Question

A 5.0 kg hoop of ring with diameter 50cm rolls along a horizontal floor so that hoops's center of mass has a speed of 0.50m/s. Determined the total energy of the hoop as it rolls along the floor. take the hoop to be a ring rotating about its center of gravity.

Answer 1

the moment of inertia for the ring is

$I=m\cdot r^{2}$

where:

$m=5kg$ mass

$r=50cm/2=25cm\cdot \left ( 1m/100cm \right )=0.25m$ radius

evaluand numerically

$I=m\cdot r^{2}=5kg\cdot \left ( 0.25m \right )^{2}=0.3125kg\cdot m^{2}$

the total energy is equal to the translational and rotational energy

$E_{T}=K_{rot}\dotplus K_{trans}$

$K_{rot}=\left ( 1/2 \right )I\cdot w^{2}$

but the angular velocity is equal to $w=V/R$

now $K_{rot}=\left ( 1/2 \right )I\cdot \left ( V/R \right )^{2}=\left ( 1/2 \right )I\cdot V^{2}/R^{2}$

evaluated numerically $K_{rot}=\left ( 1/2 \right )0.3125kg \cdot m^{2}\cdot \left ( 0.5m/s \right )^{2}/\left ( 0.25m \right )^{2}=0.625J$

$K_{rot}=0.625J$

translational energy

$K_{trans}=\left ( 1/2 \right )mV^{2}=\left ( 1/2 \right )5kg \left ( 0.5m/s \right )^{2}=0.625J$

remember

$E_{T}=K_{rot}\dotplus K_{trans}=0.625J\dotplus 0.625J=1.25j$