SOLUTION:
Starting from\begin{equation} \theta _{0}\left ( b\right ) =\int _{r_{\min }}^{\infty }\frac{dr}{r^{2}\sqrt{\frac{2Em}{l^{2}}-\frac{2mU}{l^{2}}-\frac{1}{r^{2}}}}\tag{1} \end{equation} But \begin{align*} l & =b\sqrt{2mE}\\ l^{2} & =b^{2}\left ( 2mE\right ) \end{align*}
Hence (1) becomes\begin{align} \theta _{0}\left ( b\right ) & =\int _{r_{\min }}^{\infty }\frac{dr}{r^{2}\sqrt{\frac{1}{b^{2}}-\frac{U}{b^{2}E}-\frac{1}{r^{2}}}}\nonumber \\ & =\int _{r_{\min }}^{\infty }\frac{b}{r^{2}\sqrt{1-\frac{U}{E}-\frac{b^{2}}{r^{2}}}}dr\tag{1A} \end{align}
In this problem, since \(F\left ( r\right ) =\frac{k}{r^{3}}\), therefore since \(F\left ( r\right ) =-\nabla U\) \begin{align*} U\left ( r\right ) & =-\int \frac{k}{r^{3}}dr\\ & =\frac{k}{2r^{2}} \end{align*}
Then (1A) becomes\begin{equation} \theta _{0}\left ( b\right ) =\int _{r_{\min }}^{\infty }\frac{b}{r^{2}\sqrt{1-\frac{k}{2r^{2}E}-\frac{b^{2}}{r^{2}}}}dr\tag{1B} \end{equation} Let \(z=\frac{1}{r}\) then \(\frac{dr}{dz}=-\frac{1}{z^{2}}\). When \(r=\infty \) then \(z=0\) and when \(r=r_{\min }\) then \(z=\frac{1}{r_{\min }}\). Now we need to find \(r_{\min }\). We know that when \(E=U_{effective}\) then \(r=r_{\min .}\) But \begin{align*} U_{effective} & =\frac{l^{2}}{2mr^{2}}+U\left ( r\right ) \\ & =\frac{l^{2}}{2mr^{2}}+\frac{k}{2r^{2}} \end{align*}
Hence\begin{align*} E & =U_{effective}\\ & =\frac{l^{2}}{2mr_{\min }^{2}}+\frac{k}{2r_{\min }^{2}}\\ & =\frac{l^{2}+mk}{2mr_{\min }^{2}} \end{align*}
Solving for \(r_{\min }\)\begin{align} r_{\min }^{2} & =\frac{l^{2}+mk}{2mE}\nonumber \\ & =\frac{l^{2}}{2mE}+\frac{k}{2E}\tag{2} \end{align}
But \(l^{2}=b^{2}\left ( 2mE\right ) \) then (2) becomes\begin{align*} r_{\min }^{2} & =\frac{b^{2}\left ( 2mE\right ) }{2mE}+\frac{k}{2E}\\ & =b^{2}+\frac{k}{2E} \end{align*}
Therefore \begin{equation} r_{\min }=\sqrt{b^{2}+\frac{k}{2E}}\tag{3} \end{equation} Now we can finish the limits of integration in (1B). When \(r=r_{\min }\) then \(z=\frac{1}{r_{\min }}=\frac{1}{\sqrt{b^{2}-\frac{k}{2E}}}\), now (1B) becomes (where we now replace \(r^{2}\) by \(\frac{1}{z^{2}}\))\begin{align*} \theta _{0}\left ( b\right ) & =\int _{r_{\min }}^{\infty }\frac{b}{r^{2}\sqrt{1-\frac{k}{2r^{2}E}-\frac{b^{2}}{r^{2}}}}dr\\ & =\int _{\frac{1}{\sqrt{b^{2}+\frac{k}{2E}}}}^{0}\frac{z^{2}b}{\sqrt{1-\frac{kz^{2}}{2E}-b^{2}z^{2}}}\left ( -\frac{1}{z^{2}}dz\right ) \\ & =b\int _{0}^{\frac{1}{\sqrt{b^{2}+\frac{k}{2E}}}}\frac{1}{\sqrt{1-\frac{kz^{2}}{2E}-b^{2}z^{2}}}dz\\ & =b\int _{0}^{\frac{1}{\sqrt{b^{2}+\frac{k}{2E}}}}\frac{dz}{\sqrt{1-z^{2}\left ( \frac{k}{2E}+b^{2}\right ) }} \end{align*}
Using CAS, it gives \(\int \frac{dz}{\sqrt{1-az^{2}}}=\frac{1}{\sqrt{a}}\sin ^{-1}\left ( z\sqrt{a}\right ) \). Using this result above, where \(a=\left ( \frac{k}{2E}+b^{2}\right ) \) gives\begin{align*} \theta _{0}\left ( b\right ) & =\frac{b}{\sqrt{\frac{k}{2E}+b^{2}}}\left ( \left . \sin ^{-1}\left ( z\sqrt{\frac{k}{2E}+b^{2}}\right ) \right \vert _{0}^{\frac{1}{\sqrt{b^{2}+\frac{k}{2E}}}}\right ) \\ & =\frac{b}{\sqrt{\frac{k}{2E}+b^{2}}}\left [ \sin ^{-1}\left ( \frac{1}{\sqrt{b^{2}+\frac{k}{2E}}}\sqrt{\frac{k}{2E}+b^{2}}\right ) -\sin ^{-1}\left ( 0\right ) \right ] \\ & =\frac{b}{\sqrt{\frac{k}{2E}+b^{2}}}\left [ \sin ^{-1}\left ( 1\right ) -0\right ] \\ & =\frac{b}{\sqrt{\frac{k}{2E}+b^{2}}}\frac{\pi }{2} \end{align*}
