Question

An airplane is flying in a straight line with a velocity of 200 mi/h and an acceleration of 3 mi/h2. If the propeller has a diameter of 6 ft and is rotating at an angular rate of 120 rad/s, determine the magnitudes of velocity and acceleration of a particle located on the tip of the propeller.

Answer 1

464.373 ft / s, 43200 ft/s²

Step-by-step explanation:

200 mi/h = 293.333 ft/s

3 mi/h² = 3 × (1 ft / s² / 2454.5454) = 0.0012222 = acceleration of the airplane

velocity of the rotating propeller Vr = ωr where ω is the angular rate = 120 rad/s and the radius = 6 ft /2 = 3 ft = 120 × 3 = 360 ft / s

Velocity of the particle = resultant of the velocities = √ ( 293.333² + 360² ) = 464.373 ft / s

centripetal acceleration = Vr² / r = 360² / 3 = 43200 ft/s²

acceleration of the particle = resultant accelerations = √ ( 0.001222² + 43200²) = 43200 ft/s²