How much work must be done to stop a rolling hoop?

What is the calculation to determine the work done on a hoop to stop it?

As the mass of the hoop is 120 kg and the hoop's center of mass has a speed of 0.180 m/s, how can we calculate the work required to stop the hoop?

Calculation of Work Done to Stop a Rolling Hoop

The work that must be done on the hoop to stop it is calculated to be -3.888 J.

When a 120 kg hoop rolls along a horizontal floor with its center of mass moving at a speed of 0.180 m/s, the work required to stop it can be determined by analyzing its kinetic energy and rotational inertia.

The rotational inertia (I) of the hoop can be calculated using the formula I = mr^2, where m is the mass of the hoop and r is the radius of the hoop. In this case, I = 120r^2.

By considering both the linear kinetic energy and rotational kinetic energy of the hoop, the total kinetic energy (K) can be determined using the formulas K = (1/2mv^2) + (1/2Iw^2), where w is the angular velocity of the hoop given by w = (v/r)^2.

After performing the calculations, the total kinetic energy of the hoop is found to be 3.888 J. To bring the hoop to a complete stop, the final kinetic energy must be reduced to zero, resulting in a work done of -3.888 J.

Therefore, a work of -3.888 J must be done on the hoop to stop its motion completely.