Question

A merry-go-round rotates from rest with an angular acceleration of $\mathbf{1}\mathbf{.}\mathbf{50}\mathbf{}\mathbf{rad}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}\mathbf{}$. How long does it take to rotate through (a) the firstrole="math" localid="1660899350963" $\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{rev}\mathbf{}$ and (b) the next $\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{rev}\mathbf{}$?

Short answer

1. The time required for the first 2.00 revolutions,$\mathrm{t}=4.09\text{s}$
2. The time required for the next 2.00 revolutions,${\mathrm{t}}_2=1.7\text{\hspace{0.17em}s}$

Step-by-step solution

Listing the given quantities

1. The angular acceleration of a merry-go-round,$\mathrm{\alpha }=1.5\frac{\text{rad}}{{\text{s}}^2}$
2. The angular displacement of interval, $\mathrm{\theta }=2.00\text{\hspace{0.17em}rev}=12.6\text{rad}$.

Understanding the kinematic equations

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:

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

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

(a) Calculate the time required for first two revolutions

The kinematic equation for angular motion is obtained as:

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

Substitute the values and solve as:

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

Solve for the time required.

$\begin{array}{c}\mathrm{t}=\sqrt{\frac{2\times 12.6}{1.5}}\\ =4.09\text{\hspace{0.17em}s}\end{array}$

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

(b) Calculate the time required for next two revolutions 

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

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

Substitute the values and solve as:

$\begin{array}{c}\mathrm{\omega }=0+1.5\times 4.09\\ =6.14\frac{\text{rad}}{\text{s}}\end{array}$

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

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

Substitute the values and solve as:

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

Consider the quadratic equation with variable ‘t.’ it gives two roots as:$\mathrm{t}=-9.9\text{s}\text{\hspace{0.17em}and}\mathrm{t}=1.7\text{s}$

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

The time required forthenext 2.00 revolutions,${\mathrm{t}}_2=1.7\text{\hspace{0.17em}s}$