A road has a hill with a top in the shape of a circular arc of radius 32.0 m. How fast can a car go over the top of the hill without losing contact with the ground?
Solution) Radius r = 32 m
Speed V = ?
For circular path it is centripetal force Fc that acts
F = Fc
ma = (m(V^2))/(r)
a = (V^2)/(r)
V^2 = ar
V = (ar)^(1/2)
Here a = g = 9.8 m/s^2
V = (9.8×32)^(1/2)
V = 17.70 m/s
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