How to Calculate the Linear Speed of a Hula Hoop?

What linear speed must a 0.065-kg hula hoop have if its total kinetic energy is to be 0.12 J?

Assume the hoop rolls on the ground without slipping.

Answer:

The hula hoop must have a linear speed of approximately 1.03 m/s to have a total kinetic energy of 0.12 J.

In order to determine the linear speed of the hula hoop with a total kinetic energy of 0.12 J, we can use the equation for kinetic energy and the relationship between linear speed and angular speed for a rolling object without slipping.

The kinetic energy of a rotating object is given by:

KE = (1/2) * I * ω^2

For a solid hoop rolling without slipping, the moment of inertia can be expressed as:

I = m * R^2

Now, to find the linear speed, we first need to find the angular speed (ω):

ω = √[(2 * KE) / (m * R^2)]

Substitute the given values into the equation to find the linear speed (v):

v = √[(2 * 0.12 J) / 0.065 kg]

v ≈ 1.03 m/s

This calculation shows that the hula hoop must have a linear speed of approximately 1.03 m/s to have a total kinetic energy of 0.12 J. This speed ensures that the hoop is rotating without slipping on the ground, allowing for a fun and energetic playtime experience!