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Homework & Assignments

HW# 7 | 6 | 5 | 4 | 3 | 2 | 1

Assignment #7

chapter 7 - 20, 21, 22
chapter 8 - 1, 7, 12, 18

Notes / answers

7.20


$$\begineq {\mathcal L} =&T-U=\frac 12m(\dot x^2+\dot y^2+\dot z^2)-mgz\\ =&\frac12m(R^2\sin^2\phi\,\dot\phi^2 +R^2\cos^2\phi\,\dot\phi^2 +\dot z^2)-mgz\\ =&\frac12m(R^2({\color{red}\sin^2\phi} + {\color{red}\cos^2\phi}) \,{\color{blue}\dot\phi^2} +\dot z^2)-mgz\\ =&\frac12m(R^2({\color{red}1}) \,{\color{blue}\frac{\dot z^2}{\lambda^2}} +\dot z^2)-mgz\\ {\mathcal L}=&\frac12m\left(\frac{R^2}{\lambda^2}+1\right)\dot z^2-mgz\\ \endeq $$ Calculating the two sides of the Euler-Lagrange equation: $$\begineq \frac{\del {\mathcal L}}{\del z}=&\frac{d}{dt}\frac{\del {\mathcal L}}{\del \dot z}\\ -mg =&\frac{d}{dt}m\left(\frac{R^2}{\lambda^2}+1\right)\dot z\\ -g =&\left(\frac{R^2}{\lambda^2}+1\right) \ddot z\\ \endeq $$ Solving for the acceleration:

$$\ddot z=\frac{-g}{\left(\frac{R^2}{\lambda^2}+1\right)}$$

According to the formula, as $R\to 0$ we see that $\ddot z\to -g$. In the limit $R\to 0$, the helical wire is turning into a straight wire down the $z$ axis, and a bead dropped straight down the $z$-axis will drop with the gravitational acceleration $g$.

7.21

The most general solution involves two constants. (You're solving a second-order differential equation). As usual, you can fix the two constants by using the initial conditions, if you have, for example, $r(t=0)$ and $\dot{r}(t=0)$. In the last part, show that part of the solution should vanish in the limit $t\to\infty$, leaving you with just a single exponential term.

There's no potential energy, so ${\cal{L}=T}=(1/2)m(v_r^2+v_\phi^2)=(1/2)m[\dot{r}^2 +(r\omega)^2]$, where angular speed $\omega$ is constant. Applying the E-L equation leads to the differential equation which most of you successfully reached: $$\ddot{r}(t)=\omega^2 r(t)$$ The general solution is $$r(t)=Ae^{\omega t}+Be^{-\omega t}$$ There are two boundary conditions. If the bead is released from rest, that means $$\dot{r}(t=0)=0=A\omega-B\omega \Rightarrow A= B$$ The other condition is the starting position, which we call $r_0$: $$r(0)=r_0=A+B$$ Together with the condition that $A=B$, this implies that $A=B=r_0/2$. $$r(t)=r_0/2\left(e^{\omega t}+e^{-\omega t}\right).$$ So, if $r_0=0$, then $r(t)=0$ for all time.

[But this is an unstable equilibrium: if you get just slightly away from $r=0$, the acceleration is $\ddot{r}=\omega^2 r$ which is positive if $r>0$ and negative if $r<0$. In both cases, the bead will get further away from the equilibrium position.

If $r_0 \gt 0$, the solution is as given above. But we want to show that this approaches pure exponential growth. We notice that as $r\to \infty$ the term $e^{-\omega t}\to 0$, and so $r(t)\to \frac{r_0}2 e^{\omega t}$: a pure exponential. This is because the centrifugal force is not constant as a function of $r$, but increases as the position of the bead increases.

7.22

You can use the angle of the pendulum, $\phi$, as your generalized coordinate. But you need to write the Lagrangian from the point of view of an inertial frame--say, the Earth. (OK, Earth is rotating. So it's not precisely inertial. But it's rotating s l o w l y compared to the pendulum motion). In that frame, the point of support of the pendulum has position $(\frac 12 at^2)$, and velocity $(0,at)$. Figure out the $x$ and $y$ (vertical coordinate) position and speed relative to the earth, in terms of $\phi$.

