3.A roller coaster has a vertical loop with radius 27.1 m. With what minimum speed should the roller coaster car be moving at the top of the loop so that the passengers do not lose contact with the seats?

Set the centripetal acceleration V^2/R at the top equal to g. That will assure that the car does not leave the track.

To find the minimum speed required for the roller coaster car to maintain contact with the seats at the top of the loop, we need to consider the force exerted on the passengers at that point.

At the top of the loop, the passengers experience two forces: the downward force due to gravity (mg) and the upward normal force (N) exerted by the seats. For the passengers to stay in contact with the seats, the net force acting on them at the top of the loop must be equal to or greater than zero.

At the top of the loop, the net force can be calculated as the difference between the normal force and the force due to gravity:

Net force = N - mg

To ensure that the net force is greater than or equal to zero, we set the net force equal to zero:

N - mg ≥ 0

Solving for N, we get:

N ≥ mg

The normal force N can be expressed in terms of the velocity (v), mass (m), and radius (r) of the loop:

N = m * (v^2 / r)

Substituting this expression for N into the inequality, we have:

m * (v^2 / r) ≥ mg

Canceling the mass (m) on both sides, we get:

v^2 / r ≥ g

Solving for the minimum speed (v) at the top of the loop, we have:

v^2 ≥ rg

Taking the square root of both sides, we get:

v ≥ √(rg)

Substituting the given values for radius (r = 27.1 m) and acceleration due to gravity (g = 9.8 m/s^2), we can now calculate the minimum speed:

v ≥ √(27.1 * 9.8)

v ≥ √(265.18)

v ≥ 16.28 m/s

Therefore, the minimum speed at the top of the loop should be 16.28 m/s to prevent the passengers from losing contact with the seats.