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
[tex]F_{net} = N + mg[/tex]
By Newton's Second Law, this force is equal to the centripetal force, because the car is making circular motion in the loop.
[tex]F_{net} = ma = \frac{mv^2}{R}\\N + mg = \frac{mv^2}{R}[/tex]
The critical speed is the minimum speed at which the car does not fall. So, at the critical speed the normal force is zero.
[tex]0 + mg = \frac{mv_c^2}{R}\\v_c = \sqrt{gR}[/tex]
If the car is moving twice the critical speed, then
[tex]N + mg = \frac{m(2v_c)^2}{R} = \frac{m4gR}{R} = 4mg\\N = 3mg[/tex]
Finally, the ratio of the normal force to the gravitational force is
[tex]\frac{3mg}{mg} = 3[/tex]
