91Ó°ÊÓ

1. The angular acceleration of a merry-go-round,$\mathrm{Î\pm }=1.5\text{ }\frac{\text{rad}}{{\text{s}}^2}$
2. The angular displacement of interval, $\mathrm{θ}=2.00\text{ r±ð±¹}=12.6\text{ }\text{rad}$.

Use the kinematic equation for constant angular acceleration to calculate the time forthefirst 2.00 revolutions. To calculate the time forthenext 2.00 revolutions, first calculate the final angular velocity of first 2.00 revolutions, and use it astheinitial angular velocity for the next 2.00 revolutions.

Formula:

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

$\mathrm{θ}={\mathrm{Ó\neg }}_0\mathrm{t}+\frac{1}{2}{\mathrm{Î\pm ³Ù}}^2$

The kinematic equation for angular motion is obtained as:

$\mathrm{θ}={\mathrm{Ó\neg }}_0\mathrm{t}+\frac{1}{2}{\mathrm{Î\pm ³Ù}}^2$

Substitute the values and solve as:

$12.6=0+\frac{1}{2}×1.5×{\mathrm{t}}^2$

Solve for the time required.

$\begin{array}{c}\mathrm{t}=\sqrt{\frac{2×12.6}{1.5}}\\ =4.09\text{ s}\end{array}$

The time required for the first 2.00 revolutions,$\mathrm{t}=4.09\text{ }\text{s}$

Calculate the time for the next two revolutions, and need the angular velocity after first 2.00 revolutions. Calculate it by using the kinematic equation

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

Substitute the values and solve as:

$\begin{array}{c}\mathrm{Ó\neg }=0+1.5×4.09\\ =6.14\text{ }\frac{\text{rad}}{\text{s}}\end{array}$

Use this velocity as the initial angular velocity for the next 2.00 revolution.

$\mathrm{θ}={\mathrm{Ó\neg }}_0\mathrm{t}+\frac{1}{2}{\mathrm{Î\pm ³Ù}}^2$

Substitute the values and solve as:

$12.6=6.14×\mathrm{t}+\frac{1}{2}×1.5×{\mathrm{t}}^2$

Consider the quadratic equation with variable ‘t.’ it gives two roots as:$\mathrm{t}=−9.9\text{s}\text{ a²Ô»å}\mathrm{t}=1.7\text{s}$

. By considering the positive root the obtained value is ${\mathrm{t}}_2=1.7\text{s}$.

The time required forthenext 2.00 revolutions,${\mathrm{t}}_2=1.7\text{ s}$