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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 12.1 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.

Respuesta :

Answer:

[tex]v = 10.89\ m/s[/tex]

Explanation:

given,                          

radius of loop = 12.1 m                              

to find the minimum speed transverse by the rider to not to fall out upside down                                                                

centripetal force = [tex]\dfrac{mv^2}{r}[/tex]

gravitational force  = m g

computing both the equation]

[tex]mg = \dfrac{mv^2}{r}[/tex]

[tex]v = \sqrt{rg}[/tex]

[tex]v = \sqrt{12.1 \times 9.8}[/tex]

[tex]v = \sqrt{118.58}[/tex]

[tex]v = 10.89\ m/s[/tex]

So, [tex]v=10.895 m/s[/tex] the rider does not fall out.

In the case of the loop:

The acceleration will at least be equal to the acceleration due to gravity [tex]g[/tex] such that the top of the loop and the centripetal acceleration is up and [tex]g[/tex] is down

Given radius [tex](R)=12.1 m[/tex]

Then,

[tex]mg=m\frac{v^2}{R} \\V=\sqrt{gR}[/tex]

Now, substituting the given values into the above formula we get,

[tex]v=\sqrt{9.81\times12.1}\\ =10.895 m/s[/tex]

Learn more about the case of the loop:

https://brainly.com/question/26421376

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