Question

A propeller is modeled as five identical uniform rods extending radially from its axis. The length and mass of each rod are 0.701 m and 2.81 kg , respectively. When the propellor rotates at 577 rpm (revolutions per minute), what is its rotational kinetic energy?

Answer 1

`K_r=5992J`

Step-by-step explanation:

The rotational kinetic energy is given by:

`K_r=(1)/(2)I*\omega^2`

the moment of inertia for this case is given by:

`I=(1)/(3)*m*l^2\\I=0.460kg.m^2`

`\omega=(V_(rpm))/(60)*2\pi\\\\\omega=(577)/(60)*2\pi\\\\\omega=60.4rad/s`

So the total kinetic energy is:

`K_r=5*(1)/(2)*0.460kg.m^2*(60.4rad/s)^2\\K_r=4195J`