91影视

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" $$\begin{gathered} \mathrm{胃}=36.0掳 \\ 鈭�\mathrm{l}=\left(0.422\mathrm{mm}/\mathrm{kg}\right)\mathrm{m} \end{gathered}$$

(a)

Now, cross-sectional area of wire is,

$\begin{array}{r}\begin{array}{rl}\mathrm{A}& =\mathrm{蟺}{\left(0.430脳{10}^{鈭�3}\mathrm{m}\right)}^2\\ & =5.808脳{10}^{鈭�6}{\mathrm{m}}^2\\ & \mathrm{Also},\mathrm{we}\mathrm{know}\mathrm{that}\mathrm{Young}鈥�\mathrm{s}\mathrm{modulus}\mathrm{is}\mathrm{given}\mathrm{by},\end{array}\\ \mathrm{Y}=\frac{{\mathrm{FI}}_0}{\mathrm{础螖滨}}\end{array}$ (2)

In our case, F = w = mg

Now, equation (2) becomes,

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

Now, we have,

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

Hence, the Young鈥檚 modulus of wire is$8.795脳{10}^{11}\mathrm{Pa}$

(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$U_1=mg{l}_0(1-\cos 胃)$

Final kinetic energy is given by,

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

S0, equation (1) becomes

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

Substitute all numerical values in above equation we get,

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

Consdier the free body diagram鈥�

Now, by newton鈥檚 second law,

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

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

Now, by equation 2 we have,

$Y=\frac{F{l}_0}{A鈭�l}$

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

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

Hence, the wire will stretch 0.127 m .