Modern roller coasters have vertical loops like the one shown in the figure. The

Question

Modern roller coasters have vertical loops like the one shown in the figure. The radius of curvature is smaller at the top than on the sides so that the downward centripetal acceleration at the top will be greater than the acceleration due to gravity, keeping the passengers pressed firmly into their seats.

What is the speed of the roller coaster in m/s at the top of the loop if the radius of curvature there is 17 m and the downward acceleration of the car is 1.4 g?

How high above the top of the loop must the roller coaster start from rest for it to be moving this fast when it reaches the top of the loop, assuming negligible friction? Give your answer in meters.

If it actually starts 3 m higher than your answer to the previous part (yet still reaches the top of the loop with the same velocity), how much energy, in joules, did it lose to friction? Its mass is 1100 kg.

Explanation / Answer

Part 1

v^2/r = 1.4*g

or, v = (1.4g*r)^0.5 = 15.43 m/s

Part 2

Equating potential energy to kinetic energy

mgh = 0.5*m*v^2

or, h =0.5*v^2/g = 11.9 m

Part 3

Energy lost = extra potenial energy it started with as final velocity is same = mg(h1-h2) = 1100*10*3 J = 33 kJ

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