Problems

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Solution:

The rocket principle states that the thrust generated by the rocket is \(-\frac{\d m}{\d t}u = \alpha \beta u\) This force is acting at a distance \(2a\) from \(O\) and therefore is generating a torque of \(2a \alpha \beta u\) on the system. Let's also consider the moments of inertia about \(O\). The fixed rod will have moment of inertia \(\frac13 M (2a)^2 = \frac43 M a^2\). The rocket will have moment of inertia \(I_{G} + md^2 = \frac1{12}m(t)(2l(t))^2 + m(t) ((2a)^2 + l(t)^2)= \frac43 ml^2+ 4ma^2\). Since our final equation doesn't involve \(m\), lets replace all the \(m\) with \(2al\) to obtain a total \(\displaystyle I = \frac43 Ma^2 + \frac83 \alpha l^3 + 8\alpha la^2\). Since \(\tau\) is constant, we can note that \(I\omega = 2a \alpha \beta u t\) (by integrating) and so \begin{align*} && \dot{\omega} &= \frac{\d }{\d t} \left ( \frac{2a \alpha \beta u t}{ \frac43 Ma^2 + \frac83 \alpha l^3 + 8\alpha la^2} \right) \\ &&&= \frac{\d }{\d t} \left ( \frac{3a \alpha \beta u t}{ 2Ma^2 +4\alpha l^3 + 4 \cdot 3 \cdot \alpha la^2} \right) \\ &&&= \frac{\d }{\d t} \left ( \frac{3a \alpha \beta u t}{ 2[Ma^2 +2\alpha l(l^2 + 3 a^2)]} \right) \\ \end{align*} This is, close, but not quite what they are after since the denominator also has a dependency on \(t\) we wont get exactly what they've asked for

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Solution: Suppose we divide a sphere of radius \(r\) up into slices of thickness \(\delta x\). Then the force acting on \(P\) will be: \begin{align*} F &= \sum_{\text{slices}} 2\pi mk\rho\delta x\left(1-\frac{b}{(a^{2}+b^{2})^{\frac{1}{2}}}\right) \\ &= \sum_{i=-r/\delta x}^{r/\delta x} 2\pi mk\frac{M}{\frac43 \pi r^3}\delta x\left(1-\frac{i \delta x}{((1-(i\delta x)^2)+(i \delta x)^{2})^{\frac{1}{2}}}\right) \\ &\to \int_{-r}^r \frac{1}{2} \frac{mkM}{r^3}(1-t) \d t \\ &=\frac{mkM}{r^2} \end{align*} We can see that the particle will have a force attracting it towards the centre, with magnitude \(\frac{kmM}{r^2}\), therefore and since \(\frac{\d v}{\d t} = \frac{\d v}{\d r} \frac{\d r}{\d t}\) we must have: \(v \frac{\d v}{\d r}m = - \frac{kmM}{r^2}\) and dividing by \(m\) we get exactly the result we seek. \begin{align*} && v \frac{\d v}{\d r} &= \frac{-kM}{r^2} \\ \Rightarrow && \frac{v^2}{2}+C &= \frac{kM}{r} \\ r = R, v =0: && C &= \frac{kM}{R} \\ \Rightarrow && v^2&= 2kM\left ( \frac1r - \frac1R\right ) \\ \Rightarrow && \frac{\d r}{\d t} &= -\sqrt{2kM\left ( \frac1r - \frac1R\right )} \\ \Rightarrow && -\sqrt{2kM}T &= \int_{r=R}^{r=R\cos^2 \alpha} \frac{1}{\sqrt{\frac1r-\frac1R}} \d r \\ r = R\cos^2 \theta: && -\sqrt{2kM}T &= \int_{\theta = 0}^{\theta = \alpha} \frac{\sqrt{R}}{\sqrt{\sec^2 \theta - 1}} \cdot R \cdot 2 \cdot (-\cos \theta) \cdot \sin \theta \d \theta \\ \Rightarrow && T &= \sqrt{\frac{R^3}{2kM}} \int_0^\alpha \frac{2 \cos \theta \sin \theta}{\sqrt{\sec^2 \theta - 1}} \d \theta \\ &&&= \sqrt{\frac{R^3}{2kM}} \int_0^\alpha \frac{2 \cos \theta \sin \theta}{\tan \theta} \d \theta \\ &&&= \sqrt{\frac{R^3}{2kM}} \int_0^\alpha 2\cos^2 \theta \d \theta \\ &&&= \sqrt{\frac{R^3}{2kM}} \int_0^\alpha 1 + \cos 2 \theta\d \theta \\ &&&= \sqrt{\frac{R^3}{2kM}} \left [1 + \frac12 \sin 2 \theta \right]_0^\alpha \\ &&&= \sqrt{\frac{R^3}{2kM}} \left (\alpha + \frac12 \sin 2 \alpha \right) \\ \end{align*}