2. A 550 kg roller coaster car moves frictionlessly along tracks which contains
ID: 1411537 • Letter: 2
Question
2. A 550 kg roller coaster car moves frictionlessly along tracks which contains one complete vertical loop as sketched below.
a) The coaster car is just about to leave the tracks at the top of the vertical loop when the normal force exerted by the tracks on the wheels drops to zero. Make a free-body diagram of the coaster car at the top of the vertical loop and use Newton’s second law to find the speed at which the coaster car just about leaves the tracks.
b) If the coaster car goes slower at the top of the vertical loop than the speed you calculated in #2(a), the normal force will (1) become greater than zero, (2) remain zero, or (3) become negative and the car will (1) leave the tracks, or (2) remain in contact with the tracks.
c) If the coaster car goes faster at the top of the vertical loop than the speed you calculated in #2(a), the normal force will (1) become greater than zero, (2) remain zero, or (3) become negative and the car will (1) leave the tracks, or (2) remain in contact with the tracks.
d) Use energy methods to find the minimum initial height from which the coaster car must be released with zero velocity so it will not leave the tracks at the top of the loop? e) If you release the coaster car at a higher elevation than you calculated in #2(d), the normal force will (1) become greater than zero, (2) remain zero, or (3) become negative and the car will (1) leave the tracks, or (2) remain in contact with the tracks.
f) If you release the coaster car at a lower elevation than you calculated in #2(d), the normal force will (1) become greater than zero, (2) remain zero, or (3) become negative and the car will (1) leave the tracks, or (2) remain in contact with the tracks.
g) If the coaster car starts out at the height you found is #2(d) with an initial speed of 1.0 m/s, find its speed at the top of the vertical loop using energy methods and find the normal force exerted by the tracks on the coaster’s wheels.
Explanation / Answer
Let R be radius of vertical loop
a) N= 0
mv2/R = mg => v = sqrt(Rg)
b) normal force remains zero
car will leave the track
c) normal force becomes greater than zero
car will remain in contact with the track
d) mgh = mg(2R) + 0.5 m(Rg)
=> h = 2.5R
e) normal force becomes greater than zero
car will remain in contact with the track
f) normal force remains zero
car will leave the track
g) 0.5 m x 12 + mg(2.5R) = mg(2R) + 0.5 mv2
v = sqrt(1 + gR)
N = mv2/R - mg = 550(1/R+ g - g) = 550/R newton
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