Chapter 15.1, Problem 15.3P

The motion of an oscillation flywheel is defined by the relation $\theta ={\theta }_0{e}^{-7\pi t/6}$ sin $4\pi t$ , where $\theta$ is expressed in radians and t in seconds. Knowing that ${\theta }_0=0.4$ rad, determine the angular coordinate, the angular velocity, and the angular acceleration of the flywheel when (a) $t=0.125s$ , (b) $t=\infty$ .

To determine

(a)

The angular coordinate, the angular velocity and the angular acceleration of the flywheel when $t=0.125 s$.

Answer to Problem 15.3P

The angular coordinate of the flywheel $=\theta =0.253 rad.$

The angular velocity of the flywheel $=\stackrel{•}{\theta }=-0.927 rad s^{-1}.$

The angular acceleration of the flywheel $=\stackrel{••}{\theta }=-36.55 rad s^{-2}.$

Explanation of Solution

Given information:

The motion of the flywheel is defined by, $\theta ={\theta }_0{e}^{-\frac{7\pi t}{6}}\sin 4\pi t$, where $\theta$ is expressed in radians and $t$ in seconds.

${\theta }_0=0.4 rad$

$t=0.125 s$

Concept used:

The angular velocity,

$\stackrel{•}{\theta }=\frac{d\theta }{dt}$

The angular acceleration,

$\stackrel{••}{\theta }=\frac{d\stackrel{•}{\theta }}{dt}$

Calculation:

The motion of the flywheel is defined by,

$\theta ={\theta }_0{e}^{-\frac{7\pi t}{6}}\sin 4\pi t$

Thus, at $t=0.125 s$, the angular coordinate of the flywheel,

$\theta =0.4 rad\times e^{-\frac{7\pi \times 0.125}{6}}\times \sin \frac{\pi }{2}$

$\underset{\_}{\underset{\_}{\theta =0.253 rad}}$

The angular velocity,

$\stackrel{•}{\theta }=\frac{d\theta }{dt}=\pi {\theta }_0{e}^{-\frac{7\pi t}{6}}(4\cos 4\pi t-\frac{7}{6}\sin 4\pi t)$

Thus, at $t=0.125 s$, the angular velocity of the flywheel,

$\stackrel{•}{\theta }=\frac{d\theta }{dt}=\pi \times 0.4 rad\times e^{-\frac{7\pi \times 0.125}{6}}(4\cos \frac{\pi }{2}-\frac{7}{6}\sin \frac{\pi }{2})$

$\underset{\_}{\underset{\_}{\stackrel{•}{\theta }=-0.927 rad s^{-1}}}$

The angular acceleration,

$\stackrel{••}{\theta }=\frac{d\stackrel{•}{\theta }}{dt}=-{\pi }^2{\theta }_0{e}^{-\frac{7\pi t}{6}}(\frac{527}{36}\sin 4\pi t+\frac{28}{3}\cos 4\pi t)$

Thus, at $t=0.125 s$, the angular acceleration of the flywheel,

$\stackrel{••}{\theta }=\frac{d\stackrel{•}{\theta }}{dt}=-{\pi }^2\times 0.4 rad\times e^{-\frac{7\pi \times 0.125}{6}}\times (\frac{527}{36}\sin \frac{\pi }{2}+\frac{28}{3}\cos \frac{\pi }{2})$

$\underset{\_}{\underset{\_}{\stackrel{••}{\theta }=-36.55 rad s^{-2}}}$

The angular coordinate of the flywheel $=\theta =0.253 rad.$

The angular velocity of the flywheel $=\stackrel{•}{\theta }=-0.927 rad s^{-1}.$

The angular acceleration of the flywheel $=\stackrel{••}{\theta }=-36.55 rad s^{-2}.$

Conclusion:

The angular coordinate of the flywheel $=\theta =0.253 rad.$

The angular velocity of the flywheel $=\stackrel{•}{\theta }=-0.927 rad s^{-1}.$

The angular acceleration of the flywheel $=\stackrel{••}{\theta }=-36.55 rad s^{-2}.$

To determine

(b)

Determine the angular coordinate, the angular velocity and the angular acceleration of the flywheel when $t=\infty$.

Answer to Problem 15.3P

The angular coordinate of the flywheel, $=\theta =0 rad.$

The angular velocity of the flywheel $=\stackrel{•}{\theta }=0 rad s^{-1}.$

The angular acceleration of the flywheel $=\stackrel{••}{\theta }=0 rad s^{-2}.$

Explanation of Solution

Given information:

The motion of the flywheel is defined by, $\theta ={\theta }_0{e}^{-\frac{7\pi t}{6}}\sin 4\pi t$, where $\theta$ is expressed in radians and $t$ in seconds.

${\theta }_0=0.4 rad$

$t=\infty$

Concept used:

The angular velocity,

$\stackrel{•}{\theta }=\frac{d\theta }{dt}$

$t=\infty$

The angular velocity,

$\stackrel{•}{\theta }=\frac{d\theta }{dt}$

The angular acceleration,

$\stackrel{••}{\theta }=\frac{d\stackrel{•}{\theta }}{dt}$

Calculation:

The motion of the flywheel is defined by,

$\theta ={\theta }_0{e}^{-\frac{7\pi t}{6}}\sin 4\pi t$

Thus, at $t=\infty$, the angular coordinate of the flywheel,

$\theta =0.4 rad\times e^{-\infty }\times \sin \infty$

$\underset{\_}{\underset{\_}{\theta =0 rad}}$

The angular velocity,

$\stackrel{•}{\theta }=\frac{d\theta }{dt}=\pi {\theta }_0{e}^{-\frac{7\pi t}{6}}(4\cos 4\pi t-\frac{7}{6}\sin 4\pi t)$

Thus, at $t=\infty$, the angular velocity of the flywheel,

$\stackrel{•}{\theta }=\frac{d\theta }{dt}=\pi \times 0.4 rad\times e^{-\infty }(4\cos \infty -\frac{7}{6}\sin \infty )$

$\underset{\_}{\underset{\_}{\stackrel{•}{\theta }=0 rad s^{-1}}}$

The angular acceleration,

$\stackrel{••}{\theta }=\frac{d\stackrel{•}{\theta }}{dt}=-{\pi }^2{\theta }_0{e}^{-\frac{7\pi t}{6}}(\frac{527}{36}\sin 4\pi t+\frac{28}{3}\cos 4\pi t)$

Thus, at $t=\infty$, the angular acceleration of the flywheel,

$\stackrel{••}{\theta }=\frac{d\stackrel{•}{\theta }}{dt}=-{\pi }^2\times 0.4 rad\times e^{-\infty }\times (\frac{527}{36}\sin \infty +\frac{28}{3}\cos \infty )$

$\underset{\_}{\underset{\_}{\stackrel{••}{\theta }=0 rad s^{-2}}}$

The angular coordinate of the flywheel, $=\theta =0 rad.$

The angular velocity of the flywheel $=\stackrel{•}{\theta }=0 rad s^{-1}.$

The angular acceleration of the flywheel $=\stackrel{••}{\theta }=0 rad s^{-2}.$

Conclusion:

The angular coordinate of the flywheel, $=\theta =0 rad.$

The angular velocity of the flywheel $=\stackrel{•}{\theta }=0 rad s^{-1}.$

The angular acceleration of the flywheel $=\stackrel{••}{\theta }=0 rad s^{-2}.$