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Grade: 11

                        

A fighter pilot dives his plane towards the ground at 230m/s^2. He pulls out the dive on vertical circle. What is minimum radius og circle so that normal force exerted in the pilot by his seat never exceeds three times his weight??

4 years ago

Answers : (3)

Manas Shukla
102 Points
							
We have to equal the centripetal acceleration = 3g
Since the reaction on the pilot seat = centripetal acceleration for the pilot
So
\frac{v^{2}}{R} = 3g
R = 1763.33 m
4 years ago
Vaibhavi
12 Points
							But      the      answer     is        given      2700m    ????       ??????     ????       ??????????.............................................
						
4 years ago
Manas Shukla
102 Points
							
Oops , sorry I didnt see it was a vertical circle so I didnt take into account the weight of the pilot himself.
We have to keep in mind that the pilot will also exert force Mg on the seat at the lowest point and there will be centripetal acceleration, M being the mass of the pilot.
So we will equate \frac{v^{2}}{R} = 2g
Solving we get R =2696.228 m , if we take gravity = 9.81m/s2
 
4 years ago
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