A door starts from rest and swings with an angular acceleration, alpha, α = 6 rad/s2, where t is in seconds. Determine the angle in radians through which the door turns in the first 3 s.
Hint: kinematics equation for angular position (i.e., angle) is exactly the same as for (linear) position from translational kinematics (x = x_i + v_i + 1/2at^2) with Greek letters instead of Latin letters: θ = θ_i + omega_i*t + 1/2alpha*t^2.
To determine the angle in radians through which the door turns in the first 3 seconds, we will use the given kinematics equation for angular position.
The equation for angular position is: θ = θ_i + omega_i*t + 1/2*alpha*t^2
Here, θ is the final angle, θ_i is the initial angle (which is usually taken to be 0 in most problems), omega_i is the initial angular velocity (which is also 0 in this case as the door starts from rest), alpha is the angular acceleration, and t is the time.
Given:
alpha = 6 rad/s^2 (angular acceleration)
t = 3 s (time)
We can plug in these values into the kinematics equation to find the angle θ:
θ = θ_i + omega_i*t + 1/2*alpha*t^2
Since the door starts from rest, the initial angular velocity (omega_i) is 0, and the initial angle (θ_i) is also 0, we can simplify the equation to:
θ = 1/2*alpha*t^2
Now we can substitute the given values:
θ = 1/2 * 6 * (3)^2
Simplifying further:
θ = 1/2 * 6 * 9
θ = 3 * 6
θ = 18 radians
Therefore, the door turns through an angle of 18 radians in the first 3 seconds.