Abstract
A body is said to be moving in a uniform circular motion if it is moving in a circle with a constant speed. The main objective of this experiment was to find the net force that acts on an object of mass m which moves in a circle of radius r at constant speed v and to confirm that the magnitude of that force is given by mv2/r.An experiment was done using centripetal force apparatus by hanging masses on it and computing the force required to stretch the string, the masses were then subjected to a rotational frequency and the average frequency determined. The force required were then determined from the formula and compared with the initially measured value. The results obtained showed that there is a partial agreement of the experimental and the measured value.
Theory
A body is said to be moving in a uniform circular motion if it is moving in a circle with a constant speed. Although the speed is constant, the direction of motion changes continuously and hence the velocity keeps on changing. The change in velocity shows that the body has acceleration. The acceleration is given by:
a = v2/r, toward the centre of the circle (1)
Where r is the radius of the path, and v is the velocity of the body. From the second law of motion, the amount of force that is required to produce this acceleration is be given by the formula: F = ma. Therefore, the magnitude of the force is:
F = mv2/r
The magnitude of the force can be expressed in terms of the frequency f of rotation, from
v= ω r = 2∏fr, since ω= 2∏f (2)
Therefore:
F = mr(2∏f)2 (3)
Combining equation (1) and (2),
a= ω2r = (2∏f)2r (4)
In the apparatus that are used in this experiment (Fig. 1), the acceleration of the rotating mass is caused by the force that is exerted by a stretched spring. At a given speed of rotation the pointer will be in the opposite direction of the index. This rotational speed is maintained at a constant value and it is measured by the help of the frequency meter that is mounted on the apparatus. The force that is exerted by the spring with the mass at the index position can be obtained by statically measuring the weight that is needed to cause the displacement of the spring by the same extent.
Every time a body is moving in a circular motion, a force directed to the centre of the circle must act on the body to keep it in the direction of motion along that path. The main objective of this experiment is to find the net force that acts on an object of mass m which moves in a circle of radius r at constant speed v and to confirm that the magnitude of that force is given by mv2/r.
Equipment list
- Centripetal force apparatus (rotating assembly, chuck, frequency meter, drive)
- Weight hanger and weights
- Vernier caliper
Procedure
- The weight force that was applied to stretch the spring in order for the index of the rotating apparatus and the pointer to coincide was determined by hanging the apparatus from the provided stand with the cylindrical mass M hanging downward. The cylindrical mass (placed inside the apparatus) had a string attached to its face that was used to attach a mass hanger. The masses were then started to be applied to stretch the spring.
Figure 1. Rotating assembly. Y-rotor frame; M-cylindrical mass: Z-spring; P-pointer; I-indicator
- The masses were added until the spring was extended to the point where the needle-like indicator coincided with the index. The amount of mass required was then recorded. The magnitude of the force required to stretch the spring was computed (i.e., the total weight in Newtons of all the masses that were supported by the spring).
- The Vernier calipers was used to take the measurements of the radius of rotation of mass M. this was the distance between the line that was inscribed on the side of the mass M and the central line of the apparatus.
- The rotor was inserted into the chuck jaws of the apparatus and then tightened with the chuck key.
- The speed control was turned to the lowest setting and then the motor was turned on. The readout was set to measure RPM (revolutions per minute)
- The speed was then increased gradually until the pointer of the indicator was exactly opposite the index. This happens when the spring is stretched by the same amount as when the masses were hanging from it. Care was taken to avoid taken the machine to excessive speeds.
- The frequency of rotation in RPM was indicated by a meter mounted at the base of the motor assembly.
- Five sets of data were taken and marked as high f, low f. the average or most likely value f for each set of data was determined. The uncertainty ∆f as standard deviation for the measured frequencies f was calculated.
- For each set of data, the radius of rotation and the frequency f were calculated from the value of mass M. the magnitude of the force exerted by the spring during the rotation was calculated using formula (3). The RPM was converted to revolutions per second.
- For each set of data, the uncertainty in the spring force magnitude was calculated by finding the spring force that corresponded to the maximum value of frequency f high :
∆F=Fhigh-F= Mr(2∏f high)2- Mr(2∏f)2
- The magnitude of the force exerted by the spring on the swiftly rotating small mass with that exerted by the masses that were hanged from the spring as indicated in procedure step 2 were compared. The absolute discrepancy DF for the forces that were measured in step 2 and that calculated with (3) was determined. Their agreement were compared using standard rule:
Discrepancy ≤ Uncertainty Agreement
Uncertainty < Discrepancy≤ 3 times Uncertainty Marginal Agreement
Discrepancy > 3 times Uncertainty Disagreement
Discussion and Analysis
The frequency of rotation in RPM as indicated by a meter at the base of the motor assembly is as shown in table 1.
The average value of all the values is given by the average of the two averages above
Average value = (556.6+550.4)/2 = 553.5 RPM
Average can also be given by (sum of average of low and high value)/5 = 2767.5/5
= 553.5 RPM
Converting the revolution per minute to revolutions per second = 553.5/60 = 9.225 rev/s
F = mr(2∏f)2
The uncertainty ∆f = the standard deviation
= 2553.5-553.5)2+(553.5-552)2+(553.5-554.5)2+(553.5-554)2
Therefore ∆f= 1.87 RPM
The measured weight in step one = 48.4 kg
The force in Newtons = 9.81×48 = 475 N
The calculated force as determined from the experimental values is given by:
F = mr(2∏f)2
= 0.47×0.3 ×(2∏×9.225)2
= 478N
The discrepancy = measured force-calculated force
= 478-475= 3N
Comparing the uncertainty as calculated from above and the value of the discrepancy it can be seen that Uncertainty < Discrepancy≤ 3 times Uncertainty. It can therefore be concluded that there is marginal agreement of the results.
Questions
Question 1: what was the magnitude of the acceleration experienced by the hanging masses in part 2 of the procedure? What was the magnitude of the acceleration experienced by the rotating mass M in part 5? Compute the second acceleration with (4).
The magnitude of the acceleration experienced by the hanging masses in part 2 is given by:
a= ω2r = (2∏f)2r
Where f = 0
Therefore a= (2∏×0)2r =0
The second acceleration in part 5 of the experiment is also given by:
a= ω2r = (2∏f)2r
Where f is the average value of rotational speed in rev/s, r is the radius of rotation which in this case was 30cm
Therefore a= (2∏f)2r = (2∏×9.225)2 × 0.3
= 9297.8m/s2
Question 2: explain in terms of Newton’s second law, how the magnitude of the forces applied in the two cases could be the same while the accelerations were different.
Answer: this is because from Newton’s second law, a body is said to be moving in a uniform circular motion if it is moving in a circle with a constant speed. Although the speed is constant, the direction of motion changes continuously and hence the velocity keeps on changing. The change in velocity shows that the acceleration of the body changes with the changes in velocity though the force may be the same.
Works Cited
Keynes, Milton. Circular Motion. England: Open U, 2005. Print.
