If the jet is moving at a speed of 1140 km/h at the lowest point of the loop, determine the minimum radius of the circle so that the centripetal acceleration at the lowest point does not exceed 6.0 g's.

Respuesta :

Answer:

The radius is  [tex]r =1705.44 \ m[/tex]

Explanation:

From the question we are told that

   The speed is [tex]v = 1140 \ km /h = \frac{1140 * 1000}{3600} = 316.67 \ m/s[/tex]

   The centripetal acceleration is  [tex]a = 6 g's = 6 * 9.8 = 58.8 \ m/s^2[/tex]

Generally the centripetal force acting on the jet is mathematically represented as  

        [tex]F_c = \frac{m * v^2 }{r}[/tex]

Generally this centripetal force is equal to the net force acting which according to Newton's third law is mathematically  represented as

      [tex]F = m * a[/tex]

So

      [tex]\frac{m * v^2 }{r} = ma[/tex]

=>   [tex]\frac{ v^2 }{r} = a[/tex]

=>  [tex]r = \frac{v^2}{a}[/tex]

=>  [tex]r = \frac{316.67^2}{58.8}[/tex]

=>  [tex]r =1705.44 \ m[/tex]