Question

If the jet is moving at a speed of 1040 km/h at the lowest point of the loop, determine the minimum radius of the circle so that the centripetal acceleration at the lowest point does not exceed 6.8 g's.

Answer 1

r=1,252.3528m

Step-by-step explanation:

`F_c=ma_c=(mv^2)/(r)\\a_c=(v^2)/(r)\\r=(v^2)/(a_c)\\r=(((1040) (1000)m/(60)(60)s)^2)/((6.8g)m/s^2)\\r=(((1040000)m/(3600)s)^2)/((6.8 \cdot 9.8)m/s^2)\\r=((288.889m/s)^2)/(66.64m/s^2)\\r=(83,456.790m^2/s^2)/(66.64m/s^2)\\r=(83,456.790m)/(66.64)\\r=1,252.3528m`