At the top of the loop, the net force is directed toward the center of the circular path and is given by the centripetal force:
Fc = mv² / r
where Fc is the centripetal force, m is the mass of the coaster, v is the velocity, and r is the radius of the loop.
Given:
m = 2000 kg
v = 10 m/s
r = 5 m
Plugging in these values into the formula:
Fc = (2000 kg)(10 m/s)² / 5 m
= (2000 kg)(100 m²/s²) / 5 m
= 400,000 N
Therefore, the magnitude of the normal force that the track applies to the coaster is 400,000 N.
A roller coaster heads into a circular loop with radius 5m .
If the 2000 kg coaster is moving at 10 m/s at the top of the loop, what is the magnitude of the normal force that the track applies to the coaster in N?
1 answer
1 answer