Question

A new roller coaster contains a loop-the-loop in which the car and rider are completely upside down. If the radius of the loop is 13.2 m, with what minimum speed must the car traverse the loop so that the rider does not fall out while upside down at the top? Assume the rider is not strapped to the car.

Answer 1

To solve this problem it is necessary to apply the concepts related to gravitational force and centripetal force,

By definition the gravitational force on earth is defined under Newton's second law where,

$F_g = mg$

Where,

m = mass

g = Gravity acceleration

At the same time we know that the Centripetal force is equivalent to

$F_c = \frac{mv^2}{r}$

Where,

m = mass

v = Velocity

r = Radius

Since there is a balance between the two, you have to

$F_g = F_c$

$mg = \frac{mv^2}{r}$

Re-arrange to find the velocity we have,

$v = \sqrt{rg}$

$v = \sqrt{13.2*9.8}$

$v = 11.37m/s$

Therefore the minimum speed must be 11.37m/s