Question

A roller coaster has a small vertical loop with the lowest point 5 meters above the ground. The initial elevation from which the cart was released is 62.5 m. What should be radius of that loop to make sure the vertical acceleration at the bottom point does not exceed 2g?

Answer 1

Given data

*The given initial elevation is h = 62.5 m

*The lowest point above the ground is H = 5.0 m

The formula for the final speed of the roller coaster is given by the conservation of energy as

`\begin{gathered} U_i=U_f \\ mgh+(1)/(2)mv_i^2=mgH+(1)/(2)mv^2_f^{} \end{gathered}`

*Here U_i is the initial energy

*Here U_f is the final energy

*Here v_i = 0 m/s is the initial speed

*Here g is the acceleration due to the gravity

Substitute the known values in the above expression as

`\begin{gathered} mg(62.5)+(1)/(2)m(0)^2=mg(5.0)+(1)/(2)mv^2_f \\ v=\sqrt[]{115g} \\ =33.57\text{ m/s} \end{gathered}`

As from the given data, the vertical acceleration at the bottom point does not exceed (2g) as

`\begin{gathered} a\leq2g \\ (v_f^2)/(r)\leq2g \\ r\ge(v_f^2)/(2g) \\ r\ge((33.57)^2)/(2(9.8)) \\ r\ge57.5\text{ m} \end{gathered}`

Hence, the radius of the loop is 57.5 m