(B\) is the initial speed, and \(C=-g/2\)

Thus, we can find the gravitational acceleration in the second coefficient