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
