A roller coaster car mass 988 kg is about to roll down a track with a circular l

Question

A roller coaster car mass 988 kg is about to roll down a track with a circular loop at the bottom. The diameter of the circular loop is 20 m and the car start out from rest 40m above the lowest point of the track. Ignore resistance and assume rolls without slipping. A) at what speed does the car reach the top of the loop. B) what is the force exerted on the car by the track at the top of the loop? C) from what minimum height above the bottom of the loop can the car be released so that it does not lose contact with the track at the top of the loop?

Explanation / Answer

a) conservation of energy

mgh = 1/2mv2, mass cancels out, and the h that it falls is only 20 m ( started at 40m off ground, ended at top of loop which is 20 m off ground, so 40-20 = 20 )

( 9.8)(20) = 1/2 v2

v = 19.8 m/s

b) we use centripetal:

-mg - Fn = -mv2/r

-( 988 )(9.8) - Fn = -( 988 )(19.8)2/10

Fn = 29051.15 N

c) at a minimum, Fn = 0, so:

-mg - 0 = -mv2/r

g = v2/r

9.8 = v2/10

v = 9.9 m/s

Now use mgh = 1/2mv2

( 9.8)(h) = 1/2 (9.9)2

h = 5m

so we need to start 5 m above the top of loop, so from bottom of loop is 25 m

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