A 2.0 kg hoop rolls without slipping on a horizontal surface so that its center proceeds to the right with a constant linear speed of 6.0 m/s. The radius of the hoop is 0.5 m. What is the total kinetic energy of the hoop?

72 J

Explanation:

Total kinetic energy will be tge sum of rotational and translational energy.

Rotational kinetic energy is given by where I is moment of inertia which is given by and here m is mass, r is radius. Also, hence making \omega the subject then where v is the velocity.

Rotational kinetic energy=

This is same as the formula for translational kinetic energy which is given by

Therefore, total kinetic energy=

Substituting m with 2 kg and v with 6 m/s then total energy will be

305.7

explanation:

density =m/v

density =917/3.00

density = 305.

approximately = 305.7

in four significant figures