A jet pilot takes his aircraft in a vertical loop Solution (a) The centripetal a

Question

A jet pilot takes his aircraft in a vertical loop

Explanation / Answer

(a) The centripetal acceleration is ac = v2/R => R = v2/ac > v2/6g = (416.7 m/s)2 / [(6)(9.8 m/s2)] = 2953 m Thus, in order for the centripetal acceleration to not exceed 6.0 g's at the bottom of the loop, the radius of the circle must be at least 2953 m. (b) At the bottom of the loop, the pilot experiences two forces, the force of gravity mg downward, and the normal force N that his seat exerts on him upward. According to Newton's second law, F = ma, N - mg = mac = mv2/R The force that the seat exerts on the pilot is N = mg + mv2/R = m(g + v2/R) = (80 kg)[9.8 m/s2 + (416.7 m/s)2/(2953 m)] = 5488 N At the top of the loop, the pilot experiences two forces, the force of gravity mg downward, and the normal force N that his seat exerts on him downward. According to Newton's second law, N + mg = mac = mv2/R The force that the seat exerts on the pilot is N = mv2/R - mg = m(v2/R - g) = (80 kg)[(416.7 m/s)2/(2953 m) - 9.8 m/s2] = 3920 N

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.