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

AI Thread Summary
A jet pilot experiences a force five times her normal weight at the bottom of a vertical loop with a radius of 1.50 km. The centripetal force equation, F(c) = (mv^2)/r, is applied to determine the jet's speed. The calculation involves using the relationship F(c) = F(n) - F(g), leading to the equation (v^2)/1500 = (9.8*5) - 9.8. The resulting speed calculated is 242 m/s, confirming the approach is correct. This method effectively illustrates how to analyze forces in vertical loops for jet speed determination.
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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