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The radius of curvature is smaller at the top than on the sides so that the downward centripetal acceleration at the top will be greater than the acceleration due to gravity, keeping the passengers pressed firmly into their seats. What is the speed of the roller coaster at the top of the loop (in m/s) if the radius of curvature there is 15.0 m and the downward acceleration of the car is 1.50 g?

Respuesta :

Answer:

[tex]v = 14.86 m/s[/tex]

Explanation:

As we know that the force equation at the top is given as

[tex]\frac{mv^2}{R} = ma[/tex]

now we know that

[tex]a_c = 1.5 g[/tex]

so we have

[tex]\frac{v^2}{R} = 1.5 g[/tex]

[tex]v = \sqrt{1.5 Rg}[/tex]

so we will have

[tex]v = \sqrt{1.5(15)(9.81)}[/tex]

[tex]v = 14.86 m/s[/tex]

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