Question

A disk rotates about its central axis starting from rest and accelerates with constant angular acceleration. At one time it is rotating at $\mathbf{10}\mathbf{}\frac{\mathbf{rev}}{\mathbf{s}}\mathbf{;}\mathbf{}\mathbf{60}\mathbf{}$revolutions later, its angular speed is $\mathbf{15}\mathbf{}\frac{\mathbf{rev}}{\mathbf{s}}$.

Calculate

(a) the angular acceleration,

(b) the time required to complete the $\mathbf{\text{60}}$ revolutions,

(c) the time required to reach the $\mathbf{10}\mathbf{}\frac{\mathbf{rev}}{\mathbf{s}}$angular speed, and

(d) the number of revolutions from rest until the time the disk reaches the $\mathbf{10}\mathbf{}\frac{\mathbf{rev}}{\mathbf{s}}$ angular speed

Short answer

1. Angular acceleration ofthedisk is $1.04\frac{\text{rev}}{{\text{s}}^2}$
2. The time required to complete 60 revolutions is $4.8\text{s}$
3. The time required to reach the 10 rev/s angular speed is $9.6\text{s}$
4. The number of revolutions from rest until the disk speed is 10 $\frac{\text{rev}}{\text{s}}$, $\mathrm{\theta }=48\text{rev}$

Step-by-step solution

Listing the given quantities

The initial angular speed of a disk,${\mathrm{\omega }}_0=10\frac{\text{rev}}{\text{s}}$

The final angular speed of the engine,$\mathrm{\omega }=15\frac{\text{rev}}{\text{s}}$

The angular displacement, $\mathrm{\theta }=60\text{\hspace{0.17em}rev}$

Understanding the kinematic equations

Use the kinematic equation for constant angular acceleration to calculate the time and angular acceleration. For the calculation of the number of revolutions, use the initial angular velocity as zero, as it starts from rest.

Consider the formulas:

$\mathrm{\omega }={\mathrm{\omega }}_0+\mathrm{\alpha t}$

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

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

$\mathrm{\theta }=\frac{1}{2}(\mathrm{\omega }+{\mathrm{\omega }}_0)\mathrm{t}$

(a) Calculate the angular acceleration

Consider the kinematic equation for angular motion as:

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

By rearranging it for angular acceleration

$\mathrm{\alpha }=\frac{{\mathrm{\omega }}^2-{\mathrm{\omega }}_0^2}{2\mathrm{\theta }}$

Substitute the values and solve as:

$\begin{array}{c}\mathrm{\alpha }=\frac{{15}^2-{10}^2}{2\times 60}\\ =1.04\frac{\text{rev}}{{\text{s}}^2}\end{array}$

Angular acceleration of the disk is $1.04\frac{\text{rev}}{{\text{s}}^2}$.

(b) Calculate the time required to complete 60 revolutions

Consider the formula for the time required as:

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

Solve further as:

$\begin{array}{c}\mathrm{t}=\frac{2\times 60}{15+10}\\ =4.8\text{s}\end{array}$

The time required to complete 60 revolutions is $4.8\text{s}$.

(c) Calculate the time required to reach 10  revs angular speed

Consider the initial angular speed as zero.

So, using the kinematic equation:

$\mathrm{\omega }={\mathrm{\omega }}_0+\mathrm{\alpha t}$

$\mathrm{t}=\frac{\mathrm{\omega }-{\mathrm{\omega }}_0}{\mathrm{\alpha }}$

Solve further as:

$\begin{array}{c}\mathrm{t}=\frac{10-0}{1.04}\\ =9.6\text{s}\end{array}$

The time required to reach the $10\frac{\text{rev}}{\text{s}}$ angular speeds is $9.6\text{s}$.

(d) Calculate the number of revolutions

Consider the formula for the number of revolutions:

$\begin{array}{l}{\mathrm{\omega }}^2={\mathrm{\omega }}_0^2+2\mathrm{\alpha \theta }\\ \mathrm{\theta }=\frac{{\mathrm{\omega }}^2-{\mathrm{\omega }}_0^2}{2\mathrm{\alpha }}\end{array}$

Substitute the values and solve as:

$\begin{array}{c}\mathrm{\theta }=\frac{{10}^2-0}{2\times 1.04}\\ =48\text{\hspace{0.17em}rev}\end{array}$

The number of revolutions from rest until the disk speed is $10\frac{\text{rev}}{\text{s}}=48\text{rev}$.