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!