Starting from rest, a disk rotates about its central axis with constant angular acceleration. In $\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{s}$, it rotates $\mathbf{25}\mathbf{}\mathbf{rad}$. During that time, what are the magnitudes of

(a) the angular acceleration and

(b) the average angular velocity?

(c) What is the instantaneous angular velocity of the disk at the end of the $\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{s}$?

(d) With the angular acceleration unchanged, through what additional angle will the disk turn during the next $\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{s}$?

The angular displacement for 5.0s is, $\mathrm{\theta }=25\text{\hspace{0.17em}rad}$

Use the kinematic equation for constant angular acceleration to solve for angular acceleration, instantaneous angular velocity, and angular displacement. Calculate the average angular velocity using the total displacement and total time.

Consider the required formula:

1. $\mathrm{\omega }={\mathrm{\omega }}_0+\mathrm{\alpha t}$
2. $\mathrm{\theta }={\mathrm{\omega }}_0\mathrm{t}+\frac{1}{2}{\mathrm{\alpha t}}^2$
3. ${\mathrm{\omega }}_{\mathrm{avg}}=\frac{\mathrm{\Delta \theta }}{\mathrm{\Delta t}}$

Consider the formula for the angular acceleration as:

$\mathrm{\theta }={\mathrm{\omega }}_0\mathrm{t}+\frac{1}{2}{\mathrm{\alpha t}}^2$

Substitute the values and solve as:

$\begin{array}{c}25=0\times 5.0+\frac{1}{2}\times \mathrm{\alpha }\times {5.0}^2\\ \mathrm{\alpha }=2.0\frac{\text{rad}}{{\text{s}}^2}\end{array}$

The magnitude of angular acceleration, $\mathrm{\alpha }=2.0\frac{\text{rad}}{{\text{s}}^2}$.

The equation for average velocity is

$\begin{array}{c}{\mathrm{\omega }}_{\mathrm{avg}}=\frac{\mathrm{\Delta \theta }}{\mathrm{\Delta t}}\\ =\frac{25}{5.0}\\ =5.0\frac{\text{rad}}{\text{s}}\end{array}$

The magnitude of average angular velocity is $5.0\frac{\text{rad}}{\text{s}}$.

The instantaneous angular velocity can be calculated using the kinematic equation for final angular velocity as$\mathrm{\omega }={\mathrm{\omega }}_0+\mathrm{\alpha t}$

Substitute the values and solve as:

The instantaneous angular velocity at 5.0 s, $\mathrm{\omega }=10\frac{\text{rad}}{\text{s}}$

The angular displacement for further 5.0 s from $\mathrm{t}=$5.0 s to $\mathrm{t}=$10 s is

$\mathrm{\theta }={\mathrm{\omega }}_0\mathrm{t}+\frac{1}{2}{\mathrm{\alpha t}}^2$

Solve further as:

$\begin{array}{c}\mathrm{\theta }=10\times 5.0+\frac{1}{2}\times 2.0\times {5.0}^2\\ =75\text{\hspace{0.17em}rad}\end{array}$

Here, ${\mathrm{\omega }}_0=10\frac{\text{rad}}{\text{s}}$ and$\mathrm{\Delta }\text{t}=5.0\text{s}$.

The additional angular displacement from 5.0 s to 10.0 s, $\mathrm{\theta }=75\text{rad}$.