A stunt pilot of mass 55.0 kg who has been diving her airplane vertically pulls out of the dive by changing her course to a circle in a vertical plane.If the plane’s speed at the lowest point of the circle is 95.8m/s , what is the minimum radius of the circle for the acceleration at this point not to exceed 4.00 g?

**Answer 1**

a) The centripital acceleration

a=V^2/R

V – tangential velocity

R- the radius of the tragectory

Then

R= V^2/a and for a=4g max

R= V^2/4g

R=(95.8)^2/(4 x 9.80)=234 m

b) a(lowest point)=a+g=4g+g=5g

W=mat=a(lowest point)=5mg

W= 5 x 55.0 x 9.80=2700N

**Answer 2**

centripetal acceleration =4.00 g =4 x 9.8=39.2 m/s^2

centripetal acceleration = (speed v )^2 /(radius ‘r’)

radius ‘r’= (speed v )^2 / centripetal acceleration

r = 95.8 x 95.8 /39.2= 234 meter

The radius of circle is 234 meter

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apparent weight = true weight + centripetal force

apparent weight = 55 x9.8 +55×39.2= 2695 N

Apparent weight of the pilot is 2695 N or 275 kg wt