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2 years ago

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A long thin rod of mass M and length L is situated along the y axis with one end at the origin. A small spherical mass m1 is pla

ced at the location P, which is at a distance d from the origin.
If L = 2.00 ? 104 m, d = 18.0 ? 104 m,M = 14.0 ? 106 kg,and m1 = 8.00 ? 106 kg,what is the value of this potential energy?

1 answer:

Yuri [45]2 years ago

5 0

Answer:

$= - 5.65\times 10^{-2} J$

Explanation:

Given data:

L =2.00 *10^4 m

d = 18*10^4 m

M = 18 *10^6 kg

m_1 = 8*10^6 kg

Gravitational energy is given as

$U =- G \frac{m_1 m_2}{r}$

mass per unit length is given as

$\sigma = \frac{M}{L} = \frac{18 \times 10^6}{2\times 10^4 m} = 900 kg/m$

calculating potential energy

$dU ==-G\int_{16\times 10^4}^{18\times 10^4} \frac{m_1 *dm}{r}$

$=-G\int_{16\times 10^4}^{18\times 10^4} \frac{m_1 *\sigma dr}{r}$

$=-G*m_1*\sigma\int_{16\times 10^4}^{18\times 10^4} \frac{dr}{r}$

$=-G*m_1*\sigma \left | ln r \right |_{16\times 10^4}^{18\times 10^4}$

$=-G*m_1*\sigma ln(1.125)$

$=-6.673 \times 10^{-11}*8*10^6*900*ln(1.125)$

$= - 5.65\times 10^{-2} J$

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Complete Question:

Given $\sigma = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right]$ at a point. What is the force per unit area at this point acting normal to the surface with $\b n = (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z$ ? Are there any shear stresses acting on this surface?

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There are shear stresses acting on the surface since $\tau \neq 0$

Explanation:

$\sigma = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right]$

equation of the normal, $\b n = (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z$

$\b n = \left[\begin{array}{ccc}\frac{1}{\sqrt{2} }\\0\\\frac{1}{\sqrt{2} }\end{array}\right]$

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$T_n = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right] \left[\begin{array}{ccc}\frac{1}{\sqrt{2} }\\0\\\frac{1}{\sqrt{2} }\end{array}\right]$

$T_n = \left[\begin{array}{ccc}\frac{23}{\sqrt{2} }\\0\\\frac{27}{\sqrt{33} }\end{array}\right]$

$T_n = \frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z$

To get the Force per unit area acting normal to the surface, find the dot product of the traction vector and the normal.

$\sigma_n = T_n . \b n$

$\sigma \b n = (\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z) . ((1/ \sqrt{2} ) \b e_x + 0 \b e_y +(1/ \sqrt{2}) \b e_z)\\\\\sigma \b n = 28 MPa$

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$\tau = T_n - \sigma_n \b n$

$\tau = [\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z] - 28( (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z)\\\\\tau = [\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z] - [ (28/ \sqrt{2} ) \b e_x + (28/ \sqrt{2}) \b e_z]\\\\\tau = \frac{-5}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{5}{\sqrt{2} } \b e_z$

$\tau = \sqrt{(-5/\sqrt{2})^2 + (27/\sqrt{2})^2 + (5/\sqrt{2})^2} \\\\ \tau = 19.74 MPa$

Since $\tau \neq 0$ , there are shear stresses acting on the surface.

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2 years ago