Question

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 $\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{s}$?

Short answer

1. The magnitude of angular acceleration, $\mathrm{\alpha }=2.0\frac{\text{rad}}{{\text{s}}^2}$
2. The magnitude of average angular velocity, ${\mathrm{\omega }}_{\mathrm{avg}}=5.0\frac{\text{rad}}{\text{s}}$
3. The instantaneous angular velocity at 5.0 s, $\mathrm{\omega }=10\frac{\text{rad}}{\text{s}}$
4. The additional angular displacement from 5.0 s to 10.0 s, $\mathrm{\theta }=75\text{rad}$

Step-by-step solution

Listing the given quantities

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

Determine the concept of velocity and number of revolutions

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. ${\mathrm{\omega }}_{\mathrm{avg}}=\frac{\mathrm{\Delta \theta }}{\mathrm{\Delta t}}$

(a) Calculate the angular acceleration

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, $\mathrm{\alpha }=2.0\frac{\text{rad}}{{\text{s}}^2}$.

(b) Calculate the average angular velocity as follows:

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 $5.0\frac{\text{rad}}{\text{s}}$.

(c) Calculate the instantaneous angular velocity

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:

$\begin{array}{c}\omega =0+2.0\times 5.0\\ =10\frac{\text{rad}}{\text{s}}\end{array}$

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

(d) Calculate the angular displacement

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, $\mathrm{\theta }=75\text{rad}$.