Hi there!
a.
We can use the initial conditions to solve for w₀.
It is given that:
[tex]\frac{d\theta}{dt} = w_0e^{-\sigma t}[/tex]
We are given that at t = 0, ω = 3.7 rad/sec. We can plug this into the equation:
[tex]\omega(0)= \omega_0e^{-\sigma (0)}\\\\3.7 = \omega_0 (1)\\\\\omega_0 = \boxed{3.7 rad/sec}[/tex]
Now, we can solve for sigma using the other given condition:
[tex]2 = 3.7e^{-\sigma (8.6)}\\\\.541 = e^{-\sigma (8.6)}\\\\ln(.541) = -\sigma (8.6)\\\\\sigma = \frac{ln(.541)}{-8.6} = \boxed{0.0714s^{-1}}[/tex]
b.
The angular acceleration is the DERIVATIVE of the angular velocity function, so:
[tex]\alpha(t) = \frac{d\omega}{dt} = -\sigma\omega_0e^{-\sigma t}\\\\\alpha(t) = -(0.0714)(3.7)e^{-(0.0714) (3)}\\\\\alpha(t) = \boxed{-0.213 rad\sec^2}[/tex]
c.
The angular displacement is the INTEGRAL of the angular velocity function.
[tex]\theta (t) = \int\limits^{t_2}_{t_1} {\omega(t)} \, dt\\\\\theta(t) = \int\limits^{2.5}_{0} {\omega_0e^{-\sigma t}dt\\\\[/tex]
[tex]\theta(t) = -\frac{\omega_0}{\sigma}e^{-\sigma t}\left \| {{t_2=2.5} \atop {t_1=0}} \right.[/tex]
[tex]\theta = -\frac{3.7}{0.0714}e^{-0.0714 t}\left \| {{t_2=2.5} \atop {t_1=0}} \right. \\\\\theta= -\frac{3.7}{0.0714}e^{-0.0714 (2.5)} + \frac{3.7}{0.0714}e^{-0.0714 (0)}[/tex]
[tex]\theta = 8.471 rad[/tex]
Convert this to rev:
[tex]8.471 rad * \frac{1 rev}{2\pi rad} = \boxed{1.348 rev}[/tex]
d.
We can begin by solving for the time necessary for the angular speed to reach 0 rad/sec.
[tex]0 = 3.7e^{-0.0714t}\\\\t = \infty[/tex]
Evaluate the improper integral:
[tex]\theta = \int\limits^{\infty}_{0} {\omega_0e^{-\sigma t}dt\\\\[/tex]
[tex]\lim_{a \to \infty} \theta = -\frac{\omega_0}{\sigma}e^{-\sigma t}\left \| {{t_2=a} \atop {t_1=0}} \right.[/tex]
[tex]\lim_{a \to \infty} \theta = -\frac{3.7}{0.0714}e^{-0.0714a} + \frac{3.7}{0.0714}e^{-0.0714(0)}\\\\ \lim_{a \to \infty} \theta = \frac{3.7}{0.0714}(1) = 51.82 rad[/tex]
Convert to rev:
[tex]51.82 rad * \frac{1rev}{2\pi rad} = \boxed{8.25 rev}[/tex]
