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

Question

A roller-coaster car may be represented by a block of mass 50.0kg . The car is released from rest at a height h = 60.0m above the ground and slides along a frictionless track. The car encounters 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 that the car never loses contact with the track.

Find an expression for the kinetic energy K 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.

Explanation / Answer

First recognize this is a conservation of energy problem. So TE = PE = mgH is the initial total energy at the point H = 60 m above that death defying plunge. And from the COE, that's the same TE at the top of the loop as energy is not destroyed. So there you are:

TE = PE = mgH = 1/2 mv^2 + mgD = ke + pe = TE; where D = 2R is the loop's diameter and v is the tangential speed at the top (and upside down). Solve for the kinetic energy ke = PE - pe = mg(H - D) = mg(H - 2R); where m = 50 kg, g is g, H = 60 m, and R = 20 m. ANS.

To not fall out of the track at the top, we have g = v^2/R so the weight of the car is offset by the centrifugal force on it. Just substitute v^2 above with gR and solve for H. ke = 1/2 mv^2 = mg(H - 2R) = 1/2 mgR; so then, H - 2R = R/2 and H = 5/2 R. ANS.

EX: As R = 20 m, the minimum H = 50 m and the car will still stick

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