A particularly scary roller coaster contains a loop-the-loop in which the car and rider are completely upside down.If the radius of the loop is $13.2 \mathrm {~m} \text {, }$ 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.
A)11.4 m/s
B)12.5 m/s
C)10.1 m/s
D)14.9 m/s

Question

Quizplus · Accepted Answer

For the rider not to fall out at the top of the loop, the centripetal force must be equal to or greater than the gravitational force acting on the rider. The minimum speed ($v$) at the top can be found using the equation for centripetal force $F_c = \frac{mv^2}{r}$, where $m$ is the mass, $v$ is the velocity, and $r$ is the radius of the loop. Setting this equal to the gravitational force ($mg$), where $g$ is the acceleration due to gravity (approximately $9.8 \, \text{m/s}^2$), gives $\frac{mv^2}{r} = mg$. Solving for $v$, we get $v = \sqrt{rg}$. Substituting $r = 13.2 \, \text{m}$ and $g = 9.8 \, \text{m/s}^2$, we find $v = \sqrt{13.2 \times 9.8} \approx 11.4 \, \text{m/s}$.

A Particularly Scary Roller Coaster Contains a Loop-The-Loop in Which 13.2 m, 13.2 \mathrm {~m} \text {, }13.2 m,

A Particularly Scary Roller Coaster Contains a Loop-The-Loop in Which $13.2 \mathrm {~m} \text {, }$