CP An amusement park ride consists of airplane-shaped cars attached to steel rods (Given figure). Each rod has a length of 15.00 mand a cross-sectional area of 8.00${\mathbf{cm}}^{\mathbf{2}}$.

1. How much is each rod stretched when it is vertical and the ride is at rest? (Assume that each car plus two people seated in it has a total weight of 1900 N.)
2. When operating, the ride has a maximum angular speed of $\mathbf{12}\frac{\mathbf{rev}}{\mathbf{min}}$. How much is the rod stretched then?

Consider the formula for deformation length.

$\mathbf{∆}\mathbf{l}\mathbf{=}\frac{{\mathbf{Wl}}_{\mathbf{0}}}{\mathbf{AY}}$ (1)

Here, $\mathbf{∆}\mathbf{l}$is a deformation of length, ${\mathit{l}}_{\mathbf{0}}$is identical length, Wis weight, Ais cross sectional area and Yis Young’s modulus of object respectively.

Consider the Young’s modulus of rod is $Y=20\times {10}^{10}Pa$ (For steel rod)

Cross-sectional area is $A=8\times {10}^{-4}{m}^2$

Weight is W = 1900 N

Identical length$l_0=15m$

Angular speed is $12\frac{\mathrm{rev}}{\min }$

$\begin{array}{r}\omega =12\times \frac{2\pi }{60}\frac{\mathrm{rad}}{\sec }\\ =0.4\pi \frac{\mathrm{rad}}{\sec }\end{array}$

(a)

Substitute the values in equation (1) and solve as:

$\begin{array}{r}\mathrm{\Delta }l=\frac{\left(1900\mathrm{N}\right)\left(15\mathrm{m}\right)}{\left(20\times {10}^{10}\mathrm{Pa}\right)\left(8\times {10}^{-4}{\mathrm{m}}^2\right)}\\ =1.7812\times {10}^{-4}\mathrm{m}\end{array}$

Hence, when it is vertical and the ride is at rest rod stretched at $1.7812\times {10}^{-4}\mathrm{m}$.

(b)

Free body diagram is given by,

Now, by applying Newton’s second law to determine net force,

$F\sin \theta =mr{\omega }^2$

Now,$\mathrm{r}={\mathrm{l}}_0\mathrm{sin\theta }\mathrm{a}\mathrm{and}\mathrm{m}=\frac{\mathrm{W}}{\mathrm{g}}$and

Derive the expression for the force as:

$\begin{array}{r}\mathrm{Fsin}\mathrm{\theta }=\left(\frac{\mathrm{W}}{\mathrm{g}}\right){\mathrm{I}}_0\sin {\mathrm{\theta \omega }}^2\\ \mathrm{F}=\frac{\mathrm{W}}{\mathrm{g}}{\mathrm{I}}_0{\mathrm{\omega }}^2\\ \mathrm{Now},\mathrm{by}\mathrm{substituting}\mathrm{all}\mathrm{the}\mathrm{numerical}\mathrm{values}\mathrm{in}\mathrm{above}\mathrm{equation},\\ \begin{array}{rl}\mathrm{F}& =\frac{\left(1900\mathrm{N}\right)}{\left(9.80\frac{\mathrm{m}}{{\mathrm{s}}^2}\right)}\left(15\mathrm{m}\right){\left(0.4\mathrm{\pi }\frac{\mathrm{rad}}{\sec }\right)}^2\\ & =4592.39\mathrm{N}\end{array}\end{array}$

Substitute the values in the equation for deformation and solve as:

$\begin{array}{r}\mathrm{\Delta }I=\frac{\left(4592.39\mathrm{N}\right)\left(15\mathrm{m}\right)}{\left(20\times {10}^{10}\mathrm{Pa}\right)\left(8\times {10}^{-4}{\mathrm{m}}^2\right)}\\ =4.3054\times {10}^{-4}\mathrm{m}\end{array}$

Hence, when the ride has a maximum angular speed of 12rev/min the rod is stretched at$4.3054\times {10}^{-4}m$.