PHYS 30 LESSONS��Unit 3: Electromagnetism
Lesson 2: Curve Straightening
Weighted Averages
When experiments are done, random error is unavoidable. Some values will be higher than the true value, while other values are lower than the true value.
Thus, in order to eliminate this rounding error, we compute the average.
There are many different types of averages.
e.g. mean mode median
In Physics 30, we use the weighted average. This is determined using the line of best-fit.
A line of best-fit is straightforward when the data is linear. You need only draw a straight line and the data can be analyzed (using the slope and y-intercept).
But what if the data is non-linear?
How can we use a line of best-fit approach?
Curve Straightening
When the data is non-linear (curved), we can use an approach called “curve straightning”.
In essence, this procedure will take a non-linear relationship and make it linear.
Once it does so, we can use simple straight-line theory to analyze the data.
There are 4 steps to curve-straightening:
1. Determine the relationships between the variable.
How?
∙ Identify the manipulated (x), responding (y), and controlled variables.
∙ Get the y-variable (y) by itself in the equation.
∙ State the proportion.
2. Establish a direct relationship between the variables.
How?
If y ∝ x2, then y has a direct relationship with x2.
If , then y has a direct relationship with .
Why does this work?
Consider the equation
We can see that
That is, Ek has an square relationship with v .
Ek
v
But notice what happens when we make the following substitution:
Let y = Ek
Let x = v2
Let y = Ek
x = v2
Let y = Ek
x = v2
We can see that y has a direct relationship with x.
Let y = Ek
x = v2
We can see that y has a direct relationship with x.
It follows that Ek has a direct relationship with v2 .
Graphically, this can be shown as:
Notice that we have now straightened the curve!!
It follows that Ek has a direct relationship with v2 .
Ek
v2
Ek
v
3. Make a new table and graph the direct relationship.
If y ∝ x2, then you would graph .
If , then you would graph .
Essentially, you graph the direct proportion
4. Determine the average value.
When you have a straight-line graph,
∙ Calculate the slope (including the units)
∙ Using the significance of the slope, determine the desired (average) value
Circular Motion Experiment
A student systematically changed
the radius and measured the centripetal
force needed to keep it in circular motion.
The mass and the circular speed were
held constant.
The known equation is
m
r
This example is shown on pp. 262 - 264 of your workbook
Data: (speed and mass held constant)
r (m) 0.10 0.15 0.20 0.25 0.30 0.35 0.40
Fc (N) 1450 1200 720 580 490 415 360
If the mass of the object is 5.0 kg, determine the speed using
the significance of the slope.
∙ Identify the MV, RV, and controls
r (m) 0.10 0.15 0.20 0.25 0.30 0.35 0.40
Fc (N) 1450 1200 720 580 490 415 360
r (m) 0.10 0.15 0.20 0.25 0.30 0.35 0.40
Fc (N) 1450 1200 720 580 490 415 360
MV (x): r
The radius is being changed regularly by the experimenter.
r (m) 0.10 0.15 0.20 0.25 0.30 0.35 0.40
Fc (N) 1450 1200 720 580 490 415 360
MV (x): r
RV (y): Fc
The centripetal force is responding to changes in the radius.
r (m) 0.10 0.15 0.20 0.25 0.30 0.35 0.40
Fc (N) 1450 1200 720 580 490 415 360
MV (x): r
RV (y): Fc
Controls: m and v
∙ State the relationship between the variables
Which variable do you get by itself?
MV: r
RV: Fc
y (RV) is always isolated
Thus, it follows that Fc has an inverse relationship with r.
Fc
r
∙ Establish a direct relationship between the variables
We know that Fc has an inverse relationship with r
But, Fc has a direct relationship with
Proof:
Let y = Fc and x =
then,
y ∝ x (direct relationship)
Since Fc has a direct relationship with
if we graph Fc vs , we will get a straight line.
Fc
∙ Make a new table (of the direct relationship)
Fc (N) 1450 1200 720 580 490 415 360
r (m) 0.10 0.15 0.20 0.25 0.30 0.35 0.40
We need to graph Fc vs 1/r
New table: (Fc vs 1/r)
Fc (N) 1450 1200 720 580 490 415 360
r (m) 0.10 0.15 0.20 0.25 0.30 0.35 0.40
Change the unit as well!
New table: (Fc vs 1/r)
Fc (N) 1450 1200 720 580 490 415 360
r (m) 0.10 0.15 0.20 0.25 0.30 0.35 0.40
10.0
1
0.10
New table: (Fc vs 1/r)
Fc (N) 1450 1200 720 580 490 415 360
r (m) 0.10 0.15 0.20 0.25 0.30 0.35 0.40
10.0 6.67 5.00 4.00 3.33 2.86 2.50
Do the reciprocal for each distance r
This leads to a straight-line relationship:
Fc (N)
Fc as a function of the reciprocal of radius
2
4
6
8
10
400
800
1200
1600
Fc (N)
Fc as a function of the reciprocal of radius
2
4
6
8
10
400
800
1200
1600
Draw a line of best-fit (average line)
Outlier (ignore)
Choose two points on the line. Label them clearly!
Fc (N)
Fc as a function of the reciprocal of radius
2
4
6
8
10
400
800
1200
1600
(2.50 1/m, 360 N)
(7.00 1/m, 1000 N)
Calculate the slope:
m = y2 - y1 = 1000 N - 360 N
x2 - x1 7.00 1/m - 2.50 1/m
= 142 N⋅m
Include the units
Determine the equation of the line of best-fit:
y = m x + b
Fc
y = m x + b
Fc = (slope)
Fc
∙ State the significance of the slope
- compare your equation with the known equation
Find the speed from the slope.
Your equation:
Known equation:
Find the speed:
Conceptual Example
In a circular motion experiment, the period of rotation (T) is varied and the resulting orbital speed (v) is measured. All other variables are held constant.
∙ State the relationship between these variables and sketch the straightened curve
∙ Determine how to determine the radius using the significance of the slope.
Equation:
First, identify the MV, RV, and controls
MV: Period (T), since this was varied
RV: Speed (v), since this was measured
Controls: Radius (r)
Before you state the relationship, you must get the RV by itself:
MV: Period (T)
RV: Speed (v)
Thus, speed has an inverse relationship with T.
v
T
Speed has an inverse relationship with T ...
but v has a direct relationship with 1 / T.
v
T
v
To determine the significance of the slope, we first have to determine the equation of the “best-fit” line.
The equation of any straight line through the origin is
v
y
x
Since the y-variable is v and the x-variable is 1 / T, it follows that
v
y
x
or
Now, we can compare the two equations:
Derived equation
Known equation
Finally, we can find the radius: