Chapter 15.1, Problem 15.3P

The motion of an oscillating flywheel is defined by the relation $\theta ={\theta }_0{e}^{-7\pi t/6}\sin 4\pi t$, where θ is expressed in radians and t in seconds. Knowing that θ₀ = 0.4 rad, determine the angular coordinate, the angular velocity, and the angular acceleration of the flywheel when (a) t = 0.125 s, (b) t = ∞.

Fig. P15.2 and P15.3

To determine

Find the angular coordinate, angular velocity, and angular acceleration of the flywheel at time $t=0.125\text{s}$.

Answer to Problem 15.3P

The angular coordinate, angular velocity, and angular acceleration of the flywheel at time $t=0.125\text{s}$ are $\underset{\_}{0.253\text{rad}}$, $\underset{\_}{-0.927\text{rad/s}}$, and $\underset{\_}{-36.55{\text{rad/s}}^2}$.

Explanation of Solution

Given information:

Show the expression for the motion of the flywheel as follows:

$\theta ={\theta }_0{e}^{-7\pi t/6}\sin 4\pi t$ (1)

Here, $\theta$ is in radians, and t is in seconds.

Consider the angular coordinate, angular velocity, and angular acceleration of the flywheel are denoted by $\theta$, $\omega$, and $\alpha$ respectively.

The value of ${\theta }_0$ is 0.4.

Calculation:

Modify Equation (1).

Substitute 0.4 for ${\theta }_0$.

$\theta =0.4e^{-7\pi t/6}\sin 4\pi t$ (2)

Calculate the angular coordinate at time $t=0.125\text{s}$ using the relation:

Substitute $0.125$ for t in Equation (2).

$\begin{array}{c}\theta =0.4e^{-7\pi \left(0.125\right)/6}\sin 4\pi \left(0.125\right)\\ =0.4\times 0.63245\sin \left(\frac{\pi }{2}\right)\\ =0.4\times 0.63245\times 1\\ =0.253\text{rad}\end{array}$

Thus, the angular coordinate of the flywheel at time $t=0.125\text{s}$ is $\underset{\_}{0.253\text{rad}}$.

Calculate the angular velocity at time $t=0.125\text{s}$ using the relation:

Differentiate Equation (2) with respect to time as follows:

$\begin{array}{c}\omega =\frac{d\theta }{dt}\\ =\frac{d}{dt}\left(0.4e^{-7\pi t/6}\sin 4\pi t\right)\\ =0.4\frac{d}{dt}\left(e^{-7\pi t/6}\sin 4\pi t\right)\\ =0.4\left(-\frac{7\pi }{6}{e}^{-7\pi t/6}\sin 4\pi t+4\pi e^{-7\pi t/6}\cos 4\pi t\right)\end{array}$ (3)

Substitute $0.125$ for t in Equation (3).

$\begin{array}{c}\omega =0.4\left(\begin{array}{l}-\frac{7\pi }{6}{e}^{-7\pi \left(0.125\right)/6}\sin 4\pi \left(0.125\right)\\ +4\pi e^{-7\pi \left(0.125\right)/6}\cos 4\pi \left(0.125\right)\end{array}\right)\\ =0.4\left(\begin{array}{l}-\frac{7\pi }{6}\times 0.63245\times 1\\ +4\pi \times 0.63245\times 0\end{array}\right)\\ =0.4\left(-\frac{7\pi }{6}\times 0.63245\times 1\right)\\ =-\frac{7\pi }{6}\times 0.25298\text{rad/s}\end{array}$

$\omega =-0.927\text{rad/s}$

Thus, the angular velocity of the flywheel at time $t=0$ is $\underset{\_}{-0.927\text{rad/s}}$.

