Calculating Jet Speed in Vertical Loop: Centripetal Force Question 1 Explained

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SUMMARY

The discussion focuses on calculating the speed of a jet at the bottom of a vertical loop where the pilot experiences 5.00 times her normal weight. Using the centripetal force equation F(c) = (mv^2)/r, the user derives the speed as 242 m/s. The calculation involves the relationship F(c) = F(n) - F(g), where F(n) is the normal force and F(g) is the gravitational force. The radius of the loop is specified as 1.50 km, confirming the accuracy of the derived speed.

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TheExibo
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1. A jet pilot feels 5.00 times her normal weight at the bottom of a vertical loop. The loop has a radius of 1.50 km. What is the speed of the jet at the bottom of the loop?2. F(centripetal)=(mv^2)/r3. The problem I'm having is with this equation that I've made based on the equation: F(c)=F(n)-F(g). With it, I get the answer of 242m/s (mass is not necessary to solve the question). Would (v^2)/1500=(9.8*5)-(9.8) be the correct step to solving this problem? Thanks!
 
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TheExibo said:
1. A jet pilot feels 5.00 times her normal weight at the bottom of a vertical loop. The loop has a radius of 1.50 km. What is the speed of the jet at the bottom of the loop?2. F(centripetal)=(mv^2)/r3. The problem I'm having is with this equation that I've made based on the equation: F(c)=F(n)-F(g). With it, I get the answer of 242m/s (mass is not necessary to solve the question). Would (v^2)/1500=(9.8*5)-(9.8) be the correct step to solving this problem? Thanks!
Yes. It is correct.
 

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