A roller coaster has a vertical loop with radius 18.0 m. What minimum speed should the roller-coastercar be moving at the top of the loop so that the passengers do not lose contact with the seats?
The correct answer and explanation is:
To prevent passengers from losing contact with their seats at the top of a vertical loop, the minimum speed occurs when the normal force becomes zero—this is the point where gravity alone provides the required centripetal force to keep the car moving in a circular path.
Correct Answer:
The minimum speed at the top of the loop is approximately 13.3 m/s.
Explanation (300 words):
At the top of a vertical loop, two forces act downward on the roller-coaster car and its passengers:
- The gravitational force (
mg) - The normal force (
N), which presses the passengers against the seat
For passengers to just maintain contact, the normal force becomes zero (N = 0). In this case, gravity provides all the centripetal force needed to keep the car moving in a circular loop.
The formula for centripetal force is: Fc=mv2rF_c = \frac{mv^2}{r}
Setting this equal to the gravitational force at the top: mv2r=mg\frac{mv^2}{r} = mg
Canceling the mass mm from both sides: v2r=g\frac{v^2}{r} = g
Solving for vv: v=grv = \sqrt{gr}
Given:
- g=9.8 m/s2g = 9.8 \, \text{m/s}^2
- r=18.0 mr = 18.0 \, \text{m}
v=9.8×18.0=176.4≈13.3 m/sv = \sqrt{9.8 \times 18.0} = \sqrt{176.4} \approx 13.3 \, \text{m/s}
Conclusion:
The minimum speed at the top of the loop must be about 13.3 m/s. At this speed, the force of gravity alone supplies the needed centripetal force, and the normal force is zero, meaning passengers are on the verge of losing contact. Any slower, and they would fall out unless properly restrained. For safety, real roller coasters travel faster than this minimum to maintain a positive normal force and keep passengers firmly in their seats.