Question

He angular position of a swinging door is described by θ = 5.00 + 10.0 t + 2.00 t2 [rad]. (a) determine the angular position, angular speed, and angular acceleration of the door at t = 3.00 s. (b) if the door starts from rest and its angular acceleration is 0.560 rad/s2, through what angle does it have to turn to reach an angular speed of 0.750 rad/s?

Answer 1

(a) The angular position of the door is described by
$\theta(t)=5+10t+2t^2 [rad]$

The angular velocity is given by the derivative of the angular position:
$\omega(t)=10+4t [rad/s]$

While the angular acceleration is given by the derivative of the angular velocity:
$\alpha(t)=4 [rad/s^2]$

We want to find the values of these quantities at time t=3.00 s, so we must substitute t=3.00 s into the expressions for $\theta, \omega, \alpha$:
$\theta(3.00 s)=5+(10)(3.00 s)+2(3.00s)^2 = 53 rad$
$\omega(3.00 s)=10+4(3.00s)=22 rad/s$
$\alpha(3.00s)=4 rad/s^2$

(b) The door starts from rest, so its initial angular velocity is $\omega_i=0 rad/s$, and it reaches a final angular velocity of $\omega_f=0.750 rad/s$ with an angular acceleration of $\alpha=0.560 rad/s^2$. We can find the angular distance covered by the door by using the following relationship:
$2 \alpha \theta = \omega_f^2 - \omega_i^2$
from which we find
$\theta= \frac{\omega_f^2}{2 \alpha}= \frac{(0.750 rad/s)^2}{2 \cdot 0.560 rad/s^2} =0.502 rad$