The position of the mass on the pendulum (length $L$) in the Earth reference frame (which we're approximating as an inertial reference frame for this problem) is $$(x,y)=(L\sin \phi,\frac12at^2-L\cos\phi)$$ Therefore, the speed of the pendulum is: $$(\dot x, \dot y)=(L\dot\phi \cos\phi,at+L\dot\phi\sin\phi).$$ The Lagrangian is: $$\begineq {\cal L}=&T-U=\frac12m(\dot x^2+\dot y^2)-mgy\\ =&\frac12m\left[ L^2\dot\phi^2(\sin^2\phi+\cos^2\phi) +a^2t^2+2L\dot\phi at \sin\phi \right]-mg\left( \frac12at^2-L\cos\phi\right)\\ =&\frac12m\left[ L^2\dot\phi^2 +a^2t^2+2L\dot\phi at \sin\phi \right]+mg\left( L\cos\phi-\frac12at^2 \right) \endeq$$ The E-L equation for the $\phi$ coordinate is: $$\begineq\frac{\del {\cal L}}{\del \phi} =&\frac d{dt}\frac{\del {\cal L}}{\del \dot\phi}\\ {\color{red}mL\dot\phi at\cos\phi}-mgL\sin\phi=&\frac d{dt}(mL^2\dot\phi+mLat\sin\phi)\\ =&mL^2\ddot\phi+mLa\sin\phi+\color{red}{mLat\dot\phi\cos\phi}\\ \endeq $$ Cancelling the red terms, dividing both sides by $m$, and re-arranging for $\ddot\phi$: $$\ddot \phi=-\frac{(g+a)}{L}\sin\phi$$ For small angles, $\sin\phi\approx\phi$, and then this equation is of the form $\ddot\phi=-\omega^2\phi$, and the solution is simple harmonic motion with frequency $\omega= \sqrt{\frac{g+a}{L}}$. This the angular frequency of a pendulum subject to an "effective" gravitational acceleration of $g+a$.

Here, we're writing the Lagrangian from the point of view of an inertial reference frame. But see our notes on Non-inertial frames for the answer :-> derived using the "inertial force" approach instead.

8.7

a.) That is, use $F=ma=m_ev^2/r=m_e\omega^2 r$ where $m_e$ is the mass of earth, $r$ is the distance from sun to earth, and $F=G\frac{m_em_s}{r^2}$, with $m_s=$ sun's mass.

Set centripetal force = gravitational force: $$\begineq m_e \omega^2 r =& G\frac{m_e m_s}{r^2}\\ \omega^2 =& G\frac{m_s}{r^3}\\ \omega =& \sqrt{\frac{Gm_s}{r^3}}\\ \endeq$$ So the period is $T=2\pi/\omega=2\pi\sqrt{\frac{r^3}{G m_s}}$

b.) Start with $F=\mu a$ instead. [What's the fractional difference between the period in a) and in b)?]

$$\begineq \mu \omega^2 r =& G\frac{m_e m_s}{r^2}\\ \frac{m_e m_s}{m_s+m_e}\omega^2 =& G\frac{m_e m_s}{r^3}\\ \omega^2 =& G\frac{m_s+m_e}{r^3}\\ \omega^2 =& G\frac{m_s+m_e}{ r^3}\\ \omega =& \sqrt{G\frac{m_s+m_e}{r^3}}\\ \endeq$$ So the period is $T=2\pi/\omega=2\pi\sqrt{\frac{r^3}{G(m_s+m_e)}}$. And, if $m_s$ grows much larger than $m_e$: $T\to 2\pi\sqrt{\frac{r^3}{Gm_s}}$, which is the result in a.)

The fractional difference in periods (period=$T$) between a) and b): $$\Delta T\approx\frac{\sqrt{1/m_s}-\sqrt{1/(m_s+m_e)}}{\sqrt{1/m_s}}=1-\sqrt{ \frac{m_s}{m_s+m_e}} $$ with $m_s\approx 300,000m_e$, $\Delta T/T=0.00000167$ or 1 part in 600,000.

c.) Figure out what the period is in years (or days) so as to be able to compare the answer with our actual year.

$$\begineq \mu \omega^2 r =& G\frac{m_s m_s}{r^2}\\ \frac{m_s m_s}{m_s+m_s}\omega^2 =& G\frac{m_s m_s}{r^3}\\ \omega^2 =& G\frac{m_s+m_s}{r^3}\\ \omega =& \sqrt{\frac{2Gm_s}{r^3}} \endeq$$ So the period is $$T=2\pi/\omega=2\pi\sqrt{\frac{r^3}{2Gm_s}} =\frac{1}{\sqrt 2}\left(2\pi\sqrt{\frac{r^3}{Gm_s}}\right)=0.707\left(\text{1 year}\right).$$ About 71% of the existing year.

8.12