Chapter 5, Problem 48P

(a)

To determine

The angle of rotation.

Answer to Problem 48P

The angle of rotation is $\underset{\_}{270°}$.

Explanation of Solution

Write the equation of angle from angular kinematics.

$\left(\Delta \theta \right)=\omega \left(\Delta t\right)+\frac{1}{2}\alpha {\left(\Delta t\right)}^2$ (I)

Here, the angle is $\left(\Delta \theta \right)$, the angular speed is $\omega$, the time is $\left(\Delta t\right)$, the angular acceleration is $\alpha$.

Conclusion:

Substitute, $90°$ for $\left(\Delta \theta \right)$, $0$ for $\omega$ , $1\text{s}$ for $\left(\Delta t\right)$ in equation (I)

$\begin{array}{l}\left(90\right)=\left[\left(0\right)\left(\Delta t\right)\right]+\frac{1}{2}\alpha {\left(1\text{s}\right)}^2\\ \alpha =180°{\text{/s}}^{\text{2}}\end{array}$

Substitute, $180°{\text{/s}}^{\text{2}}$ for $\alpha$, $\alpha \left(\Delta t\text{'}\right)$ for $\omega$ , $1\text{s}$ for $\left(\Delta t\right)$, $1\text{s}$ for $\left(\Delta t\text{'}\right)$ in equation (I)

$\begin{array}{c}\left(\Delta \theta \right)=\left[\alpha \left(\Delta t\text{'}\right)\right]\left(\Delta t\right)+\frac{1}{2}\alpha {\left(\Delta t\right)}^2\\ =\left[\left(180{\text{°/s}}^{\text{2}}\right)\left(1\text{s}\right)\left(\text{1}\text{s}\right)\right]+\left[\frac{1}{2}\left(180{\text{°/s}}^{\text{2}}\right){\left(1\text{s}\right)}^2\right]\\ =270°\end{array}$

Thus, the angle of rotation is $\underset{\_}{270°}$.

(b)

To determine

The angle of rotation.

Answer to Problem 48P

The angle of rotation is $\underset{\_}{450°}$.

Explanation of Solution

Write the equation of angle from angular kinematics.

$\left(\Delta \theta \right)=\omega \left(\Delta t\right)+\frac{1}{2}\alpha {\left(\Delta t\right)}^2$ (I)

Here, the angle is $\left(\Delta \theta \right)$, the angular speed is $\omega$, the time is $\left(\Delta t\right)$, the angular acceleration is $\alpha$.

Conclusion:

Substitute, $90°$ for $\left(\Delta \theta \right)$, $0$ for $\omega$ , $1\text{s}$ for $\left(\Delta t\right)$ in equation (I)

$\begin{array}{l}\left(90\right)=\left[\left(0\right)\left(\Delta t\right)\right]+\frac{1}{2}\alpha {\left(1\text{s}\right)}^2\\ \alpha =180°{\text{/s}}^{\text{2}}\end{array}$

Substitute, $180°{\text{/s}}^{\text{2}}$ for $\alpha$, $\alpha \left(\Delta t\text{'}\right)$ for $\omega$ , $1\text{s}$ for $\left(\Delta t\right)$, $2\text{s}$ for $\left(\Delta t\text{'}\right)$ in equation (I)

$\begin{array}{c}\left(\Delta \theta \right)=\left[\alpha \left(\Delta t\text{'}\right)\right]\left(\Delta t\right)+\frac{1}{2}\alpha {\left(\Delta t\right)}^2\\ =\left[\left(180{\text{°/s}}^{\text{2}}\right)\left(2\text{s}\right)\left(\text{1}\text{s}\right)\right]+\left[\frac{1}{2}\left(180{\text{°/s}}^{\text{2}}\right){\left(1\text{s}\right)}^2\right]\\ =450°\end{array}$

Thus, the angle of rotation is $\underset{\_}{450°}$.