Question

A roller coaster car with a mass 700 Kg starts from rest at h1 above the ground and slides along a track. The car encounters a loop of radius 12 m. The bottom of the loop is a height h2 = 5 m from the ground. What would be the max height of release h1 for the roller coaster car if the amount of thermal energy produced between the point of release and the top of the loop should not exceed 15% of the initial mechanical energy and the normal force at the top should be no more than 580 N? answer in m

Answer 1

Given that,

Mass of roller coaster car = 700 kg

Radius = 12 m

Height from the ground h₂= 5 m

Normal force = 580 N

We need to calculate the speed of roller coaster at top

Using balance equation

$N+mg=\dfrac{mv^2}{r}$

Put the value into the formula

$580+700\times9.8=\dfrac{700v^2}{12}$

$v^2=\dfrac{12(580+700\times9.8)}{700}$

$v=\sqrt{\dfrac{12(580+700\times9.8)}{700}}$

$v=11.29\ m/s$

We need to calculate the value of maximum height

Using conservation of energy

$mgh_{1}=\dfrac{1}{2}mv^2+0.15(mgh_{1})+mgh_{2}$

$mgh_{1}-0.15(mgh_{1})-mgh_{2}=\dfrac{1}{2}mv^2$

Put the value into the formula

$9.8(h_{1}-0.15h_{1}-5)=\dfrac{1}{2}\times(11.29)^2$

$0.85h_{1}-5=\dfrac{(11.29)^2}{9.8}$

$h_{1}=\dfrac{13.00+5}{0.85}$

$h_{1}=21.18\ m$

Hence, The max height of release h₁ for the roller coaster car is 21.18 m