Lab (Centripetal Force)-2 PDF

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Islam 1 Name: Professor: Class: Physics 215 Date: 10/16/2020. Lab: Centripetal Force Introduction: Centripetal Forces means “center-seeking”, it is provided by gravitational and electrical interactions, correspondingly, for each of these cases. For example, the Earth revolves around the Sun. Another example is the electrons move around the nucleus. Therefore, the centripetal force is what is keeping these objects in orbit. The object has a constant speed when it moves around, but velocity is changing in terms of direction because it is moving in circular motion, so the velocity of the direction is perpendicular to the circular motion. This change is velocity results from centripetal acceleration because of the centripetal force. Objectives: Our objective in this lab is to describe why the centripetal force is necessary for the circular motion. Also, our objective is to explain how the frequency of rotation of the object, mass, and radius affects the magnitude of the centripetal force to form a constant circular motion. Procedures: Manual Centripetal Force Apparatus: 1. First, determine the mass of the bob. Then, modify the position of the vertical pointer rod to the smallest radius as possible. Finally measure the distance of the radius and record it on Data Table 1.

Islam 2 2. Now, attach the bob on the horizontal support along with spring. Then, practice rolling the rotor between your thumb and fingers so that the bob revolves around and passes over the pointer. 3. Next, simultaneously we are going to revolve the bob about 25 revolutions and measure the time. Then, record on table 1. 4. We are going to repeat step 3 three times. After that, we must calculate the time per revolution of the bob for each trial and find the average time per revolution of the three trials: v = 2 πr / T. T is the average time per revolution (period). 5. Afterward, we are going to find the centripetal force by using this equation Fc= mv2 / r 6. Subsequently, we are going to attach a string to the bob on opposite side with a weight hanger. Then, we are going to add weights to the hanger until the bob is directly aligned with the pointer. Later, we are going to measure of the total weight we added on the hanger and record it. This weight is directly measure of the centripetal force supplied by the string during rotation. 7. Then, calculate the percent difference between the computed value of centripetal force and the direct measurement of centripetal force. 8. Variation of mass – unscrew the nut that is on top of bob and place 100g, then lock the screw very tightly. We are going to repeat step 3-7. Finally, record it on Data Table 2. 9. Variation of the radius – remove the 100g that was on top of the bob and lock it. Then, we are going to move the horizontal support arm farther away to provide a larger radius. We are going to repeat step 3-7. Finally record it on Data table 3.

Data Table:

Islam 3 Data Table 1: To determine period of revolution for computation of centripetal force. Trial 1

Trial 2

Trial 3

25rev

25rev

25rev

Total Time (s)

26.43s

26.16s

26.40s

Time/ Revolution

1.06 s/rev

1.05 s/rev

1.06 s/rev

Number of revolutions

(s/rev)

Mass of bob: 0.4547kg Radius of circular path: 0.150m Average time per revolution: 1.06 s/rev Average speed of bob (v): 0.89m/s Computed value of centripetal force: 2.40N Direct measurement of centripetal force: 2.21N Percent difference: 8.2% Data Table 2: To observe the effect of varying mass. Trial 1

Trial 2

Trial 3

25rev

25rev

25rev

Total Times(s)

28.13s

28.60s

28.09s

Time/Revolution

1.13 s/rev

1.14 s/rev

1.12 s/rev

Number of revolutions

(s/rev)

Mass of bob: 0.5547kg Radius of circular path: 0.150m Average time per revolution: 1.13 s/rev

Islam 4 Average speed of bob (v): 0.83m/s Computed value of centripetal force: 2.50N Direct measurement of centripetal force: 2.20N Percent difference: 12.8% Data Table 3: To observe the effect of varying radius. Trial 1

Trial 2

Trial 3

25rev

25rev

25rev

Total Times(s)

22.50s

22.14s

22.18s

Time/Revolution

0.90 s/rev

0.89 s/rev

0.89 s/rev

Number of revolutions

(s/rev)

Mass of bob: 0.4547kg Radius of circular path: 0.185m Average time per revolution: 0.89 s/rev Average speed of bob (v): 1.31m/s Computed value of centripetal force: 4.22N Direct measurement of centripetal force: 4.02N Percent difference: 4.85% Calculations: Data Table 1: Average time per revolution: (1.06+1.05+1.06) / 3 = 1.06 s/rev Average speed of the bob: v = 2πr / T 2π (0.150m) / 1.06 s/rev = 0.89 m/s

Islam 5 Computed value of the centripetal force: Fc = mv2/r {(0.4547kg) (0.89m/s)} / (0.150m) = 2.40 N Directed measurement of centripetal force: F = m  a (0.225kg) (9.81m/s) = 2.21 N Percent Difference: {| x1 – x2 | / (x1 + x2) / 2} *100 | 2.21 - 2.40 | / {(2.21 + 2.40)/2} *100 = 8.2% Conclusion: In conclusion, we can see that radius, mass, and frequency of the rotation affected the centripetal force. For example, when we kept the radius was in short distance the velocity kept changing direction. However, when we added more mass on the bob the velocity decreased. But what if we move the radius farther away and remove the extra mass we recently added? Then, the velocity is going to increase because we added more force, and the more force we add in the center, the more acceleration we are going to get. Therefore, the centripetal force is happening in the center, just like in space, the more mass you have in space the more stuff you are going to attract and orbit around you because you have that force, and you are pulling to its center.

Lab questions and answer: 1. Give two examples on objects that perform circular motion and determine the source of centripetal force for each. Ans: 1. A rock tied to a string and whirled in horizontal circle. The tension force on the rock provides the centripetal force.

Islam 6 2. A car moving about a roundabout is another example. If the roundabout is flat, then friction provides the centripetal force otherwise normal force and friction provide the force. 2. Suppose you tie a rock to a string and spin it in a vertical plane. In what direction will the rock move if the string breaks exactly when the rock is passing the lowest point. Ans: When the rock is at the lowest point the velocity of the rock is a horizontal direction. If the string is cut at this point the rock will move in horizontal direction. the rock will be in projectile motion and follow parabolic path. 3. Using v = 2πRf, derive a formula for Fc in terms of M, R, and frequency f. Simplify the formula after you substitute for v in the centripetal force formula. Ans: Fc = Mv2/R V = 2πRf Fc = M (2πRf) ^2 /R F =4 π^2MRf^2...