Question

A 356kg roller coaster starts at rest a height of 5R above the ground and goes around a loop-the-loop of diameter 2.3R. a. What is its speed at the top of the loop, if the work done by friction is zero? b. What is its speed at the end of the track at height of zero, if the work done by friction remains zero? c. Explain what the maximum work done by friction could be if the roller coaster is to still complete the trip along the entire track.

Answer 1

(a)

use conservation of energy

mgh_i = mgh_f + 1/2mv²

cancel 'm' on both sides, we have

g * 5R = g * 2.3R + 1/2v²

g ( 5R - 2.3R) = 1/2v²

2.7gR = 1/2v²​​​​​​​

so,

v = sqrt ( 5.4gR)

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(b)

again use conservation of energy, this time final height is zero

so,

g * 5R = 1/2v²

so,

v = sqrt ( 10gR)

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(c)

for rolled coaster to complete the trip, it has to have minimum velocity at the top which is found as

mv² / R = mg

v = sqrt ( rg)

v = sqrt ( 1.15Rg)

so,

the maximum work done should be equal to this kinetic energy.