A roller-coaster car may be represented by a block of mass 50.0 kg. The car is r

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.

a. Find an expression for the kinetic energy K of the car at the top of the loop.
Express the kinetic energy numerically, in joules.


b.Find 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 the top of the loop.
Express your answer numerically, in meters.

Explanation / Answer

Total energy = potential energy in the beginning = mgh = 50*9.81*48 = 23544 J

P.E at the top of the loop = mg*(2R) = 50*9.81*(2*16) = 15696 J

a) Hence, KE at the top of loop = 23544 - 15696 = 7848 J

b) For the car to stay just in contact with loop, normal force from loop on car should be zero.

Balancing the forces on the car, centrifugal force = weight + normal force,

mv2/R = mg + 0

v2 = Rg

KE = 1/2*mv2 = 1/2*mRg

PE at the point = mg*(2R)

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

But total energy = mgH where H is inital height of fall.

Thus, mgH = 5/2*mgR

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

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