Question

Review: Why is the following situation impossible? A new high-speed roller coaster is claimed to be so safe that the passengers do not need to wear seat belts or any other restraining device. The coaster is designed with a vertical circular section over which the coaster travels on the inside of the circle so that the passengers are upside down for a short time interval. The radius of the circular section is 12.0 m, and the coaster enters the bottom of the circular section at a speed of 22.0 m/s. Assume the coaster moves without friction on the track and model the coaster as a particle.

Short answer

The centripetal acceleration of each passenger as the coaster passes over the top of the circle is $a_C=1.13m/s^2.$Since this is less than the acceleration due to gravity, the unrestrained passengers will fall out of the cars.

Step-by-step solution

Step 1

Aforce acting radially inward onabody moving on a circular path is called a centripetal force. This force is responsible for the continuous change in direction of abody in a circular motion.

The potential energy of the system is:

$\mathbf{U}\mathbf{=}\mathbf{mgR}$

The kinetic energy of the system is:

$\mathbf{K}\mathbf{=}\frac{1}{2}{\mathbf{mv}}^2$

Here:

$U=$Potential energy

$K=$Kinetic energy

$v=$Speed of body

$g=$Gravitational acceleration

$R=$ Displacement in meter

$m=$ Mass of a body

Given Data

Speed of the roller coaster$v=22\text{\hspace{0.17em}m}/\text{s}$

Radius of the circular section$R=12.0\text{\hspace{0.17em}m}$

At the bottom, the speed of the roller coaster is $v=22\text{\hspace{0.17em}m}/\text{s}$.This slows down with an increase in the altitude. At the top of the track, the centripetal acceleration must be at least equal to the acceleration due to gravity $\left(g\right)$, to remain on the track. Using the law of conservation of energyat the bottom and the top of the circular path of the roller coaster.

$\Delta K+\Delta U=0$

Replacethe values for the change in kinetic and potential energy:

$\begin{array}{l}\frac{1}{2}m{v}_{top}^2-\frac{1}{2}m{v}_{bottom}^2+(mg{y}_{top}-mg{y}_{bottom})=0\\ \frac{1}{2}m{v}_{top}^2-\frac{1}{2}m{v}_{bottom}^2+(mg2R-0)=0\\ v_{top}^2=v_{bottom}^2-4gR\end{array}$

Here, $y$is the vertical distance from the ground.

For the given values, the above equation becomes:

$\begin{array}{l}{v}_{top}^2={(22.0\text{m/s})}^2-4(9.8{\text{m/s}}^2)(12.0\text{m})\\ =13.6{\text{m}}^{\text{2}}{\text{/s}}^2.\end{array}$

$\begin{array}{c}{a}_C=\frac{v_{top}^2}{R}\\ =\frac{13.6{\text{m}}^{\text{2}}{\text{/s}}^{\text{2}}}{12.0\text{m}}\\ =1.13{\text{m/s}}^{\text{2}}.\end{array}$

The centripetal acceleration of each passenger as the coaster passes over the top of the circle is $a_C=1.13{\text{m/s}}^{\text{2}}.$Since this is less than the acceleration due to gravity, the unrestrained passengers will fall out of the cars.