Question

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?

Answer 1

Answer:

72 J

Explanation:

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

Rotational kinetic energy is given by $0.5I\omega^{2}$ where I is moment of inertia which is given by $mr^{2}$ and here m is mass, r is radius. Also, $v=\omega r$ hence making \omega the subject then $\omega=\frac {v}{r}$ where v is the velocity.

Rotational kinetic energy=$0.5I\omega^{2}=0.5mr^{2}\times(\frac {v}{r})^{2}= 0.5mv^{2}$

This is same as the formula for translational kinetic energy which is given by $0.5mv^{2}$

Therefore, total kinetic energy= $0.5mv^{2}+0.5mv^{2}=mv^{2}$

Substituting m with 2 kg and v with 6 m/s then total energy will be $2\times 6^{2}=72 J$