Question

A particularly scary roller coaster contains a loop-the-loop inwhich the car and rider are completely upside down. If the radiusof the loop is 22.7 m with what minimum speed must the cartraverse the loop so that the rider does not fall out while upsidedown at the top? Assume the rider is not strapped to the car.

Answer 1

The forces on the car at the top of the loop are the normal force and the weight, they both points downward. Also the accereation at that point points downward then Newton's second law is:

$-F_N-F_g=-ma$

Now, since this is a circular motion the acceleration is a centripetal one and that is given as:

$a=\frac{v^2}{R}$

where v is the velocity and R is the radius of the circle. Plugging this in Newton's second law we have:

$-F_N-F_g=-m\frac{v^2}{R}$

If the car has the minimum speed to reamin in contact then it would be on the verge of losing contact , this means that Fn=0 at the toop of the loop. Then we have that the equation above:

$\begin{gathered} -F_N-F_g=-m\frac{v^2}{R} \\ 0-mg=-m\frac{v^2}{R} \\ v=\sqrt[]{gR} \end{gathered}$

Plugging the values of the acceleration of gravity and the radius of the loop we have:

$\begin{gathered} v=\sqrt[]{(9.8)(22.7)} \\ v=14.92 \end{gathered}$

Therefore the minimum velocity is 14.92 meters per second.