Now we solve for \(b\). Squaring both sides\[ \theta _{0}^{2}=\frac{b^{2}}{\frac{k}{2E}+b^{2}}\frac{\pi ^{2}}{4}\] Using \(E=\frac{1}{2}mv^{2}\) then\begin{align} \theta _{0}^{2} & =\frac{b^{2}}{\left ( \frac{k}{mv^{2}}+b^{2}\right ) }\frac{\pi ^{2}}{4}\nonumber \\ 4\theta _{0}^{2}\left ( \frac{k}{mv^{2}}+b^{2}\right ) & =b^{2}\pi ^{2}\nonumber \\ \frac{k4\theta _{0}^{2}}{mv^{2}}+4\theta _{0}^{2}b^{2}-b^{2}\pi ^{2} & =0\nonumber \\ b^{2}\left ( 4\theta _{0}^{2}-\pi ^{2}\right ) & =-\frac{k4\theta _{0}^{2}}{mv^{2}}\nonumber \\ b^{2}\left ( \pi ^{2}-4\theta _{0}^{2}\right ) & =\frac{k4\theta _{0}^{2}}{mv^{2}}\nonumber \\ b^{2} & =\frac{k4\theta _{0}^{2}}{mv^{2}\left ( \pi ^{2}-4\theta _{0}^{2}\right ) }\nonumber \\ b & =\frac{2\theta _{0}}{v}\sqrt{\frac{k}{m\left ( \pi ^{2}-4\theta _{0}^{2}\right ) }}\tag{4} \end{align}
But \(\theta _{0}\left ( b\right ) =\frac{\pi }{2}-\frac{\theta _{s}}{2}\), where \(\theta _{s}\) is the scattering angle. Therefore the above becomes \begin{align} b & =\frac{2\left ( \frac{\pi }{2}-\frac{\theta _{s}}{2}\right ) }{v}\sqrt{\frac{k}{m\left ( \pi ^{2}-4\left ( \frac{\pi }{2}-\frac{\theta _{s}}{2}\right ) ^{2}\right ) }}\nonumber \\ b & =\frac{\pi -\theta _{s}}{v}\sqrt{\frac{k}{m\left ( \pi ^{2}-\left ( \theta _{s}^{2}-2\pi \theta _{s}+\pi ^{2}\right ) \right ) }}\nonumber \\ b & =\frac{\pi -\theta _{s}}{v}\sqrt{\frac{k}{m\left ( 2\pi \theta _{s}-\theta _{s}^{2}\right ) }}\tag{5} \end{align}
Now we are ready to find \(\sigma \left ( \theta _{s}\right ) \)\[ \sigma \left ( \theta _{s}\right ) =\frac{b}{\sin \theta _{s}}\left \vert \frac{db}{d\theta _{s}}\right \vert \] From (5)\[ \frac{db}{d\theta _{s}}=-\frac{\pi ^{2}\sqrt{\frac{k}{m\left ( 2\pi \theta _{s}-\theta _{s}^{2}\right ) }}}{v\left ( 2\pi \theta _{s}-\theta _{s}^{2}\right ) }\] Therefore\begin{align*} \sigma \left ( \theta _{s}\right ) & =\frac{b}{\sin \theta _{s}}\left \vert \frac{db}{d\theta _{s}}\right \vert \\ & =\frac{\frac{\pi -\theta _{s}}{v}\sqrt{\frac{k}{m\left ( 2\pi \theta _{s}-\theta _{s}^{2}\right ) }}}{\sin \theta _{s}}\frac{\pi ^{2}\sqrt{\frac{k}{m\left ( 2\pi \theta _{s}-\theta _{s}^{2}\right ) }}}{v\left ( 2\pi \theta _{s}-\theta _{s}^{2}\right ) }\\ & =\frac{\frac{\pi -\theta _{s}}{v}\frac{k}{m\left ( 2\pi \theta _{s}-\theta _{s}^{2}\right ) }}{\sin \theta _{s}}\frac{\pi ^{2}}{v\left ( 2\pi \theta _{s}-\theta _{s}^{2}\right ) }\\ & =\frac{\left ( \pi -\theta _{s}\right ) k}{mv\sin \theta _{s}}\frac{\pi ^{2}}{v\left ( 2\pi \theta _{s}-\theta _{s}^{2}\right ) ^{2}}\\ & =\frac{k\pi ^{2}\left ( \pi -\theta _{s}\right ) }{mv^{2}\sin \theta _{s}\left ( 2\pi \theta _{s}-\theta _{s}^{2}\right ) ^{2}} \end{align*}
Or\[ \sigma \left ( \theta _{s}\right ) =\frac{k\pi ^{2}\left ( \pi -\theta _{s}\right ) }{mv^{2}\theta _{s}^{2}\left ( 2\pi -\theta _{s}\right ) ^{2}\sin \theta _{s}}\] Hard problem. Time taken to solve: 6 hrs.
