CP A 12.0kgmass, fastened to the end of an aluminium wire with an unstretched length of, is whirled in a vertical circle with a constant angular speed of. The cross-sectional area of the wire is 120rev/min. Calculate the elongation of the wire when the mass is

1. at the lowest point of the path and
2. at the highest point of its path.

Stress, which quantifies the force per unit area applied to a material, is a physical quantity. Which is given by,

$\mathit{\sigma }\mathbf{=}\frac{\mathbf{F}}{\mathbf{A}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$ (1)

Where$\mathit{\sigma }$ is stress, is force and is area.

Mass is given by m = 12.0 kg

Length of aluminum wire is r = 0.7 m
Now, angular speed is r = 120$\frac{\mathrm{rev}}{\min }$

role="math" localid="1668141796113" $\begin{array}{r}\omega =120\times \frac{2\pi }{60}\frac{\mathrm{rad}}{\sec }\\ =4\pi \frac{\mathrm{rad}}{\sec }\end{array}$

Area of wire is A = 0.014${\mathrm{cm}}^2$

(a)

The force working on aluminium rod will be due to gravity and angular acceleration will be

$F=mg\pm m{\omega }^{2}r$ (2)

Now, we have for lowest point. So, consider,

$\begin{array}{r}F=mg-m{\omega }^{2}r\\ =\left(12\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)-\left(12\mathrm{kg}\right)(4\pi \mathrm{rad}/\sec {)}^2(0.7\mathrm{m})\\ =1208\mathrm{N}\end{array}$

Now, from equation (1),

$\begin{array}{r}\sigma =\frac{F}{A}\\ =\frac{1208\mathrm{N}}{1.4\times {10}^{-6}{\mathrm{m}}^2}\\ =862.8\mathrm{MPa}\\ \approx 863\mathrm{MPa}\end{array}$

Now, we know that young modulus of aluminium rod is 70000MP a

Now, elongation is given by,

$\begin{array}{r}\delta =\frac{\sigma }{Y}\times I\\ =\frac{863\mathrm{MPa}}{7000\mathrm{MPa}}\times 70\mathrm{cm}\\ =0.86\mathrm{cm}\end{array}$

The elongation of the wire when the mass is at the lowest point of the path is 0.86 cm.

(b)

From equation (2) we have$\mathrm{F}=\mathrm{mg}\pm {\mathrm{m\omega }}^2\mathrm{r}$

Now, we have for lowest point. So, consider,

$\begin{array}{r}\mathrm{F}=\mathrm{mg}+{\mathrm{m\omega }}^2\mathrm{r}\\ =\left(12\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)+\left(12\mathrm{kg}\right)(4\mathrm{\pi rad}/\sec {)}^2(0.7\mathrm{m})\\ =1443\mathrm{N}\\ \mathrm{Now},\mathrm{from}\mathrm{equation}\left(1\right),\\ \mathrm{\sigma }=\frac{\mathrm{F}}{\mathrm{A}}\\ =\frac{1443\mathrm{N}}{1.4\times {10}^{-6}{\mathrm{m}}^2}\\ =1030.7\mathrm{MPa}\\ \approx 1031\mathrm{MPa}\end{array}$

Now, we know that young modulus of aluminium rod is 70000MP a

Now, elongation is given by,

$\begin{array}{r}\delta =\frac{\sigma }{Y}\times I\\ =\frac{1031\mathrm{MPa}}{7000\mathrm{MPa}}\times 70\mathrm{cm}\\ =1\mathrm{cm}\end{array}$

The elongation of the wire when the mass is at the highest point of the path is 1 cm