Question

A propeller is modeled as five identical uniform rods extending radially from its axis. The length and mass of each rod are 0.827 m and 2.89 kg, respectively. When the propeller rotates at 507 rpm (revolutions per minute), what is its rotational kinetic energy K

Answer 1

Answer:

3241.35J

Explanation:

No. Of rods = 5

Mass = 2.89kg

Length (L) = 0.827m

W = 507rpm

Kinetic energy of rotation = ½I*ω²

For each rod, the moment of inertia (I) = ML² / 3

I = ML² / 3

I = [2.89*(0.827)²] / 3

I = 1.367 / 3 = 0.46kgm²

ω = 507 rev/min. Convert rev/min to rev/sec.

507 * 2Πrads/60s = 53.09rad/s

ω = 53.09rad/s

k.e = ½ I * ω²

K.E = ½ * 0.46 * (53.09)²

K.E = 648.27.

But there five (5) rods, so kinetic energy is equal to

K.E = 5 * 648.27 = 3241.35J