SOLUTION:
Using\begin{align} x & =\frac{1}{3}\omega gt^{3}\cos \lambda -\omega t^{2}\left ( \dot{z}_{0}\cos \lambda -\dot{y}_{0}\sin \lambda \right ) +\dot{x}_{0}t+x_{0}\nonumber \\ y & =\dot{y}_{0}t-\omega t^{2}\dot{x}_{0}\sin \lambda +y_{0}\tag{1}\\ z & =\dot{z}_{0}t-\frac{1}{2}gt^{2}+\omega t^{2}\dot{x}_{0}\cos \lambda +z_{0}\nonumber \end{align}
Where \(\left \{ \dot{x}_{0},\dot{y}_{0},\dot{z}_{0}\right \} \) are the initial speeds in each of the body frame directions and \(\left \{ x_{0},y_{0},z_{0}\right \} \) are the initial position of the projectile at \(t=0\). Let \(v_{0}=800\ m/s^{2}\) and \(\theta =37^{0}\). We are given that \begin{align*} \dot{y}_{0} & =-v_{0}\cos \theta \\ \dot{z}_{0} & =v_{0}\sin \theta \\ \dot{x}_{0} & =0 \end{align*}
The minus sign for \(\dot{y}_{0}\) above was added since the direction is south, which is negative \(y\) direction for the local frame. And we are given that \(x_{0}=y_{0}=z_{0}=0\). Substituting these in (1) gives (where \(\lambda =50^{0}\))\begin{align} x & =\frac{1}{3}\omega gt^{3}\cos \lambda -\omega t^{2}\left ( v_{0}\sin \theta \cos \lambda +v_{0}\cos \theta \sin \lambda \right ) \nonumber \\ y & =-\left ( v_{0}\cos \theta \right ) t\tag{2}\\ z & =\left ( v_{0}\sin \theta \right ) t-\frac{1}{2}gt^{2}\nonumber \end{align}
The drift due to the Coriolis force is found from the \(x\) component. The projectile will drift west (to the right direction of its motion) since it is moving south. We can now calculate this \(x\) drift. We know that \(\omega =7.3\times 10^{-5}\) rad/sec (rotation speed of earth), so we just need to find time of flight \(t\). From\begin{align*} \dot{z} & =\dot{z}_{0}-gt\\ & =v_{0}\sin \theta -gt \end{align*}
The projectile time up (when \(\dot{z}\) first becomes zero) is \(t=\frac{v_{0}\sin \theta }{g}=\frac{800\sin \left ( 37\left ( \frac{\pi }{180}\right ) \right ) }{9.81}\approx 50\) sec. Hence total time of flight is twice this which is \(t_{f}=100\) sec. Now we use this time in the \(x\) equation in (2) above\begin{align*} x & =\frac{1}{3}\left ( 7.3\times 10^{-5}\right ) \left ( 9.81\right ) \left ( 100\right ) ^{3}\cos \left ( 50^{0}\right ) -\left ( 7.3\times 10^{-5}\right ) \left ( 100\right ) ^{2}\left ( 800\sin 37^{0}\cos 50^{0}+800\cos 37^{0}\sin 50^{0}\right ) \\ & =-532 \end{align*}
So it will drift by about \(532\) meter to the west (since negative sign). In the above \(g=9.81\) was used. This does not include all the terms such as the centrifugal acceleration. But \(9.81\frac{m}{s^{2}}\) is good approximation for this problem.
