Question

George Washington Gale Ferris, Jr., a civil engineering graduate from Rensselaer Polytechnic Institute, built the original Ferris wheel for the$\mathbf{1893}$ World’s Columbian Exposition in Chicago. The wheel, an astounding engineering construction at the time, carried $\mathbf{\text{36}}$wooden cars, each holding up to passengers, around a circle $\mathbf{\text{76}}\mathbf{}\mathbf{\text{\hspace{0.17em}m}}\mathbf{}$in diameter. The cars were loaded $\mathbf{6}$at a time, and once all $\mathbf{36}$cars were full, the wheel made a complete rotation at constant angular speed in about$\mathbf{2}\mathbf{}\mathbf{\text{\hspace{0.17em}min}}$ . Estimate the amount of work that was required of the machinery to rotate the passengers alone.

Short answer

Amount of work required of the machinery to rotate the passengers alone is$2.0\times {10}^5\text{J}\le \text{W}\le 4.0\times {10}^5\text{J}$

Step-by-step solution

 Step 1: Given

i) Total cars are 36.

ii) Each car carries 60 passengers.

iii) Diameter of circle is$\text{d}=76\text{\hspace{0.17em}m}$.

iv) Time is $\text{T}=2\text{\hspace{0.17em}min}$.

Understanding the concept

Use the concept of work done related to change in angular kinetic energy.First, find the angular velocity and moment of inertia.Thenplugging it in the equation, find the work done.Also, find the range of work done by plugging the average value of mass of the person.

Formulae:

$\text{W}=\frac{1}{2}\text{I}{\omega }_f^2-\frac{1}{2}\text{I}{\omega }_i^2$

$\text{I}={\text{MR}}^2$

$\omega =\frac{2\pi }{\text{T}}$

Calculate the amount of work that was required of the machinery to rotate the passengers alone

Amount of work required of the machinery to rotate the passengers alone:

First, find the total number of people in 36 cars.

$36\times 60=2160$

Angular velocity can be found from time

$\begin{array}{c}\omega =\frac{2\pi }{\text{T}}\\ =\frac{2\times 3.14}{120}\\ =0.05\text{\hspace{0.17em}rad}/\text{s}\end{array}$

Initiallytheangular velocity is zero, so we can write,

role="math" localid="1661513224896" $\begin{array}{c}\text{W}=\frac{1}{2}\text{I}{\omega }_f^2-\frac{1}{2}\text{I}{\left(0\right)}^2\\ =\frac{1}{2}\text{I}{\omega }_f^2\end{array}$

Radius of circle is,

$\begin{array}{c}\text{R}=\frac{d}{2}\\ =\frac{76}{2}\\ =38\text{\hspace{0.17em}m}\end{array}$

Moment of inertia of the wheel is$\text{I}={\text{MR}}^2$

Plugging these values in equation of W, we get

$\begin{array}{c}\text{W}=\frac{1}{2}\text{M}{\left(38\right)}^2{\left(0.05\right)}^2\\ =1.805\text{\hspace{0.17em}M}\end{array}$

This is work done for one person.Forall people, we can write,

$\text{W}=2.0\text{\hspace{0.17em}M}\times 2160$

Average mass of a person varies from 50kg to 100kg. We get the range of work done:

$2.0\times 50\times 2160\le \text{W}\le 2.0\times 100\times 2160$

$2.0\times {10}^5\text{J}\le \text{W}\le 4.0\times {10}^5\text{J}$