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

Answer:

$K_r=5992J$

Explanation:

The rotational kinetic energy is given by:

$K_r=\frac{1}{2}I*\omega^2$

the moment of inertia for this case is given by:

$I=\frac{1}{3}*m*l^2\\I=0.460kg.m^2$

$\omega=\frac{V_{rpm}}{60}*2\pi\\\\\omega=\frac{577}{60}*2\pi\\\\\omega=60.4rad/s$

So the total kinetic energy is:

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