Taking Latitude as \(42^{0}\) (New York). Therefore \(\lambda =42^{0}\) and \(\theta =15^{0}\). Initial conditions are \begin{align*} \dot{y}_{0} & =V_{0}\cos \theta \\ \dot{z}_{0} & =V_{0}\sin \theta \\ \dot{x}_{0} & =0 \end{align*}
Where \(V_{0}\) is the initial speed the ball was hit with (which we do not know yet), and \(x_{0}=y_{0}=z_{0}=0\). Using\begin{align} x & =\frac{1}{3}\omega gt^{3}\cos \lambda -\omega t^{2}\left ( \dot{z}_{0}\cos \lambda -\dot{y}_{0}\sin \lambda \right ) +\dot{x}_{0}t+x_{0}\nonumber \\ y & =\dot{y}_{0}t-\omega t^{2}\dot{x}_{0}\sin \lambda +y_{0}\tag{1}\\ z & =\dot{z}_{0}t-\frac{1}{2}gt^{2}+\omega t^{2}\dot{x}_{0}\cos \lambda +z_{0}\nonumber \end{align}
Then applying initial conditions the above reduces to\begin{align} x & =\frac{1}{3}\omega gt^{3}\cos \lambda -\omega t^{2}\left ( V_{0}\sin \theta \cos \lambda -V_{0}\cos \theta \sin \lambda \right ) \nonumber \\ y & =\left ( V_{0}\cos \theta \right ) t\tag{2}\\ z & =\left ( V_{0}\sin \theta \right ) t-\frac{1}{2}gt^{2}\nonumber \end{align}
From \(y\left ( t_{f}\right ) =\left ( V_{0}\cos \theta \right ) t_{f}\) then, since we are told that \(y\left ( t_{f}\right ) =200\) ft,\begin{equation} 200\left ( 0.3048\right ) =\left ( V_{0}\cos \theta \right ) t_{f} \tag{3} \end{equation} Where \(t_{f}\) is time of flight. But time of flight is also found\begin{align*} \dot{z} & =\dot{z}_{0}-gt\\ & =V_{0}\sin \theta -gt \end{align*}
And solving for \(\dot{z}=0\), which gives \(\frac{V_{0}\sin \theta }{g}\). So time of flight is twice this or\[ t_{f}=\frac{2V_{0}\sin \theta }{g}\] Substituting the above into (3) to solve for \(V_{0}\) gives\begin{align*} 200\left ( 0.3048\right ) & =\left ( V_{0}\cos \theta \right ) \frac{2V_{0}\sin \theta }{g}\\ 60.96 & =\frac{2}{9.81}V_{0}^{2}\left ( \cos 15^{0}\right ) \left ( \sin 15^{0}\right ) \\ V_{0}^{2} & =\frac{\left ( 60.96\right ) \left ( 9.81\right ) }{2\cos 15^{0}\sin 15^{0}}\\ & =1196.0 \end{align*}
Hence\[ V_{0}=34.583\ \ \ \ \text{m/s}\] Now we can go back and solve for time of flight \(t_{f}\). From\begin{align*} 200\left ( 0.3048\right ) & =\left ( V_{0}\cos \theta \right ) t_{f}\\ t_{f} & =\frac{200\left ( 0.3048\right ) }{34.583\text{ }\left ( \cos 15^{0}\right ) }\\ & =1.825\text{ sec} \end{align*}
Using (2) we solve for \(x\), the drift due to Coriolis forces. \begin{align*} x & =\frac{1}{3}\omega gt^{3}\cos \lambda -\omega t^{2}\left ( V_{0}\sin \theta \cos \lambda -V_{0}\cos \theta \sin \lambda \right ) \\ & =\frac{1}{3}\left ( 7.3\times 10^{-5}\right ) \left ( 9.81\right ) \left ( 1.825\right ) ^{3}\cos 42^{0}-\left ( 7.3\times 10^{-5}\right ) \left ( 1.825\right ) ^{2}\left ( 34.58\sin 15^{0}\cos 42^{0}+34.58\cos 15^{0}\sin 42^{0}\right ) \\ & =4.897\times 10^{-3}\text{ meter} \end{align*}
So the ball will drift about \(5mm\). This is too small and the ball player can therefore ignore Coriolis forces when hitting the ball.
