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7.5 Quizz 5

PDF (letter size)

PDF (letter size)

7.5.1 Problems

7.5.2 Problem 1 solution

Angular speed of earth around sun is

$$\dot{\theta }=\frac{2\pi }{\left ( 364\right ) \left ( 24\right ) \left ( 60\right ) \left ( 60\right ) }$$ Force on earth is therefore $mr\dot{\theta }^{2}$. Equating this to $F=G\frac{m_{e}m_{s}}{r^{2}}$ and solving for $r$ gives $$r\dot{\theta }^{2}=G\frac{m_{s}}{r^{2}}$$ One equation with one unknown $r$. Solving gives (taking the positive root) $$r=149.26\times 10^{9}\text{ meter}$$

7.5.3 Problem 2 solution

$$y\left ( x\right ) =8\sin \left ( \pi x\right )$$ We want to solve for $v$ in $$\frac{v^{2}}{\rho }=g$$ But $\rho =\frac{\left ( 1+y^{\prime }\left ( x\right ) ^{2}\right ) ^{\frac{3}{2}}}{\left \vert y^{\prime \prime }\left ( x\right ) \right \vert }$. To find what $x$ to use, since at top of hill, then we want $\sin \left ( \pi x\right ) =1$ or $x=\frac{1}{2}$. Plugging this into $\rho$ gives $$\rho =0.0126651$$ Hence $$\frac{v^{2}}{0.0126651}=9.81$$ Or $$v=0.3525\text{ m/s}$$