You are to use a long, thin wire to build a pendulum in a science museum. The wire has an unstretched length of 22.00mand a circular cross-section of diameter 0.860 mm; it is made of an alloy that has a large breaking stress. One end of the wire will be attached to the ceiling, and a 9.50 kgmetal sphere will be attached to the other end. As the pendulum swings back and forth, the wire’s maximum angular displacement from the vertical will be$\mathbf{36}\mathbf{.}\mathbf{0}\mathbf{°}$. You must determine the maximum amount the wire will stretch during this motion. So, before you attach the metal sphere, you suspend a test mass (mass m) from the wire’s lower end. You then measure the increase in lengthof the wire for several different test masses. Given figure, a graph ofversusshows the results and the straight line that gives the best fit to the data. The equation for this line is$\mathbf{∆}\mathbf{l}\mathbf{=}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{422}\mathbf{m}\mathbf{/}\mathbf{kg}\mathbf{)}\mathbf{m}$.

1. Assume that g = 9.8 m/localid="1668148375172" ${\mathbf{s}}^{\mathbf{2}}$, and use given figure to calculate Young’s modulusfor this wire.
2. You remove the test masses, attach the 9.50 kgsphere, and release the sphere from rest, with the wire displaced by 36.localid="1668148446283" $\mathbf{0}\mathbf{°}$. Calculate the amount the wire will stretch as it swings through the vertical. Ignore air resistance.

Consider the formula for the energy conservation:

${\mathbf{U}}_{\mathbf{1}}\mathbf{+}{\mathbf{K}}_{\mathbf{1}}\mathbf{+}{\mathbf{E}}_{\mathbf{other}}\mathbf{=}{\mathbf{U}}_{\mathbf{2}}\mathbf{+}{\mathbf{K}}_{\mathbf{2}}$ (1)

Here, ${\mathit{U}}_{\mathbf{1}}\mathbf{,}{\mathit{U}}_{\mathbf{2}}$is initial and final potential energy respectively.

The${\mathit{K}}_{\mathbf{1}}$ and${\mathit{K}}_{\mathbf{2}}$ is initial and final kinetic energy respectively.

${\mathbf{E}}_{\mathbf{o}\mathbf{t}\mathbf{h}\mathbf{e}\mathbf{r}}$is other energies.

Here we have, length of wire is${\mathrm{I}}_0=22.0\mathrm{m}$

Diameter of wire is d = 0.860 mm .

Mass of sphere is m = 9.5 kg

role="math" $$\begin{gathered} \mathrm{\theta }=36.0° \\ ∆\mathrm{l}=\left(0.422\mathrm{mm}/\mathrm{kg}\right)\mathrm{m} \end{gathered}$$

(a)

Now, cross-sectional area of wire is,

$\begin{array}{r}\begin{array}{rl}\mathrm{A}& =\mathrm{\pi }{\left(0.430\times {10}^{-3}\mathrm{m}\right)}^2\\ & =5.808\times {10}^{-6}{\mathrm{m}}^2\\ & \mathrm{Also},\mathrm{we}\mathrm{know}\mathrm{that}\mathrm{Young}’\mathrm{s}\mathrm{modulus}\mathrm{is}\mathrm{given}\mathrm{by},\end{array}\\ \mathrm{Y}=\frac{{\mathrm{FI}}_0}{\mathrm{A\Delta I}}\end{array}$ (2)

In our case, F = w = mg

Now, equation (2) becomes,

$\begin{array}{r}Y=\frac{mg{I}_0}{A\mathrm{\Delta }l}\\ Y=\frac{g{I}_0}{A\left(\frac{\mathrm{\Delta }l}{m}\right)}\end{array}$ (3)

Now, we have,

$\begin{array}{r}\mathrm{\Delta I}=\left(0.422\frac{\mathrm{mm}}{\mathrm{kg}}\right)\mathrm{m}\\ \frac{\mathrm{\Delta l}}{\mathrm{m}}=0.422\frac{\mathrm{mm}}{\mathrm{kg}}\\ \mathrm{By}\mathrm{substituting}\mathrm{the}\mathrm{numerical}\mathrm{values}\mathrm{in}\mathrm{equation}\left(3\right),\mathrm{we}\mathrm{get},\\ \mathrm{Y}=\frac{\left(9.8\frac{\mathrm{m}}{{\mathrm{s}}^2}\right)(22.0\mathrm{m})}{\left(5.808\times {10}^{-6}{\mathrm{m}}^2\right)\left(0.422\frac{\mathrm{mm}}{\mathrm{kg}}\right)}\\ =8.795\times {10}^{11}\mathrm{Pa}\end{array}$

Hence, the Young’s modulus of wire is$8.795\times {10}^{11}\mathrm{Pa}$

(b)

Consider the diagram for the condition:

From equation (1), we have,

$U_1+K_1+E_{other}=U_2+K_2$

Take position 2 as zero gravitational potential energy level.

