Question

A helicopter propeller blade starts from rest and experiences a constant angular acceleration of 22.0 rad/s2. Find the speed of a point on the blade 2.30 m from the axis of rotation when the blade has completed 15 revolutions

Answer 1

here,

the constant angular acceleration , alpha = 22 rad/s^2

the distance from the axis , r = 2.3 m

the number of revolutions , N = 15 rev

the angle covered , theta = N * 2pi

theta = 15 * 2pi = 94.2 rad

let the final angular velocity be w

using third equation of motion

w^2 - w0^2 = 2 * alpha * theta

v^2 - 0 = 2 * 22 * 94.2

solving for w

w = 64.4 rad/s

the final speed , v = r * w

v = 2.3 * 64.4 m/s = 148.1 m/s