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Challenge Experiment: Tracker and Vertical Loops
Equipment: Video Capture, Tracker Analysis program, string, washers, rulers, plastic cups
or lids, water
Challenge: Take a plastic cup full of water, hold it with a straight arm, and whirl
it repeatedly in a vertical circle around your head. Try whirling the cup at
different speeds. What do you expect to happen as you slow down the rotation
speed. What determines whether you end up with wet or dry hair.
If you spin your arm slowly enough, then the water’s natural trajectory – a parabola – is
below the circular arc and the water has time to fall out of the cup while the cup travels
further along the circle. If you spin your arm quickly, the natural parabolic trajectory of the
water is upwards above the circle, but the cup traps the water and pushes it along the circular
arc so it can’t fall from the cup.
We don’t need to use a cup of water. A string with a mass on it can be used instead.
However, a mass on a string rotates with Non-Uniform Circular Motion. The mass is moving
at a higher speed at the bottom of the circle than at the top.
Let’s analyse this case of vertical motion. As
the water rotates in the cup, it would like to
follow a quadratic-type trajectory. Some
combination of the acting forces is keeping the
water rotating in a circle.
mg
NL
Nt
Nr
mg
mg
Nb
mg
There are only two forces acting on the water …
a downwards weight force W=mg and the
inwardly directed force variously labelled the
tension or normal force N. The only net force
on the water is constructed from these two
forces. In the diagram, there are four different
tension forces, at the bottom Nb, at the right Nr,
at the top Nt, and on the left NL.
Circular motion then requires that the centrally directed net force, or centripetal force Fcent, is
given by Fcent = mv2/r.
Write down four relations at the bottom, right, top and left side of the circles giving the
centrally directed force (noting there are different velocities at every point):
Bottom:
Top:
mvb2/r = Fcent =
mvt2/r = Fcent =
Right:
Left:
mvr2/r = Fcent =
mvL2/r = Fcent =
Rearrange these relations to give the tension in the string at various points in the circle:
Nb =
Nr =
Nt =
NL =
At what point in the circle is the tension the largest:
If the string is going to break, at what point in the rotation would it break:
At what point is the tension in the string the smallest:
If we keep our hand still and don’t put energy into the mass as it rotates, and if it loses a
negligible amount of energy, then energy is conserved. The total energy Etot = KE + PE is
always a constant. Write down the total energy at the bottom of the loop at height h=0, at the
top of the loop at height h=2r, and at some arbitrary point in between at height h where the
velocity is v:
Ebottom =
Etop =
Eh =
h
All of these energies are equal to each other. Let us say
that they all equal a constant E. Notice that there is a
linear relationship between v2 and h at any point. Write
down this relationship
v2 =
If we plot y=v2 and x=h, we get a straight line y=mx+c. Write down a formula for your
expected gradient and Y axis intercepts:
Gradient =
Intercept =
Rearrange your relation to make h the subject as a quadratic function of v in the form
y=ax2+bx+c. Write a formula for your expected value of “a” and the intercept with the Y
axis.
a=
Intercept =
Aim: To investigate the relationship between v and h for vertical circular motion, and to use
this to measure the acceleration due to gravity.
Hypothesis: That v2 varies linearly with h, so a determination of the gradient allows a
measurement of the acceleration due to gravity. And that h varies quadratically with v so
quadratic regression gives a measurement of the acceleration due to gravity.
Data and Analysis:
Use as large a radius as possible to slow down the motion. Measure the radius of the String:
r=
The frame rate is probably 30frames per second (unless you put the camera into “SPORT”
mode where the frame rate is doubled). This means that the time between images is probably
1/30s.
Swing a string of length r in a vertical loop. Video the loop and load the video into Tracker.
