Question

A roller coaster car crosses the top of a circular loop-the-loop at twice the critical speed. Part A What is the ratio of the normal force to the gravitational force? What is the ratio of the normal force to the gravitational force? n/FG=2n/FG=2 n/FG=3n/FG=3 n/FG=4n/FG=4 n/FG=5n/FG=5

Answer 1

Answer:

n/(FG) = 3.

Explanation:

At the top of the loop-the-loop, the normal force is directed downwards as well as the weight of the car. So, the total net force of the car is

$F_{net} = N + mg$

By Newton's Second Law, this force is equal to the centripetal force, because the car is making circular motion in the loop.

$F_{net} = ma = \frac{mv^2}{R}\\N + mg = \frac{mv^2}{R}$

The critical speed is the minimum speed at which the car does not fall. So, at the critical speed the normal force is zero.

$0 + mg = \frac{mv_c^2}{R}\\v_c = \sqrt{gR}$

If the car is moving twice the critical speed, then

$N + mg = \frac{m(2v_c)^2}{R} = \frac{m4gR}{R} = 4mg\\N = 3mg$

Finally, the ratio of the normal force to the gravitational force is

$\frac{3mg}{mg} = 3$