A roller coaster car may be approximated by a block of mass m. Thecar, which starts from rest, is released at a height h above the ground and slides along a frictionless track. The car encounters a loop of radius R. Assume that the initial height h is great enough so that the car never losses contact with the track.

Required:

a. Find an expression for the kinetic energy of the car at the top of the loop. Express the kinetic energy in terms of m, g, h, and R.

b. Find the minimum initial height h 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.

Question

Adyson Joyce

Physics

9 September, 06:02

A roller coaster car may be approximated by a block of mass m. Thecar, which starts from rest, is released at a height h above the ground and slides along a frictionless track. The car encounters a loop of radius R. Assume that the initial height h is great enough so that the car never losses contact with the track.

Required:

a. Find an expression for the kinetic energy of the car at the top of the loop. Express the kinetic energy in terms of m, g, h, and R.

b. Find the minimum initial height h 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.

+3

Answers (1)

Know the Answer?

Quinton Berry · Answer 1

At height h, potential energy of coaster car having mass m = mgh.

The car will lose potential energy and gain kinetic energy.

height lost by car when it is at the top of loop of radius R

= h - 2R

potential energy lost = mg (h - 2R)

kinetic energy gained = mg (h - 2R)

kinetic energy = 0 + mg (h - 2R)

= mg (h - 2R)

b)

For the car to remain in contact with the track, if v be the minimum velocity

centripetal force at top = mg

m v² / R = mg

v² = gR

kinetic energy = 1/2 mv²

= 1/2 m x gR

= mgR /

If h be the minimum height that can give this kinetic energy

mg (h - 2R) = mgR / 2

h - 2R = R / 2

h = 2.5 R.