SOLUTION:
Initial conditions are \begin{align*} \dot{y}_{0} & =0\\ \dot{z}_{0} & =v_{0}\\ \dot{x}_{0} & =0 \end{align*}
And \(x_{0}=y_{0}=z_{0}=0\). Using\begin{align} x & =\frac{1}{3}\omega gt^{3}\cos \lambda -\omega t^{2}\left ( \dot{z}_{0}\cos \lambda -\dot{y}_{0}\sin \lambda \right ) +\dot{x}_{0}t+x_{0}\nonumber \\ y & =\dot{y}_{0}t-\omega t^{2}\dot{x}_{0}\sin \lambda +y_{0}\tag{1}\\ z & =\dot{z}_{0}t-\frac{1}{2}gt^{2}+\omega t^{2}\dot{x}_{0}\cos \lambda +z_{0}\nonumber \end{align}
The reduce to (using initial conditions) to \begin{align} x & =\frac{1}{3}\omega gt^{3}\cos \lambda -\omega t^{2}v_{0}\cos \lambda \nonumber \\ y & =0\tag{2}\\ z & =v_{0}t-\frac{1}{2}gt^{2}\nonumber \end{align}
To find time of flight of bullet (going up and then down again), from \(\dot{z}=v_{0}-gt\), we solve for \(\dot{z}=0\), which gives \(t=\frac{v_{0}}{g}\). So time of flight is twice this amount\[ t_{f}=\frac{2v_{0}}{g}\text{ sec}\] To find the amount \(x\) the bullet moves during this time, we use (2) and solve for \(x\)\begin{align*} x\left ( t_{f}\right ) & =\frac{1}{3}\omega gt_{f}^{3}\cos \lambda -\omega t_{f}^{2}v_{0}\cos \lambda \\ & =\frac{1}{3}\omega g\left ( \frac{2v_{0}}{g}\right ) ^{3}\cos \lambda -\omega \left ( \frac{2v_{0}}{g}\right ) ^{2}v_{0}\cos \lambda \\ & =\frac{1}{3}\omega \frac{8v_{0}^{3}}{g^{2}}\cos \lambda -\omega \frac{4v_{0}^{3}}{g^{2}}\cos \lambda \\ & =\left ( \frac{8}{3}-4\right ) \left ( \omega \frac{v_{0}^{3}}{g^{2}}\cos \lambda \right ) \\ & =-\frac{4}{3}\omega \frac{v_{0}^{3}}{g^{2}}\cos \lambda \end{align*}
This means when it lands again, the bullet will be \(-\frac{4}{3}\omega \frac{v_{0}^{3}}{g^{2}}\cos \lambda \) meters relative to the original point it was fired from (the origin of the local body frame). Since the sign is negative, it means it is west.
SOLUTION:
When Ant is moving in direction of rotation:
\begin{align} \vec{r} & =b\cos \theta \vec{i}+b\sin \theta \vec{j}\nonumber \\ \vec{v} & =\vec{v}_{rel}+\vec{\omega }\times \vec{r} \tag{1} \end{align}
But\begin{align*} \vec{v}_{rel} & =\frac{d}{dt}\vec{r}\\ & =-b\dot{\theta }\sin \theta \vec{i}+b\dot{\theta }\cos \theta \vec{j} \end{align*}
And \begin{align*} \vec{\omega }\times \vec{r} & =\omega \vec{k}\times \left ( b\cos \theta \vec{i}+b\sin \theta \vec{j}\right ) \\ & =b\omega \cos \theta \vec{j}-b\omega \sin \theta \vec{i} \end{align*}
Hence (1) becomes\begin{align*} \vec{v} & =\left ( -b\dot{\theta }\sin \theta \vec{i}+b\dot{\theta }\cos \theta \vec{j}\right ) +\left ( b\omega \cos \theta \vec{j}-b\omega \sin \theta \vec{i}\right ) \\ & =\vec{i}\left ( -b\dot{\theta }\sin \theta -b\omega \sin \theta \right ) +\vec{j}\left ( b\dot{\theta }\cos \theta +b\omega \cos \theta \right ) \end{align*}
The above is the velocity of the ant, in the inertial frame, using local body unit vector \(\vec{i},\vec{j}\). Now we find the ant acceleration, given by\[ \vec{a}=\vec{a}_{rel}+2\left ( \omega \vec{k}\times \vec{v}_{rel}\right ) +\left ( \dot{\omega }\vec{k}\times \vec{r}\right ) +\omega \vec{k}\times \left ( \vec{\omega }\times \vec{r}\right ) \] But \(\dot{\omega }=0\) since disk has constant \(\omega \) then\begin{equation} \vec{a}=\vec{a}_{rel}+2\left ( \omega \vec{k}\times \vec{v}_{rel}\right ) +\omega \vec{k}\times \left ( \vec{\omega }\times \vec{r}\right ) \tag{1} \end{equation} But \begin{align*} \vec{a}_{rel} & =\frac{d}{dt}\vec{v}_{rel}\\ & =\vec{i}\left ( -b\ddot{\theta }\sin \theta -b\dot{\theta }^{2}\cos \theta \right ) +\vec{j}\left ( b\ddot{\theta }\cos \theta -b\dot{\theta }^{2}\sin \theta \right ) \end{align*}