Do the required data analysis, and then export the data from Tracker into Excel. You can
choose to get Tracker to do all the work – Right Clicking on the graph in Tracker and choose
to show velocity v and export v, h, and t into Excel. Alternatively, I export x,y, t into Excel
and then use formulas to calculate v and h.
x2, y2, t2
For two neighbouring points, the mass has moved from point (x1,y1) to
point (x2,y2). Using Pythagorus, the distance moved is
d=
This movement occurred in a time of (t2-t1) and so the velocity was
v = d/( t2-t1) =
x1, y1, t1
This velocity occurred at an average height h = (y2 + y1)/2.
Your data should look something like this. (Tracker is probably using cm as units, so don’t
forget to correct this to m.)
t
0.00
0.03
0.07
0.10
0.13
0.17
x (cm)
0.00
17.80
33.87
46.63
55.20
59.41
y (cm)
0.00
2.70
10.47
22.25
36.50
51.60
x (m)
0.00
0.18
0.34
0.47
0.55
0.59
y (m)
0.00
0.03
0.10
0.22
0.36
0.52
v
(m/s)
5.4222
5.3731
5.2295
5.0039
4.7167
4.3917
Now you have to construct the velocity
and height data (or simply export it from
Tracker). Use formulas for distance d and
v above.
Use your data to make a plot of v2 vs h.
Use the gradient to calculate g.
Gradient =
Acceleration due to gravity g =
x (m)
0.00
0.18
0.34
0.47
0.55
y (m)
0.00
0.03
0.10
0.22
0.36
d (m)
0.18
0.18
0.17
0.17
v
(m/s)
5.40
5.35
5.21
4.99
v^2
(m^2/s^2)
29.18
28.66
27.16
24.88
h
(m)
0.01
0.07
0.16
0.29
Now make a plot of h vs v, and use your plot to calculate a measured value of g. The
quadratic regression gives
a=
From this, calculate your value of the acceleration due to gravity g =
Conclusion: What do you conclude about circular motion and the acceleration due to gravity.
Answers:
E.g. Bottom Fcent = Nb-mg, Right Fcent = Nr, E.g. Top Nt=Fcent-mg, Bottom Nb=Fcent+mg,
vt2=rg, Ebottom = mvb2/2, Etop = mvt2/2 +2mgr, Eh = mv2/2 + mgh, v2 = -2gh+2E/m, gradient =
-2g, intercept=2E/m, h=(-1/2g)v2+2E/m, d=sqrt[(x2-x1)2+(y2-y1)2], v=d/t.
Theoretical Data
x (m)
0.00
0.18
0.34
0.47
0.55
0.59
0.60
0.57
0.52
0.45
0.37
0.29
0.21
0.13
0.05
-0.03
-0.11
-0.19
y (m)
0.00
0.03
0.10
0.22
0.36
0.52
0.66
0.79
0.91
1.00
1.07
1.12
1.16
1.19
1.20
1.20
1.19
1.17
d (m)
0.18
0.18
0.17
0.17
0.16
0.15
0.13
0.12
0.11
0.10
0.10
0.09
0.09
0.08
0.08
0.08
0.08
v
(m/s)
5.40
5.35
5.21
4.99
4.70
4.38
4.04
3.71
3.41
3.13
2.90
2.71
2.57
2.48
2.43
2.43
2.47
Quadratic equation is
h = (-1/2g) v2 + E/(mg)
Quadratic regression gives
-0.0753= -1/(2g) s2/m
This rearranges to give
g = 6.64m/s/s.
v^2
(m^2/s^2)
29.18
28.66
27.16
24.88
22.12
19.19
16.36
13.80
11.61
9.82
8.42
7.36
6.62
6.14
5.91
5.89
6.09
h
(m)
0.01
0.07
0.16
0.29
0.44
0.59
0.73
0.85
0.95
1.03
1.10
1.14
1.17
1.19
1.20
1.19
1.18
Linear Gradient -2g = -20.172 2/s2.
This rearranges to give acceleration
g = -20.172/-2 = 10.6m/s2.