Asked by larry
The angle an airplane propeller makes with the horizontal as a function of time is given by \theta = (115 <units>rad/s</units>)t + (45.0 <units>rad/s^2</units>)t^2. calculate the angular acceleration from 0 to 1.o sec
Answers
Answered by
drwls
Your equation is rather hard to read. Let's write is as
theta = 115 (rad/s)*t + 45 (rad/s^2)*t^2.
The angular velocity is
d(theta)/dt = omega = 115 + 90 t rad/s
The angular acceleration rate is
d^2/(theta)/dt^2 = alpha = 90 rad/s^s
The angular acceleration rate (alpha) is twice the second term, or 90 rad/s^2
90 will remain the acceleration rate while it is accelerating according to that formula you provided.
theta = 115 (rad/s)*t + 45 (rad/s^2)*t^2.
The angular velocity is
d(theta)/dt = omega = 115 + 90 t rad/s
The angular acceleration rate is
d^2/(theta)/dt^2 = alpha = 90 rad/s^s
The angular acceleration rate (alpha) is twice the second term, or 90 rad/s^2
90 will remain the acceleration rate while it is accelerating according to that formula you provided.
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