Since Bug moves with constant speed, then \(\ddot{\theta }=0\) and the above becomes\[ \vec{a}_{rel}=\vec{i}\left ( -b\dot{\theta }^{2}\cos \theta \right ) +\vec{j}\left ( -b\dot{\theta }^{2}\sin \theta \right ) \] Now the Coriolis term \(2\left ( \vec{\omega }\times \vec{v}_{rel}\right ) \) is found\begin{align*} 2\left ( \vec{\omega }\times \vec{v}_{rel}\right ) & =2\left ( \omega \vec{k}\times \left ( -b\dot{\theta }\sin \theta \vec{i}+b\dot{\theta }\cos \theta \vec{j}\right ) \right ) \\ & =2\left ( -\omega b\dot{\theta }\sin \theta \vec{j}-b\omega \dot{\theta }\cos \theta \vec{i}\right ) \end{align*}
Now the \(\vec{\omega }\times \left ( \vec{\omega }\times \vec{r}\right ) \) is found\begin{align*} \vec{\omega }\times \left ( \vec{\omega }\times \vec{r}\right ) & =\omega \vec{k}\times \left ( b\omega \cos \theta \vec{j}-b\omega \sin \theta \vec{i}\right ) \\ & =-b\omega ^{2}\cos \theta \vec{i}-b\omega ^{2}\sin \theta \vec{j} \end{align*}
Hence (1) becomes\begin{align*} \vec{a} & =\vec{a}_{rel}+2\left ( \omega \vec{k}\times \vec{v}_{rel}\right ) +\omega \vec{k}\times \left ( \vec{\omega }\times \vec{r}\right ) \\ & =\vec{i}\left ( -b\dot{\theta }^{2}\cos \theta \right ) +\vec{j}\left ( -b\dot{\theta }^{2}\sin \theta \right ) +2\left ( -\omega b\dot{\theta }\sin \theta \vec{j}-b\omega \dot{\theta }\cos \theta \vec{i}\right ) -b\omega ^{2}\cos \theta \vec{i}-b\omega ^{2}\sin \theta \vec{j}\\ & =\vec{i}\left ( -b\dot{\theta }^{2}\cos \theta -2b\omega \dot{\theta }\cos \theta -b\omega ^{2}\cos \theta \right ) +\vec{j}\left ( -b\dot{\theta }^{2}\sin \theta -2\omega b\dot{\theta }\sin \theta -b\omega ^{2}\sin \theta \right ) \end{align*}
Since this is valid for all time, lets take snap shot when \(\theta =0\), which gives\[ \vec{a}=\vec{i}\left ( -b\dot{\theta }^{2}-2b\omega \dot{\theta }-b\omega ^{2}\right ) \] So when \(\theta =0\), the ant acceleration (as seen in inertial frame) is towards the center of the disk with the above magnitude. If the ant speed is \(V\) then \(V=b\dot{\theta }\) and the above can be re-written in terms of \(V\) as\[ \vec{a}=-\vec{i}\left ( \frac{V^{2}}{b}+2V\omega +b\omega ^{2}\right ) \] The ant will starts to slip, when the force preventing it from sliding radially in the outer direction equals the centrifugal force \(m\left ( \frac{V^{2}}{b}+3V\omega +b\omega ^{2}\right ) \)Hence \begin{align*} \mu mg & =m\left ( \frac{V^{2}}{b}+2V\omega +b\omega ^{2}\right ) \\ \frac{V^{2}}{b}+2V\omega +b\omega ^{2}-\mu g & =0\\ V^{2}+2Vb\omega -\left ( \mu bg+b^{2}\omega ^{2}\right ) & =0 \end{align*}
This is quadratic in \(V\), hence\begin{align*} V & =\frac{-2b\omega }{2}\pm \frac{1}{2}\sqrt{4b^{2}\omega ^{2}+4\left ( -\mu bg+b^{2}\omega ^{2}\right ) }\\ & =-b\omega \pm \sqrt{b^{2}\omega ^{2}-\mu bg+b^{2}\omega ^{2}}\\ & =-b\omega \pm \sqrt{2b^{2}\omega ^{2}-\mu bg} \end{align*}
