Centripetal Force Lab
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Centripetal Force Lab Online
Theory
Let’s consider a mass moving in a circle, with constant radius r
with a constant tangential speed |
v
|
. This means that at point
P
1
, and at time t
1
, the mass has velocity v
1
,
but later on at
point P
1
and time t
2
, the mass has velocity v
2
.
Now while the
direction of the velocity vector clearly changes as the mass
moves from the first point to the second, the magnitude of the
velocity vector (the speed) stays the same!
|
v
1
|
=
|
v
2
|
=
|
v
|
Even though the magnitudes of the velocity vectors are the
same the acceleration they are experiencing can’t be zero
because the velocity vectors are changing direction. ∆ v
=
v
2
−
v
1
≠
0
It turns out that ∆ v
points directly radially inwards to the
center of the circle, and that the difference in orientation of
the two velocity vectors is identical to ∆θ
the angular
displacement of the mass as it moved from P
1
to P
2.
If we consider this motion over an extremely small time
interval ∆t
=
t
2
−
t
1
→
0
s
, then something interesting
happens with the geometry of the situation; that being, we
get to similar triangles. Where ∆ s
is the arc length (linear displacement)
traveled by the mass during the time interval ∆t .
From the rules of simple geometry we know that
the ratios of corresponding sides of similar
triangles are equal to each other. Which allows us
to write the following equation.
∆s
r
=
∆v
v
1
Solving this for ∆ v
gives us,
∆ v
=
∆ s∙v
r
If we now divide both sides by the time interval we get,
∆v
∆t
=
∆ s∙v
∆t ∙r
Finally, invoking the definitions of acceleration and velocity we obtain a
c
=
v
2
r
a
c
is the centripetal acceleration the mass is experiencing as it moves in
a circle of radius r .
Centripetal means ‘center seeking’. The centripetal
acceleration always points towards the center of the curvature of the
path the mass is traveling on. From Newton’s Second Law we know that all forces can be written as
F
=
ma
. That means that the centripetal force acting on the mass
causing it move in a circle can be written as:
F
c
=
m a
c
=
m
v
2
r
Just like the centripetal acceleration, the centripetal force always points to the center of the curvature of the circular path the mass is traveling on. 2
Any force can act as a centripetal force, be it gravity, tension, friction or a combination thereof. The summation of the forces on the mass, acting in the radial direction, collectively is the centripetal force causing the circular motion. One method to determine the centripetal force acting on a mass is to utilize free body diagrams and force summation equations. Let us consider a mass traveling in a vertical circle attached to a string. Drawing a free body diagram of this mass when it’s at the lowest point of its circular path gives us the following. At this location in the mass’s path, the tension T
in the string points straight upwards, and the force of gravity mg
points straight downwards. From this diagram we write our force summation equation:
T
−
mg
=
F
c
In this simple case we see that the centripetal force is the difference between the tension in the string and the force of gravity. (Would this be true for any other location?)
The Tension could be replaced by a normal force from something like the track of a roller coaster as goes through a circular loop, and it would yield the same results. 3
Setup
1.
Go to the following website: https://www.walter-fendt.de/html5/phen/looping_en.htm
2.
You should now see the following:
Procedure
1.
In the Green area on the right side of your screen make the following settings.
a.
Radius, R = 0.500 m b.
Initial Height, y
1
= 1.260 m
c.
Gravitational Acceleration, g = 9.81 m/s
2
d.
Mass, m = 1.000 Kg
2.
Record these values in the Table in this worksheet. 3.
Near the bottom left of the Green area, make sure only the ‘weight, contact force’ is selected.
a.
Contact force is just another name for the Normal force. 4.
Near the bottom of the Yellow area please note that is where the value of the Contact Force can be found, as well as the velocity value. 5.
In the Green area, click on the start button to start the experiment.
a.
Once you click on the start button it will turn into a pause button. 6.
Right when the mass reaches the bottom of the loop, click on the pause button so that the
arrow representing the Contact force points straight up, or at least nearly straight up.
a.
You will most likely have to do this multiple times before you get one that is close
enough.
b.
Take a screenshot of your screen with the Contact force arrow pointing straight up, or
nearly straight up, and make sure to turn it in with this completed worksheet. 4
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