Question

You are testing a new amusement park roller coaster with an empty car with a mass of 110 kg

kg . One part of the track is a vertical loop with a radius of 12.0 m
m . At the bottom of the loop (point A
A) the car has a speed of 25.0 m/s and at the top of the loop (point B
B) it has speed of 8.00 m/s

Answer 1

The Physics question asks for the speed of a roller coaster at the top of a loop to ensure riders experience a centripetal acceleration greater than gravity, in a high school context. By using the formula for centripetal acceleration, we can solve for the required speed using the given acceleration and radius.

Step-by-step explanation:

The subject of the question is Physics, specifically dealing with the concepts of speed, gravity, and centripetal acceleration as it relates to roller coasters. When testing roller coasters, engineers must ensure that the design allows for proper forces to keep riders safely in their seats during loops. In this case, the roller coaster must achieve a speed at the top of the loop that results in a downward centripetal acceleration greater than that of gravity. Given that the downward acceleration is 1.50 times the acceleration due to gravity (g), and knowing the radius of curvature (r) is 15.0m, we can use the formula for centripetal acceleration
`ac = v^2/r` where ac is the centripetal acceleration, v is the speed at the top of the loop, and r is the radius of curvature.

Since ac here is 1.50g, we set the equation 1.50g = v^2/r and solve for v to find the required speed at the top of the loop to keep passengers safely seated. It's important to make sure we use the correct units for g (9.8 m/s^2).