Rolling Hula Hoop: Unlocking the Secrets of Kinetic Energy

What is the linear speed required for a 0.0494-kg hula hoop to have a total kinetic energy of 0.111 J?

Is there a specific formula we can use to calculate the linear speed needed for the hula hoop?

Answer:

The linear speed of the hula hoop can be obtained by equating the total kinetic energy to the summation of kinematic and rotational energies and solving for the speed.

To determine the linear speed of a 0.0494-kg hula hoop with a total kinetic energy of 0.111 J, we can use the principle of kinetic energy. When a hula hoop rolls on the ground without slipping, the total kinetic energy is the combination of translational kinetic energy and rotational kinetic energy.

For a rolling hoop, the total kinetic energy (KE) is given by: KE = 1/2*m*v² + 1/2*I*(v/r)², where m is the mass of the hoop, v is the linear speed, r is the radius of the hoop, and I is the moment of inertia of the hoop.

Given that the moment of inertia of a hoop is equal to m*r², and the angular speed (w) is equal to v/r, we can substitute these values into our equation. By equating the total kinetic energy to 0.111 J, we can solve for the linear speed v.

Therefore, the linear speed of the hula hoop can be calculated as follows:

0.111 = 1/2*0.0494*v² + 1/2*0.0494*r²*(v/r)²

Simplifying the equation, we get: 0.111 = 0.0494*v²

Solving for v, we find: v = √(0.111 / 0.0494), which gives us the linear speed of the hula hoop in meters per second.

By calculating the linear speed using this method, we can determine the exact velocity required for the hula hoop to have a total kinetic energy of 0.111 J. Understanding the relationship between kinetic energy and motion can provide insights into the dynamics of rolling objects like hula hoops.