1. The water will not fal out at the top of the vertical loop if the centripetal acceleration M V^2/R exceeds the weight M g.
Thus the requirement is
V^2 > R g
In this case, R = 1.195 m
Solve for V (the minimum required value)
2. Weff = M g - M V^2/R
The effective weight Weff is the force that the tires apply to the road. It is less than the true weight M*g because the car is accelerating downwards on the verticall curved path. A net force downwards is required. The reduced force of the road on the tires accomplishes this.
1.Suppose you wish to whirl a pail full of water in a vertical circle without spilling any of its contents. If your arm is 0.99 m long (from shoulder to fist) and the distance from the handle to the surface of the water is 20.5 cm, what minimum speed is required?
2.A 1010 kg car travels on a road that runs straight up a hill and reaches the rounded crest at 10.7 m/s. If the hill at that point has a radius of curvature (in a vertical plane) of 54.0 m, what is the effective weight of the car at the instant it is horizontal at the top of the hill?
1 answer