Chapter 16.3, Problem 1FP

Determine its constant angular acceleration and the time required.

To determine

The gear’s constant angular acceleration and the time required to achieve the angular velocity $\omega =30\text{rad/s}$

Answer to Problem 1FP

The gear’s constant angular acceleration $\alpha$ to achieve the angular velocity $\omega =30\text{rad/s}$

Is $3.58{\text{rad/s}}^2$

The time required to achieve the angular velocity $\omega =30\text{rad/s}$ is $8.38\text{s}$.

Explanation of Solution

Given:

The angular acceleration $\alpha$ is $3{\text{rad/s}}^2$.

Final angular velocity $\omega$ is $30\text{rad/s}$.

The angular displacement of gear $\theta$ is $20\text{rev}$.

Write the expression of angular displacement for gear.

$\begin{array}{c}\theta =20\text{rev}\left(\frac{2\pi \text{rad}}{1\text{rev}}\right)\\ =40\pi \text{rad}\end{array}$

Write the equation of the angular acceleration.

$\alpha =3{\text{rad/s}}^2$

Here, angular acceleration is $\alpha$.

Write the expression for the angular velocity.

$\omega ={\omega }_0+{\alpha }_{c}t$ (I)

Here, angular velocity is $\omega$, angular acceleration is ${\alpha }_c$ and time is $t$.

Write the expression for the angular velocity.

${\omega }^2={\omega }_0^2+2{\alpha }_c\left(\theta -{\theta }_0\right)$ (II)

Here, angular velocity is $\omega$, initial velocity is ${\omega }_0$, initial revolutions is ${\theta }_0$, angular acceleration is ${\alpha }_c$ and revolutions is $\theta$.

Conclusion:

Substitute   $40\pi$ for $\theta$, $0$ for ${\theta }_0$, $0$ for ${\omega }_0$ and $30\text{rad/s}$ for $\omega$ in Equation (I).

$\begin{array}{l}{\omega }^2={\omega }_0^2+2{\alpha }_c\left(\theta -{\theta }_0\right)\\ {\left(30\right)}^2={\left(0\right)}^2+2{\alpha }_c\left[40\pi -0\right]\\ {\alpha }_c=3.581{\text{rad/s}}^2\end{array}$

Thus, the gear’s constant angular acceleration $\alpha$ to achieve the angular velocity $\omega =30\text{rad/s}$ is $3.58{\text{rad/s}}^2$

Substitute $3.581{\text{rad/s}}^2$ for ${\alpha }_c$, $0$ for ${\omega }_0$ and $30\text{rad/s}$ for $\omega$ in Equation (II).

$\begin{array}{l}\omega ={\omega }_0+{\alpha }_{c}t\\ \left(30\right)=\left(0\right)+\left(3.581\right)t\\ t=8.38\text{s}\end{array}$

Thus, the time required to achieve the angular velocity $\omega =30\text{rad/s}$ is $8.38\text{s}$.