Since \(V>0\) then \begin{align*} V & =-b\omega +b\omega \sqrt{2-\frac{\mu g}{b\omega ^{2}}}\\ & =b\omega \left ( \sqrt{2-\frac{\mu g}{b\omega ^{2}}}-1\right ) \end{align*}
When Ant is moving the opposite direction of rotation, then the Coriolis term \(2\left ( \omega \vec{k}\times \vec{v}_{rel}\right ) \) will have the opposite sign from the above. Then means the final answer will be\[ \vec{a}=-\vec{i}\left ( \frac{V^{2}}{b}-2V\omega +b\omega ^{2}\right ) \] Which means\begin{align*} V & =\frac{2b\omega }{2}\pm \frac{1}{2}\sqrt{4b^{2}\omega ^{2}+4\left ( -\mu bg+b^{2}\omega ^{2}\right ) }\\ & =b\omega \pm \sqrt{b^{2}\omega ^{2}-\mu bg+b^{2}\omega ^{2}}\\ & =b\omega \pm \sqrt{2b^{2}\omega ^{2}-\mu bg} \end{align*}
Or\begin{align*} V & =b\omega +b\omega \sqrt{2-\frac{\mu g}{b\omega ^{2}}}\\ & =b\omega \left ( \sqrt{2-\frac{\mu g}{b\omega ^{2}}}+1\right ) \end{align*}
SOLUTION:
\[ \vec{g}=\vec{g}_{0}-\vec{\omega }\times \vec{\omega }\times \vec{R}\] Using \(a\times \left ( b\times c\right ) =b\left ( a\cdot c\right ) -c\left ( a\cdot b\right ) \) the above becomes\begin{align*} \vec{g} & =\vec{g}_{0}-\left ( \vec{\omega }\left ( \vec{\omega }\cdot \vec{R}\right ) -\left ( \vec{\omega }\cdot \vec{\omega }\right ) \vec{R}\right ) \\ & =\vec{g}_{0}-\left ( \vec{\omega }\left ( \vec{\omega }\cdot \vec{R}\right ) -\omega ^{2}\vec{R}\right ) \end{align*}
Then using\begin{equation} \vec{g}\times \vec{g}_{0}=gg_{0}\left ( \sin \varepsilon \right ) \vec{n}\tag{1} \end{equation} Where \(\vec{n}\) is perpendicular to plane of \(\vec{g},\vec{g}_{0}\) which is \(\hat{x}\) in this case. Then the LHS of the above is\begin{align*} \vec{g}\times \vec{g}_{0} & =\left [ \vec{g}_{0}-\left ( \vec{\omega }\left ( \vec{\omega }\cdot \vec{R}\right ) -\omega ^{2}\vec{R}\right ) \right ] \times \vec{g}_{0}\\ & =\vec{g}_{0}\times \vec{g}_{0}-\left ( \vec{\omega }\left ( \vec{\omega }\cdot \vec{R}\right ) \times \vec{g}_{0}\right ) +\left ( \omega ^{2}\vec{R}\times \vec{g}_{0}\right ) \end{align*}
But \(\vec{R}\times \vec{g}_{0}=0\) since they are in same direction, also \(\vec{g}_{0}\times \vec{g}_{0}=0\) and the above becomes\begin{equation} \vec{g}\times \vec{g}_{0}=-\vec{\omega }\left ( \vec{\omega }\cdot \vec{R}\right ) \times \vec{g}_{0}\tag{2} \end{equation} But \[ \vec{\omega }\cdot \vec{R}=\omega R\cos \left ( \frac{\pi }{2}-\lambda \right ) \] Therefore (2) becomes\[ \vec{g}\times \vec{g}_{0}=-\omega R\cos \left ( \frac{\pi }{2}-\lambda \right ) \vec{\omega }\times \vec{g}_{0}\] But \(\vec{\omega }\times \vec{g}_{0}=-\omega g_{0}\sin \left ( \frac{\pi }{2}-\lambda \right ) \hat{x}\), hence the above becomes\[ \vec{g}\times \vec{g}_{0}=\omega R\cos \left ( \frac{\pi }{2}-\lambda \right ) \omega g_{0}\sin \left ( \frac{\pi }{2}-\lambda \right ) \hat{x}\] Now we go back to (1) and apply the definition, therefore \[ \omega R\cos \left ( \frac{\pi }{2}-\lambda \right ) \omega g_{0}\sin \left ( \frac{\pi }{2}-\lambda \right ) \hat{x}=gg_{0}\left ( \sin \varepsilon \right ) \hat{x}\] Or\begin{align*} \omega R\cos \left ( \frac{\pi }{2}-\lambda \right ) \omega g_{0}\sin \left ( \frac{\pi }{2}-\lambda \right ) & =gg_{0}\left ( \sin \varepsilon \right ) \\ \sin \varepsilon & =\frac{\omega R\cos \left ( \frac{\pi }{2}-\lambda \right ) \omega g_{0}\sin \left ( \frac{\pi }{2}-\lambda \right ) }{gg_{0}}\\ & =\frac{R\omega ^{2}\cos \left ( \frac{\pi }{2}-\lambda \right ) \sin \left ( \frac{\pi }{2}-\lambda \right ) }{g} \end{align*}
