Will the Hula Hoop Make It to the Top of the Hill?

Given the speed of the hula hoop at 26 ft/s and the height of the hill at 16 ft, we can utilize the concept of energy conservation to determine the outcome. At the bottom of the hill, the hoop's energy comprises both kinetic energy (translational) and rotational energy.

The initial energy at the bottom of the hill is calculated using the formula:

Eb = 1/2mv^2 + 1/2Iω^2

Where:

m = mass of the hoop

I = moment of inertia of the hoop

ω = angular velocity

In this case, the initial energy at the bottom (Eb) is equivalent to mv^2, given that the hoop is a uniform ring. Substituting the values, we get Eb = 676m.

The total energy required to reach the top of the hill (ET) can be calculated as mgh, where m is the mass of the hoop and g is the acceleration due to gravity (32.2 ft/s^2). Thus, ET = 512.2m.

By subtracting the total energy spent to raise the hoop to the top from the initial energy at the bottom, we find the remaining energy is in the form of kinetic and rotational energy. This difference is equal to 163.8m.

Therefore, by solving for the speed (v) using the remaining energy equation (163.8m = mv^2), we find the hula hoop's translational speed at the top to be approximately 12.8 ft/s if it makes it to the top of the hill.