PHYS 2O7 - LAB 4 - CENTRIPETAL FORCE

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Physics

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Dec 6, 2023

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MOHSEN KANJ GIORGIO WIRAWAN EKENE NNEBE INSTRUCTOR: DR. ALIREZA HASHEMI LAB 4 – CENTRIPETAL FORCE INTRODUCTION: To set an object into circular motion, there needs to be a force continuously directing it towards the center of that motion. In the scenario of a ball attached to a string moving in a circle, this force is provided by the tension in the string. Similarly, for a planet revolving around the sun, it's the gravitational force that serves this purpose. We were able to relate the center-directed force to the speed and radius of the motion, by considering the centripetal acceleration, Ac = v^2 / r F = mAc = mv^2 / r The goal of the lab will be to experimentally verify this relationship. We will design an experiment using the apparatus that shows that for an object rotating in uniform circular motion, the centripetal force applied is equal to mv^2 / r.
PROCEDURE: Initially, my lab partners and I conducted a thorough inspection of all the equipment in the laboratory, including the cross-arm, counterweight, bob, background for viewing, spring, and mass/pan, among other things. Subsequently, I assembled the system and suspended the bob using a string from the cross-arm. To balance the bob, we placed the counterweight on the opposite end of the arm. We then documented the weight of the mass/pan and the counterweight, striving to make them equal the spring force, which acts as the centripetal force. Next, we rotated the rod and bob, attempting to return the bob to its balanced position, allowing it to pass above a pointer. We recorded the time taken and the number of revolutions completed. Following this, we calculated the period T and ultimately compared our findings with the provided equation. Later, we added a mass of 50g on to the block but kept the same radius and checked the same results that we were looking for and we found difference in the centripetal speed. We did the same step by adding 100g instead of 50g and the results changed. For the last step, we kept the block with no mass on it, but changed the radius two times, one for 0.178m and one for 0.157m and we realized there was a difference in the oscillations and period T.
DATA/CALCULATIONS/QUESTIONS: Mass of block = 0.468kg. 1) 8 oscillations. T = 6.65 sec. 0 mass added. R = 0.19m Centripetal speed = w = (oscillations x π) / T = 8 x π / 6.65 = 1.203 π rad/sec. V = Rw = 0.19 x 1.203 π = 0.718 m/s. m of masses added on the rope = 710g. Measurement of Fc = m x g = 710 x 10 = 7100N. Centripetal force = F = (m x v^2 ) / R = 1.26N 2) 50g added. 3 oscillations. T = 2.43 sec. R = 0.19m. Centripetal speed = w = (oscillations x π) / T = 3 x π / 2.43 = 1.23 π rad/sec. V = Rw = 0.19 x 3.87 = 0.735 m/s. m of masses added on the rope = 710g. Measurement of Fc = m x g = 710 x 10 = 7100N. Centripetal force = F = (m x v^2 ) / R = (0.518 x (0.736)^2) / 0.19 = 1.47 N. 3) 100g added. 6 oscillations. T = 4.79 sec. R = 0.19m. Centripetal speed = w = (oscillations x π) / T = 6 x π / 4.79 = 1.25 π rad/sec. V = Rw = 0.19 x 3.93 = 0.7467 m/s. m of masses added on the rope = 710g.
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Measurement of Fc = m x g = 710 x 10 = 7100N. Centripetal force = F = (m x v^2 ) / R = (0.568 x (0.7467)^2) / 0.19 = 1.66 N. 4) No masses added. 5 oscillations. T = 4.45 sec. R changed to R = 0.178m Centripetal speed = w = (oscillations x π) / T = 5 x π / 4.45 = 1.123 π rad/sec. V = Rw = 0.178 x 3.52 = 0.62656 m/s. m of masses added on the rope = 710g. Measurement of Fc = m x g = 710 x 10 = 7100N. Centripetal force = F = (m x v^2 ) / R = (0.468 x (0.62656)^2) / 0.178 = 1.0321 N. 5) No masses added. 6 oscillations. T = 6.34 sec. R changed to R = 0.157m Centripetal speed = w = (oscillations x π) / T = 6 x π / 6.34 = 0.946 π rad/sec. V = Rw = 0.157 x 2.97 = 0.4667 m/s. m of masses added on the rope = 710g. Measurement of Fc = m x g = 710 x 10 = 7100N. Centripetal force = F = (m x v^2 ) / R = (0.468 x (0.4667)^2) / 0.157 = 0.649 N.
0 0.52 0.57 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 Centipetal l Force as a function of added mass on the block Column1 Added m Fc
0.19 0.178 0.157 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Centripetal force as a function of radius Series 1 Radius R Fc
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0.72 0.74 0.75 0.63 0.47 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 Centrepital force as a function of speed Series 1 V Fc QUESTIONS: Questions 1: What is the centripetal force pulling the bob towards the center? Answer: The force of the spring pulling the bob towards the center. Question 2: How will you measure the Fc? Answer: Fc is the weight of the counter mass and pan. Question 3: Where is the center of the circle that the bob is traveling around? Answer: The bob is traveling around the rod. Question 4: How will you measure r? Answer: r is the distance from the rod to the pointer. Question 5: How will you measure v? Answer: We record the time and how many rounds it went, then we get speed.
Question 6: Can making multiple measurements help? Answer: Yes, we can change the position of the pointer and then we changed the radius. CONCLUSION: We can conclude that our experiments had different results in many ways, whether speed, mass, radius, or angular velocity. We can also realize that there was a big margin of error calculating the centripetal force. By changing the mass of the block, the angular speed was changing, and by changing the radius from the rod to the block, the angular speed also changed; however, the mass added to the rope didn’t need changing. We can see that all factors in the experiment are major factors that affect all the results we’re looking for.

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