But \(\sin \left ( \frac{\pi }{2}-\lambda \right ) =\cos \lambda \) and \(\cos \left ( \frac{\pi }{2}-\lambda \right ) =\sin \lambda \) hence the above becomes\begin{equation} \sin \varepsilon =\frac{R\omega ^{2}\sin \lambda \cos \lambda }{g}\tag{3} \end{equation} To find \(g=\left \vert \vec{g}\right \vert \), since \(\vec{g}=\vec{g}_{0}-\left ( \vec{\omega }\left ( \vec{\omega }\cdot \vec{R}\right ) -\omega ^{2}\vec{R}\right ) \), then taking dot product gives\begin{align*} \left \vert \vec{g}\right \vert & =\vec{g}\cdot \vec{g}\\ & =\left [ \vec{g}_{0}-\left ( \vec{\omega }\left ( \vec{\omega }\cdot \vec{R}\right ) -\omega ^{2}\vec{R}\right ) \right ] \cdot \left [ \vec{g}_{0}-\left ( \vec{\omega }\left ( \vec{\omega }\cdot \vec{R}\right ) -\omega ^{2}\vec{R}\right ) \right ] \\ & =g_{0}^{2}-2\vec{g}_{0}\cdot \left ( \vec{\omega }\left ( \vec{\omega }\cdot \vec{R}\right ) -\omega ^{2}\vec{R}\right ) +\overset{\text{ignore. All }\omega ^{4}\text{ powers. too small}}{\overbrace{\left ( \vec{\omega }\left ( \vec{\omega }\cdot \vec{R}\right ) -\omega ^{2}\vec{R}\right ) \cdot \left ( \vec{\omega }\left ( \vec{\omega }\cdot \vec{R}\right ) -\omega ^{2}\vec{R}\right ) }}\\ & \approx g_{0}^{2}-2\vec{g}_{0}\cdot \left ( \vec{\omega }\left ( \vec{\omega }\cdot \vec{R}\right ) -\omega ^{2}\vec{R}\right ) \\ & =g_{0}^{2}-\left ( -2g_{0}\hat{z}\right ) \cdot \left ( \left ( \omega \cos \lambda \hat{y}+\omega \sin \lambda \hat{z}\right ) \left ( \omega R\cos \left ( \frac{\pi }{2}-\lambda \right ) \right ) -\omega ^{2}R\hat{z}\right ) \\ & =g_{0}^{2}-\left ( -2g_{0}\hat{z}\right ) \cdot \left ( \left ( \omega \cos \lambda \hat{y}+\omega \sin \lambda \hat{z}\right ) \left ( \omega R\sin \lambda \right ) -\omega ^{2}R\hat{z}\right ) \\ & =g_{0}^{2}-\left ( -2g_{0}\hat{z}\right ) \cdot \left ( \omega ^{2}R\sin \lambda \cos \lambda \hat{y}+\left ( \omega ^{2}R\sin ^{2}\lambda -\omega ^{2}R\right ) \hat{z}\right ) \\ & =g_{0}^{2}-\left ( -2g_{0}\left ( \omega ^{2}R\sin ^{2}\lambda -\omega ^{2}R\right ) \right ) \\ & =g_{0}^{2}+2g_{0}\omega ^{2}R\sin ^{2}\lambda -2g_{0}\omega ^{2}R\\ & =g_{0}^{2}+2g_{0}\omega ^{2}R\left ( 1-\cos ^{2}\lambda \right ) -2g_{0}\omega ^{2}R\\ & =g_{0}^{2}+2g_{0}\omega ^{2}R-2g_{0}\omega ^{2}R\cos ^{2}\lambda -2g_{0}\omega ^{2}R\\ & =g_{0}^{2}-2g_{0}\omega ^{2}R\cos ^{2}\lambda \end{align*}
Therefore (3) becomes\[ \sin \varepsilon =\frac{R\omega ^{2}\sin \lambda \cos \lambda }{g_{0}^{2}-2g_{0}\omega ^{2}R\cos ^{2}\lambda }\] Since \(\varepsilon \) is small, then \(\sin \varepsilon \approx \varepsilon \), therefore\[ \varepsilon \approx \frac{R\omega ^{2}\sin \lambda \cos \lambda }{g_{0}^{2}-2g_{0}\omega ^{2}R\cos ^{2}\lambda }\] The solutions has an extra \(g_{0}\) in the denominator. I am not sure why. I will what is given for part(2) to plot it.
This plot shows the maximum \(\varepsilon \) is at \(\lambda =45^{0}\). Here is the code used and the plot generated