Calculate the angular acceleration at time $t=0\text{s}$ using the relation:

Differentiate Equation (3) with respect to time as follows:

$\begin{array}{c}\alpha =\frac{d\omega }{dt}\\ =\frac{d}{dt}\left[0.4\left(-\frac{7\pi }{6}{e}^{-7\pi t/6}\sin 4\pi t+4\pi e^{-7\pi t/6}\cos 4\pi t\right)\right]\\ =0.4\frac{d}{dt}\left(-\frac{7\pi }{6}{e}^{-7\pi t/6}\sin 4\pi t+4\pi e^{-7\pi t/6}\cos 4\pi t\right)\\ =0.4\left(\begin{array}{l}\frac{49{\pi }^2}{36}{e}^{-7\pi t/6}\sin 4\pi t-\frac{28{\pi }^2}{6}{e}^{-7\pi t/6}\cos 4\pi t\\ -\frac{28{\pi }^2}{6}{e}^{-7\pi t/6}\cos 4\pi t-16{\pi }^2{e}^{-7\pi t/6}\cos 4\pi t\end{array}\right)\end{array}$

$\alpha =0.4\left(\begin{array}{c}\frac{49{\pi }^2}{36}{e}^{-7\pi t/6}\sin 4\pi t-\frac{28{\pi }^2}{3}{e}^{-7\pi t/6}\cos 4\pi t\\ -16{\pi }^2{e}^{-7\pi t/6}\cos 4\pi t\end{array}\right)$ (4)

Substitute $0.125$ for t in Equation (4).

$\begin{array}{c}\alpha =0.4\left(\begin{array}{c}\frac{49{\pi }^2}{36}{e}^{-7\pi \left(0.125\right)/6}\sin 4\pi \left(0.125\right)-\frac{28{\pi }^2}{3}{e}^{-7\pi \left(0125\right)/6}\cos 4\pi \left(0.125\right)\\ -16{\pi }^2{e}^{-7\pi \left(0.125\right)/6}\cos 4\pi \left(0.125\right)\end{array}\right)\\ =0.4\left(\begin{array}{c}0-\frac{28{\pi }^2}{3}\times 0.63245\times 1\\ -16{\pi }^2\times 0.63245\times 1\end{array}\right)\\ =-36.55{\text{rad/s}}^2\end{array}$

Thus, the angular acceleration of the flywheel at time $t=0.125\text{s}$ is $\underset{\_}{-36.55{\text{rad/s}}^2}$.

To determine

Find the angular coordinate, angular velocity, and angular acceleration of the flywheel at time $t=\infty$.

Answer to Problem 15.3P

The angular coordinate, angular velocity, and angular acceleration of the flywheel at time $t=\infty$ are $\underset{\_}{0\text{rad}}$, $\underset{\_}{0\text{rad/s}}$, and $\underset{\_}{0{\text{rad/s}}^2}$.

Explanation of Solution

Given information:

Calculation:

Calculate the angular coordinate at time $t=\infty$ using the relation:

Substitute $\infty$ for t in Equation (2).

$\begin{array}{c}\theta =0.4\left(e^{-7\pi \left(\infty \right)/6}\right)\sin 4\pi \left(\infty \right)\\ =0.4\times 0\times \sin 4\pi \left(\infty \right)\\ =0\text{rad}\end{array}$

Thus, the angular coordinate of the flywheel at time $t=\infty$ is $\underset{\_}{0\text{rad}}$.

Calculate the angular velocity at time $t=\infty$ using the relation:

Substitute $\infty$ for t in Equation (3).

$\begin{array}{c}\omega =0.4\left(\begin{array}{l}-\frac{7\pi }{6}{e}^{-7\pi \left(\infty \right)/6}\sin 4\pi \left(\infty \right)\\ +4\pi e^{-7\pi \left(0.125\right)/6}\cos 4\pi \left(\infty \right)\end{array}\right)\\ =0\text{rad/s}\end{array}$

Thus, the angular velocity of the flywheel at time $t=\infty$ is $\underset{\_}{0\text{rad/s}}$.

Calculate the angular acceleration at time $t=\infty$ using the relation:

Substitute $\infty$ for t in Equation (4).

$\begin{array}{c}\alpha =0.4\left(\begin{array}{c}\frac{49{\pi }^2}{36}{e}^{-7\pi \left(\infty \right)/6}\sin 4\pi \left(\infty \right)-\frac{28{\pi }^2}{3}{e}^{-7\pi \left(\infty \right)/6}\cos 4\pi \left(\infty \right)\\ -16{\pi }^2{e}^{-7\pi \left(\infty \right)/6}\cos 4\pi \left(\infty \right)\end{array}\right)\\ =0{\text{rad/s}}^2\end{array}$

Thus, the angular acceleration of the flywheel at time $t=0.125\text{s}$ is $\underset{\_}{0{\text{rad/s}}^2}$.