Question

A roller-coaster car may be represented by a block of mass 50.0 kg . The car is released from rest at a height h = 48.0 m above the ground and slides along a frictionless track. The car encounters a loop of radius R = 16.0 m at ground level, as shown. As you will learn in the course of this problem, the initial height 48.0 m is great enough so that the car never loses contact with the track.

1.) Find the kinetic energy K of the car at the top of the loop.

2.)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

Given , m = 50 kg , h = 48 m, R = 16 m

Using conservation of energy ,

Initial Potential energy (PE1) = mgh

PE1 = 50*9.81*48 = 23544 J

Now,

Potential energy at the top of the loop (PE2)= mg*(2R)

(PE2) = 50*9.81*(2*16) = 15696 J

a) Hence, Kinetic energy at the top of loop

KE = 23544 - 15696 = 7848 J

b) normal force from loop on car should be zero, if the car should stay in contact with loop

By balancing the forces ,

centrifugal force = weight + normal force,

mv²/R = mg + 0

v^2 = Rg

Substituting the vaue of v^2 ,

KE = 1/2*mv² = 1/2*mRg

But,

PE at the point = mg*(2R)

So, total energy = 1/2*mRg + mg*(2R) = 5/2*mgR

Also, total energy = mgH

Let 'H' is inital height of fall.

Thus,

mgH = 5/2*mgR

H = 5/2*R = 5/2*16 = 40 m

Thus , minimum height = 40 m