Centripetal Force Lab Online
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Dec 6, 2023
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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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