91Ó°ÊÓ

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

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

Where$\mathit{σ}$ 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" $\begin{array}{r}\mathrm{Ó\neg }=120×\frac{2\mathrm{Ï€}}{60}\frac{\mathrm{rad}}{\sec }\\ =4\mathrm{Ï€}\frac{\mathrm{rad}}{\sec }\end{array}$

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

$F=mgÂ\pm m{\mathrm{Ó\neg }}^{2}r$ (2)

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

$\begin{array}{r}F=mg−m{\mathrm{Ó\neg }}^{2}r\\ =\left(12\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)−\left(12\mathrm{kg}\right)(4\mathrm{Ï€}\mathrm{rad}/\sec {)}^2(0.7\mathrm{m})\\ =1208\mathrm{N}\end{array}$

Now, from equation (1),

$\begin{array}{r}\mathrm{σ}=\frac{F}{A}\\ =\frac{1208\mathrm{N}}{1.4×{10}^{−6}{\mathrm{m}}^2}\\ =862.8\mathrm{MPa}\\ ≈863\mathrm{MPa}\end{array}$

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

Now, elongation is given by,

$\begin{array}{r}\mathrm{δ}=\frac{\mathrm{σ}}{Y}×I\\ =\frac{863\mathrm{MPa}}{7000\mathrm{MPa}}×70\mathrm{cm}\\ =0.86\mathrm{cm}\end{array}$

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$\mathrm{F}=\mathrm{mg}Â\pm {\mathrm{mÓ\neg }}^2\mathrm{r}$

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

$\begin{array}{r}\mathrm{F}=\mathrm{mg}+{\mathrm{mÓ\neg }}^2\mathrm{r}\\ =\left(12\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)+\left(12\mathrm{kg}\right)(4\mathrm{Ï€°ù²¹»å}/\sec {)}^2(0.7\mathrm{m})\\ =1443\mathrm{N}\\ \mathrm{Now},\mathrm{from}\mathrm{equation}\left(1\right),\\ \mathrm{σ}=\frac{\mathrm{F}}{\mathrm{A}}\\ =\frac{1443\mathrm{N}}{1.4×{10}^{−6}{\mathrm{m}}^2}\\ =1030.7\mathrm{MPa}\\ ≈1031\mathrm{MPa}\end{array}$

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

Now, elongation is given by,

$\begin{array}{r}\mathrm{δ}=\frac{\mathrm{σ}}{Y}×I\\ =\frac{1031\mathrm{MPa}}{7000\mathrm{MPa}}×70\mathrm{cm}\\ =1\mathrm{cm}\end{array}$

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