Method and Apparatus for the Demonstration and Calculation
of Centripetal Force
How can we determine the centripetal force acting on an object traveling in a circular path?
How can we establish the relationship between the centripetal force exerted on an object traveling in a circular path
and its orbital period, orbital frequency, mass, orbital velocity, orbital radius, and force of gravity?
Laboratory Exercise No. 2
Apparatus
Data Gathering and Calculations.
a. Convert the six suspended mass values into kilograms and enter into data table.
b. Calculate the weights (wt) of each of the six suspended mass values and enter into data table.
c. Along with your lab partners, use the stopwatch to measure the time it takes to complete 30 full orbits of the rubber stopper. Repeat this three times for each mass increment and average. Enter into data table.
d. Calculate the average orbital period (T) for each mass increment by dividing the average of 30 full orbits obtained in Step C above by 30. Enter into data table.
e. Calculate the orbital frequency (f) for each mass increment by calculating the inverse of the average orbital period, that is, by setting f = 1/T. Enter into data table.
f. Calculate the orbital velocity (v) for the given orbital radius and period by setting v = 2pr/T, where p = 3.14. Enter into data table.
g. Calculate the centripetal force (Fc) from the relationship Fc = mvexp2/r. Enter into data table.
h. Compare the vertical force (wt) of each suspended mass increment to the calculated centripetal force (Fc).
Questions
1. Plot a graph of centripetal force (Fc) versus orbital velocity (v) for each of the six individual values of suspended mass (hence weight). Orbital velocity should be regarded as the independent variable and centripetal force should be regarded as the dependent variable. Make sure to set-up your axes properly with each commencing numeration at zero. Create a separate table of values of Fc versus vbefore plotting your data.
2. Based on your graph of Fc versus v, briefly explain the observed relationship between centripetal force and orbital velocity of an object traveling in a circular path at a constant orbital radius.
3. Assuming that the Moon orbits the Earth in a circular path, calculate the Moon's orbital velocity (vm) and orbital period (Tm) given the following data and equations from the Reference Tables for Physics:
qMean distance from Earth to Moon: (r) = 3.8 x 10exp8 m
qMass of the Earth (me) = 6.0 x 10exp24 kg
qMass of the Moon (mm) = 7.4 x 10exp22 kg
qUniversal Gravitational Constant (G) = 6.7 x 10exp-11 N-mexp2/kgexp2
qForce of Gravitational Attraction between Earth and Moon (Fg) = G x me x mm/rexp2
qCentripetal Force Exerted on Moon by Earth (Fc) = mm x (vm)exp2/r
qOrbital Velocity of the Moon (vm) = 2pr/Tm
qOrbital Period of the Moon (Tm) = 2pr/vm
[Hint: Set Fg = Fc and solve for vm. Then solve for Tm.]
Express vm in km/s and Tm in days. How is the length of time Tm more familiarly known?
Introduction. As long as no resultant (net) force acts obliquely upon an object in motion it will travel along a linear path. If we wish to make an object travel in a circular path, however, a force acting towards the center of the circle must be applied to the object. This force is called a centripetal force (Fc). The centripetal force must change with the orbital velocity of the object if we are to keep the object traveling along a circular path of fixed radius. The orbital velocity (v), in turn, is proportional to the frequency of revolution (f) of the object and is inversely proportional to the period of revolution (T) of the object. Similarly, the centripetal acceleration is proportional to the square of the orbital velocity and inversely proportional to the orbital radius (r). Finally, following Newton's Second Law of Motion, the centripetal force acting on an object moving along a circular path is proportional to both the mass of the object (m) and its centripetal acceleration (ac). This is equivalent to the force of gravitational attraction (Fg) between itself and a central mass, much like that which exists between a planet and the Sun. This is simulated, in this particular situation, through the use of suspended masses attached to a line connected in turn to the orbital mass. This exercise is designed to illustrate the intriguing physical and mathematical relationships that exist between the centripetal force acting on an object traveling in a circular path and its period, frequency, mass, velocity, radius of travel, and force of gravity.
Cylindrical plastic ballpoint pen casing
1.5 m length of narrow gauge braided nylon cord
00 size single vent rubber stopper with a mass of 0.0045 kg
Adhesive tape
Paper clip
Meter stick
Stopwatch
Set of slotted "coin" type masses from Hooke's Law Apparatus:
1-10 gram mass
2-20 gram masses
2-50 gram masses
1-100 gram mass
Procedure. Thread one end of the nylon cord through the rubber stopper vent hole and securely attach with adhesive tape. Thread the other end of the cord through the ballpoint pen casing and tie a 4-cm diameter loop at the end. This loop will serve to connect the series of slotted masses to the apparatus. Attach the paper clip between the looped end of the cord and the bottom of the pen casing so as to establish a 0.50 m orbital radius (r) cord length between the top of the pen casing and the center of the rubber stopper. Add the masses according to the schedule listed below:
50 grams
100 grams
120 grams
150 grams
180 grams
200 grams
For each orbital velocity, the orbital radius (r) of the stopper will assume a constant value (0.50 m) and horizontal orientation, and the suspended masses will occupy a corresponding vertical orientation. Whirl the stopper over your head in a horizontal circular path at such a rate that the paper clip is just pulled up to, but not touching, the bottom of the pen casing. Here, the orbital radius of the rubber stopper should be exactly 0.50 m. Reposition the paper clip on the cord accordingly in order to ensure the 0.50 m orbital radius. Adjust the orbital velocity (faster v or slower v) so that the radius of the circular path remains at the constant 0.50 m value. This is accomplished by the "limiting" paper clip attached to the cord so that the masses remain suspended a constant distance below the bottom end of the pen casing. The centripetal force (Fc) furnished by the weight (wt) of the suspended masses is now keeping the stopper moving in a circular path of fixed radius.
It is important to keep two factors constant during the course of this exercise: the mass of the orbiting object and the radius of the orbit. The centripetal force is varied by altering theamount of mass suspended from the string. Note that the mass of the rubber stopper is 0.0045 kg and the mass of the cord is neglected.
qMean distance from Earth to Moon: (r) = 3.8 x 10exp8 m
