Read 125 Physics Projects for the Evil Genius Online
Authors: Jerry Silver
3. Holding the tube in one hand, swing the rubber stopper in a smooth, horizontal circle.
Figure 13-2
Using a spring scale to measure centripetal acceleration
.
4. Measure how many seconds it takes to make ten rotations, and then divide by ten to get the period for one rotation. Be careful to count the first rotation at the
end, rather than at the beginning
, of the rotation. It may help to count “zero” when you start, and then to count “one” when the first rotation is completed.
5. Using the marker, place a series of marks at 1 centimeter intervals, starting at the loop for the hanging mass.
First investigation: Force versus velocity (for fixed radius and fixed rotating mass)6. Using the meter stick, identify the distance between the top of the tube and the rubber stopper associated with the mark closest to the hanging weight. You can now easily measure the radius by subtracting 1 centimeter for every mark below the tube that you can count. (You can also determine the radius by measuring the length of string below the tube and subtracting from the total length of the string.) You can also use a piece of tape or a paper clip to mark the position of the string to give a radius that you measure before spinning. However you do it, make sure that nothing restricts the free movement of the string through the tube.
Second investigation: Force versus the rotating object’s mass (for fixed radius and fixed period)Radius for all trials = ____ meters
Radius for all trials = (constant) ____ meters
Third investigation: Force versus orbital radius (for fixed period and fixed rotating mass)Period for all trials = (constant) ____ seconds
This part is more complicated than the previous two investigations and will require a greater degree of skill and patience.
Expected ResultsPeriod for all trials = ____ seconds
This project leads to the following conclusions:
1. The faster the rotation (or the shorter the period of rotation), the greater the centripetal force needed to maintain circular motion.
For a 12-gram rubber stopper, the expected results are shown in
Figure 13-3
. This shows the relationship is
not
linear, but that it increases more rapidly as the velocity increases.
2. The greater the mass, the greater the force needed to keep the rubber stopper going at a given speed at a particular radius. This result is expected to be linear.
3. For a given rotational speed, the shorter the string, the greater the force needed.
For a 12-gram rubber stopper, the expected results are shown in
Figure 13-4
, which shows an inverse relationship between force and string length.
Figure 13-3
Centripetal force versus velocity
.
Figure 13-4
Force versus string length
.
The “string” that keeps an object going around in a circle is provided by a centripetal force. In this case, it is literally a string. In the case of a satellite or planet, the “string” is the gravitational force.
The faster the object goes (for a given radius), the greater the force, according to the equation:
where
F
c
is the centripetal force,
m
is the mass of the spinning object (the washer in our case),
v
is the velocity of the washer, and
r
is the radius of the circle.
Given the data shown in
Figure 13-3
, we can determine that force increases with the square of the rubber stopper’s velocity in one of two ways:
1. Use a curve-fitting program, such as Excel. From a scatter plot, with the data selected, go to the Chart menu, select Add Trendline, and then select a power fit option. Select Add Equation to the Chart from the Options tab. This displays the mathematical model for your data. The expected result is for this to be the form y = x
2
or close to it.