Question

Problem 1 You are designing a roller coaster, shown in the figure below. An empty car has a mass of 120 kg. The radius of the vertical loop in the track is 12.0 m and the bottom of the loop is 5.0 m above the ground. Assuming the roller coaster is frictionless, how high must the car initially climb above the ground (point A) such that the speed of the car is 25.0 m/s at the top of the loop (point B)? Assume that the cart has a negligible speed at point A. B A

Answer 1

given :

m = 120 kg

speed at B = v = 25 m/s

speed at A = u = 0

height of B from ground = h = 5 + 12 + 12 = 29 m

Let height of A from ground be H.

Applying conseration of mechanical energy between A and B :

potential energy at A + kinetic energy at A = Potential energy at B + kinetic energy at B

$\Rightarrow mgH + \frac{1}{2}mu^{2}=mgh + \frac{1}{2}mv^{2}$

gH +0=gh +502 2

$\Rightarrow H=h + \frac{v^{2}}{2g} = 29+\frac{25^{2}}{2*9.81}=60.86 m$[answer]