Wednesday

July 23, 2014

July 23, 2014

Posted by **Daisy** on Wednesday, May 9, 2012 at 8:06pm.

A)Determine the minimum radius of the circle so that the centripetal acceleration at the lowest point does not exceed 5.5 's

B)Calculate the 68 pilot's effective weight (the force with which the seat pushes up on him) at the bottom of the circle.

C)Calculate the 68 pilot's effective weight (the force with which the seat pushes up on him) at the top of the circle (assume the same speed)

- Physics 6A -
**bobpursley**, Wednesday, May 9, 2012 at 8:11pmWithout units on the numbers, it is impossible.

At the bottom of the loop, you have centripetal acceleartion and weight. Centripetal acceleartion = v^2/r

in g's, divide by 9.8m/s^2

- Physics 6A -
**Daisy**, Wednesday, May 9, 2012 at 8:28pmA jet pilot takes his aircraft in a vertical loop. If the jet is moving at a speed of 1100km/h at the lowest point of the loop:

A)Determine the minimum radius of the circle so that the centripetal acceleration at the lowest point does not exceed 5.5g's

B)Calculate the 68kg pilot's effective weight (the force with which the seat pushes up on him) at the bottom of the circle.

C)Calculate the 68kg pilot's effective weight (the force with which the seat pushes up on him) at the top of the circle (assume the same speed)

- Physics 6A -
**Elena**, Thursday, May 10, 2012 at 9:52amv =1100 km/h = 305.6 m/s.

a =v^2/R => R = v^2/a = (305.6)^2/5.5•9.8 = 1732 m

R > 1732 m.

The magnitude of the effective weight = the magnitude of the normal force N

bottom: N = m•g – m•a =

= m• (g - 5.5•g) = - 4.5•m•g.

W = - N = 4.5•m•g =4.5•68•9.8 =2998.8 N.

top: N = - m•g – m•a = -m• (g+a) =

= -m• (g+5.5•g),

W = - N = 6.5•m•g =6.5•68•9.8 =4331.6 N.

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