Question

a 52 kg person is on a loop-the-loop amusement park ride with a radius of 9.0 m. be sure to include units with all of your answers. a. if the ride is moving at a speed of 15.1 m/s, what is the normal force on the rider at the top of the loop?

Answer 1

The normal force on the rider at the top of the loop is 1612.68 N.

Step-by-step explanation:

To find the normal force on the rider at the top of the loop, we need to consider the forces acting on the person. At the top of the loop, there are two forces: the gravitational force (mg) and the normal force (N). The net force at the top of the loop should be equal to the centripetal force required to keep the person moving in a circle.

The centripetal force is given by:

Fc = (mv^2) / r

Where m is the mass of the rider, v is the velocity, and r is the radius of the loop. In this case, the velocity is given as 15.1 m/s and the radius is 9.0 m, so we can plug in these values to calculate the centripetal force.

Fc = (52 kg) * (15.1 m/s)^2 / 9.0 m = 1103.08 N

Since the net force at the top of the loop should be equal to the centripetal force, we can set up the following equation:

Fnet = N - mg = Fc

Substituting the values, we can solve for the normal force:

N - (52 kg) * (9.8 m/s^2) = 1103.08 N

Simplifying the equation,

N - 509.6 N = 1103.08 N

Then, solving for N:

N = 1612.68 N

Therefore, the normal force on the rider at the top of the loop is 1612.68 N.