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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??

Vaibhavi , 8 Years ago
Grade 11
anser 3 Answers
Manas Shukla

Last Activity: 8 Years ago

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

Vaibhavi

Last Activity: 8 Years ago

But the answer is given 2700m ???? ?????? ???? ??????????.............................................

Manas Shukla

Last Activity: 8 Years ago

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
 

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