Here, at rest kinetic energy is zero. So, $K_1=0$

Final potential energy is zero. So,${\mathrm{U}}_2=0$.

Here, no energy losses. So,$E_{other}$= 0

Initial potential energy$U_1=mg{l}_0(1-\cos \theta )$

Final kinetic energy is given by,

$K_1=\frac{1}{2}m{v}_2^2$

S0, equation (1) becomes

$\begin{array}{r}{\mathrm{mgI}}_0(1-\cos \mathrm{\theta })+0+0=0+\frac{1}{2}{\mathrm{mv}}_2^2\\ {\mathrm{v}}_2=\sqrt{2{\mathrm{gl}}_0(1-\cos \mathrm{\theta })}\\ \mathrm{Simplify}\mathrm{the}\mathrm{expression}\mathrm{for}\mathrm{angular}\mathrm{momentum}\mathrm{as}:\\ {\mathrm{v}}_2={\mathrm{\omega }}_2{\mathrm{I}}_0\\ {\mathrm{\omega }}_2=\frac{{\mathrm{v}}_2}{{\mathrm{I}}_0}\\ {\mathrm{\omega }}_2=\frac{\sqrt{2{\mathrm{gl}}_0(1-\cos \mathrm{\theta })}}{{\mathrm{I}}_0}\end{array}$

Substitute all numerical values in above equation we get,

$\begin{array}{r}{\omega }_2=\frac{\sqrt{2\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(22.0\mathrm{m}\right)\left(1-\cos {36.0}^{\circ }\right)}}{\left(22.0\mathrm{m}\right)}\\ =3.743\frac{\mathrm{rad}}{\mathrm{s}}\end{array}$

Consdier the free body diagram”

Now, by newton’s second law,

$\begin{array}{r}\sum {\mathrm{F}}_{\mathrm{y}}={\mathrm{ma}}_{\text{centripetal}}\\ \mathrm{T}-\mathrm{W}={\mathrm{ml}}_0{\mathrm{\omega }}_2^2\\ \mathrm{T}=\mathrm{W}+{\mathrm{ml}}_0{\mathrm{\omega }}_2^2\\ \mathrm{T}=\mathrm{mg}+{\mathrm{ml}}_0{\mathrm{\omega }}_2^2\\ \mathrm{By}\mathrm{substituting}\mathrm{all}\mathrm{the}\mathrm{numerical}\mathrm{values}\mathrm{in}\mathrm{above}\mathrm{equation}\mathrm{and}\mathrm{solve}:\\ \mathrm{T}=(9.50\mathrm{kg})\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)+(95.0\mathrm{kg})(22.0\mathrm{m})(3.743\mathrm{rad}/\mathrm{s}{)}^2\\ =2959.1\mathrm{N}\end{array}$

Now, by newton’s third law it has same magnitude and opposite direction of the force exerted by sphere on the wire.

Now, by equation 2 we have,

$Y=\frac{F{l}_0}{A∆l}$

In our case, T = F . So above equation becomes,

$\begin{array}{r}\mathrm{Y}=\frac{{\mathrm{TI}}_0}{\mathrm{A\Delta I}}\\ \mathrm{\Delta I}=\frac{{\mathrm{TI}}_0}{\mathrm{AY}}\\ \mathrm{Now},\mathrm{substitute}\mathrm{the}\mathrm{values}\mathrm{in}\mathrm{above}\mathrm{equation}.\mathrm{So},\mathrm{we}\mathrm{get},\\ \mathrm{\Delta I}=\frac{(2959.1\mathrm{N})(22.0\mathrm{m})}{\left(5.808\times {10}^{-6}{\mathrm{m}}^2\right)\left(8.795\times {10}^{11}\mathrm{Pa}\right)}\\ =0.127\mathrm{m}\end{array}$

Hence, the wire will stretch 0.127 m .