Math  /  Calculus

Question```latex \documentclass{article} \usepackage{amsmath} \usepackage{amssymb}
\begin{document}
\section*{Problem 12.83}
Assume that the particles are released from rest at r=r0r = r_{0}.
\begin{enumerate} \item[(a)] Determine the expression relating their relative position rr and time. Hint: \int \sqrt{x(1-x)} \, dx = \sin^{-1}(\sqrt{x}) - \sqrt{x(1-x)} \] \item[(b)] Determine the time it takes for the objects to come into contact if r_{0} = 3 \, \text{m},, Aand and Bhavemassesof have masses of 1.1 \, \text{kg}and and 2.3 \, \text{kg}, respectively, and \begin{enumerate} \item[(i)] The diameters of Aand and Bare are d_{A} = 22 \, \text{cm}and and d_{B} = 15 \, \text{cm}, respectively. \item[(ii)] The diameters of Aand and B$ are infinitesimally small. \end{enumerate} \end{enumerate}
\textbf{Figure P12.82 and P12.83}
\end{document} ```

Studdy Solution

STEP 1

STEP 2

The force of attraction between the two particles due to gravity is given by:
F=GmAmBr2 F = G \frac{m_A m_B}{r^2}
where G G is the gravitational constant, mA m_A and mB m_B are the masses of particles AA and BB, and rr is the distance between them.

STEP 3

Using Newton's second law, the relative acceleration r¨ \ddot{r} between the two particles is given by:
r¨=2GmA+mBr2 \ddot{r} = 2G \frac{m_A + m_B}{r^2}

STEP 4

Rewriting the above equation in terms of r¨\ddot{r}, we have:
r¨=2G(mA+mB)r2 \ddot{r} = - \frac{2G(m_A + m_B)}{r^2}
\section*{High_Level_Step 2: Solve the differential equation}

STEP 5

Multiply both sides by r˙\dot{r} and integrate with respect to rr:
r˙r¨=2G(mA+mB)r2r˙ \dot{r} \ddot{r} = - \frac{2G(m_A + m_B)}{r^2} \dot{r}

STEP 6

Integrate both sides:
r˙r¨dr=2G(mA+mB)r2r˙dr \int \dot{r} \ddot{r} \, dr = - \int \frac{2G(m_A + m_B)}{r^2} \dot{r} \, dr

STEP 7

Use the hint provided:
x(1x)dx=sin1(x)x(1x) \int \sqrt{x(1-x)} \, dx = \sin^{-1}(\sqrt{x}) - \sqrt{x(1-x)}

STEP 8

Substitute x=rr0x = \frac{r}{r_0} and dx=drr0dx = \frac{dr}{r_0}:
rr0(1rr0)drr0 \int \sqrt{\frac{r}{r_0} \left(1 - \frac{r}{r_0}\right)} \frac{dr}{r_0}

STEP 9

Simplify and integrate:
sin1(rr0)rr0(1rr0) \sin^{-1}\left(\sqrt{\frac{r}{r_0}}\right) - \sqrt{\frac{r}{r_0} \left(1 - \frac{r}{r_0}\right)}

STEP 10

Solve for tt as a function of rr and the initial conditions:
t=r02G(mA+mB)(sin1(rr0)rr0(1rr0)) t = \frac{r_0}{2G(m_A + m_B)} \left( \sin^{-1}\left(\sqrt{\frac{r}{r_0}}\right) - \sqrt{\frac{r}{r_0} \left(1 - \frac{r}{r_0}\right)} \right)
\section*{High_Level_Step 3: Substitute given initial conditions}

STEP 11

Given r0=3m r_0 = 3 \, \text{m} , mA=1.1kg m_A = 1.1 \, \text{kg} , and mB=2.3kg m_B = 2.3 \, \text{kg} , calculate the time it takes for the objects to come into contact.

STEP 12

Determine the effective radius when the particles come into contact: \begin{enumerate} \item (i) When the diameters of AA and BB are given as dA=22cmd_A = 22 \, \text{cm} and dB=15cmd_B = 15 \, \text{cm} respectively, the contact distance rf=dA+dB2=22+15100m=0.185mr_f = \frac{d_A + d_B}{2} = \frac{22 + 15}{100} \, \text{m} = 0.185 \, \text{m}. \item (ii) When the diameters are infinitesimally small, the contact distance rf0mr_f \approx 0 \, \text{m}. \end{enumerate}

STEP 13

Calculate the time it takes for the objects to come into contact for both cases. \begin{enumerate} \item (i) With rf=0.185mr_f = 0.185 \, \text{m}: t32G(1.1+2.3)(sin1(0.1853)0.1853(10.1853)) t \approx \frac{3}{2G(1.1 + 2.3)} \left( \sin^{-1}\left(\sqrt{\frac{0.185}{3}}\right) - \sqrt{\frac{0.185}{3} \left(1 - \frac{0.185}{3}\right)} \right) \item (ii) With rf0mr_f \approx 0 \, \text{m}: t32G(1.1+2.3)(sin1(0)01)=0s t \approx \frac{3}{2G(1.1 + 2.3)} \left( \sin^{-1}(0) - \sqrt{0 \cdot 1} \right) = 0 \, \text{s} \end{enumerate}
\section*{Solution}
For case (i), with the given diameters: t32G(1.1+2.3)(sin1(0.1853)0.1853(10.1853)) t \approx \frac{3}{2G(1.1 + 2.3)} \left( \sin^{-1}\left(\sqrt{\frac{0.185}{3}}\right) - \sqrt{\frac{0.185}{3} \left(1 - \frac{0.185}{3}\right)} \right)
For case (ii), with infinitesimally small diameters: t0s t \approx 0 \, \text{s}

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