Centripetal Force Lab Online

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University of Texas, San Antonio *

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2322

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Mechanical Engineering

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Dec 6, 2023

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docx

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3

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Analysis of Centripetal Force Lab Name Milandeep Batth _____________________________ Course/Section PHY 1611 Section 006______________ Instructor Erica Dykes_______________________ Tables (20 points) R(m) 0.500 m y 1 (m) 1.260 m g(m/s 2 ) 9.81 m/s^2 m(kg) 1.000 kg N b 59.194 N v b 4.970 m/s N T 0.399 N v T 2.259 m/s 1. Draw the free body diagram for the mass at the bottom of the loop, and the write the force summation equation from it. Then use it to calculate the centripetal force of the mass at the bottom of the loop. (15 points) T-mg=F c or N b -mg= F c 59.194 N – (1.000 x 9.81) = 49.384 Centripetal Force= N b -mg= 59.194 N – (1.000 x 9.81)= 49.38 N 1
2. Calculate the centripetal force for the mass at the bottom of the loop using the formula mv b 2 /R (10 points) Centripetal Force = mv b 2 /R= (1.000 x (4.970) 2 )/0.500= 49.40 3. Calculate the percent error between the values of question 1 and question 2, using the value from question 1 as the theorical value. (5 points) Percent Error= |(49.40-49.38)/49.38|x 100% =0.041% 4. Draw the free body diagram for the mass at the top of the loop, and the write the force summation equation from it. Then use it to calculate the centripetal force of the mass at the top of the loop. (15 points) Centripetal Force= mg- N t = 9.81-0.399= 9.41 N 5. Calculate the centripetal force for the mass at the top of the loop using the formula mv T 2 /R (10 points) Centripetal Force = mv T 2 /R = (1.000 x (2.259) 2 )/0.500 =10.21 N 2
6. Calculate the percent error between the values of question 1 and question 2, using the value from question 1 as the theorical value. (5 points) Percent Error = |(10.21-9.41)/9.41) x100% = 8.50 % 7. Turn in the two screenshots. (20 points) 3
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