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```
A pilot of mass 50 kg in a jet aircraft is executing a loop-the-loop with constant speed of 250 m/s. If the radius of circle is 5 km, compute the force exerted by seat on the pilot i. at the top of loop. ii. at the bottom of loop.

```
4 years ago

RUSHABH
22 Points
```								the force on top of the loop will be the normal force between the pilot and the seat, as the gravity on the pilot is downwards therefor the normal between and seat and pilot is 0 N.	the force on the top of the loop will the normal force between seat and pilot will be 500 N. (F=ma).there is no acceration in the vertical direction so no change in force.
```
3 years ago
Mahendra
11 Points
```							Let N1 be the normal force on the top and N2 be at the bottom.N1+mg= (mv²)/rN2-mg=(mv²)/rMv²/r= 50×62500÷5000         = 625 Nmg= 50×9.8     = 490 NN1=625-490=135NN2=625+490=1115N
```
2 years ago
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• Test paper with Video Solution
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• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions