Question

A roller-coastercar may be represented by a block of mass 50.0 . The car is released from rest at a height h= 60.0 above the ground and slides along a frictionless track. The carencounters a loop of radius R= 20.0m at ground level, as shown. As you will learn in the course of this problem, the initial height 60.0m is great enough so thatthe car never loses contact with the track.
Find an expression for the kinetic energy of the car at the top of the loop.
Find the minimum initial height Hmin at which the car can be released that still allows the car to stay in contact with the track at the top of the loop.

Answer 1

Potential energy at height h is P = mghWhere m = mass of roller coaster= 50 kgh = 60 mKinetic energy of the roller coaster at the bottom of the loop K = mghSince from law of conservation of energy.K = mgh( 1/ 2) mv ² = mghFrom this speed at the bottom of the loop v = √[2gh ]= 34.29 m / sRadius of the loop R = 20 mMinimum speed of the roller coaster still allows the car to stay in contact with the track at thebottom of the loop v ' = √[ 5Rg]= 31.3 m / sv > v 'So,the car never loses contact with the track.(b). the minimum initial height H _min at which the car can be released that still allows the car to stay in contact with the track at thetop of the loop H = ?Critical speed of the coaster at the bottom of the loop v ' = 31.3 m / sFrom law of conservation of energy mgH = ( 1/ 2) mv ' ²H = v' ² / 2g= 50m