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
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.
To calculate the angular acceleration from 0 to 1.0 sec, we need to find the derivative of the given function \(\theta = (115 \ \text{rad/s})t + (45.0 \ \text{rad/s}^2)t^2\) with respect to time (t). The derivative of the equation will give us the angular velocity (\(d\theta/dt\)), and taking the derivative again will give us the angular acceleration (\(d^2\theta/dt^2\)).
Let's go step by step:
1. Compute the first derivative of \(\theta\) with respect to time (t).
\(\frac{d\theta}{dt} = 115 \ \text{rad/s} + 2(45.0 \ \text{rad/s}^2)t\)
2. Substitute t = 1.0 sec into the equation to find the angular velocity at t = 1.0 sec.
\(\frac{d\theta}{dt} = 115 \ \text{rad/s} + 2(45.0 \ \text{rad/s}^2)(1.0 \ \text{sec})\)
Calculate the result:
\(\frac{d\theta}{dt} = 115 \ \text{rad/s} + 2(45.0 \ \text{rad/s}^2)(1.0 \ \text{sec}) = 205 \ \text{rad/s}\)
3. Compute the second derivative of \(\theta\) with respect to time (t), which gives the angular acceleration.
\(\frac{d^2\theta}{dt^2} = 2(45.0 \ \text{rad/s}^2)\)
Calculate the result:
\(\frac{d^2\theta}{dt^2} = 2(45.0 \ \text{rad/s}^2) = 90.0 \ \text{rad/s}^2\)
Therefore, the angular acceleration from 0 to 1.0 sec is \(90.0 \ \text{rad/s}^2\).