91Ó°ÊÓ

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

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

The angular displacement, $\mathrm{θ}=60\text{ r±ð±¹}$

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:

$\mathrm{Ó\neg }={\mathrm{Ó\neg }}_0+\mathrm{Î\pm ³Ù}$

$\mathrm{θ}=\mathrm{Ó\neg t}−\frac{1}{2}{\mathrm{Î\pm ³Ù}}^2$

${\mathrm{Ó\neg }}^2={\mathrm{Ó\neg }}_0^2+2\mathrm{Î\pm θ}$

$\mathrm{θ}=\frac{1}{2}(\mathrm{Ó\neg }+{\mathrm{Ó\neg }}_0)\mathrm{t}$

Consider the kinematic equation for angular motion as:

${\mathrm{Ó\neg }}^2={\mathrm{Ó\neg }}_0^2+2\mathrm{Î\pm θ}$

By rearranging it for angular acceleration

$\mathrm{Î\pm }=\frac{{\mathrm{Ó\neg }}^2-{\mathrm{Ó\neg }}_0^2}{2\mathrm{θ}}$

Substitute the values and solve as:

$\begin{array}{c}\mathrm{Î\pm }=\frac{{15}^2−{10}^2}{2×60}\\ =1.04\frac{\text{rev}}{{\text{s}}^2}\end{array}$

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

Consider the formula for the time required as:

$\mathrm{t}=\frac{2\mathrm{θ}}{\mathrm{Ó\neg }+{\mathrm{Ó\neg }}_0}$

Solve further as:

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

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

Consider the initial angular speed as zero.

So, using the kinematic equation:

$\mathrm{Ó\neg }={\mathrm{Ó\neg }}_0+\mathrm{Î\pm ³Ù}$

$\mathrm{t}=\frac{\mathrm{Ó\neg }−{\mathrm{Ó\neg }}_0}{\mathrm{Î\pm }}$

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 $10\text{ }\frac{\text{rev}}{\text{s}}$ angular speeds is $9.6\text{s}$.

Consider the formula for the number of revolutions:

$\begin{array}{l}{\mathrm{Ó\neg }}^2={\mathrm{Ó\neg }}_0^2+2\mathrm{Î\pm θ}\\ \mathrm{θ}=\frac{{\mathrm{Ó\neg }}^2−{\mathrm{Ó\neg }}_0^2}{2\mathrm{Î\pm }}\end{array}$

Substitute the values and solve as:

$\begin{array}{c}\mathrm{θ}=\frac{{10}^2−0}{2×1.04}\\ =48\text{ r±ð±¹}\end{array}$

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