Problems

Solution:

1. Since we have \(h'' + \omega^2 h = 0\) we must have that \(h(t) = A \cos \omega t + B \sin \omega t\). The initial conditions tell us that \(A = 0\) and \(B = L\), so \(h(t) = L \sin \omega t\).
   

   + - ↺
   Therefore we can see the angle at \(B\) is \(\omega t\) and so \(P\) has \(y\)-coordinate \((1-s)L \sin \omega t\) and \(x\)-coordinate \(sL \cos \omega t\)
2. If the position is \(\binom{sL \cos \omega t}{(1-s) L \sin \omega t}\) then the acceleration is \(-\omega^2 \binom{sL \cos \omega t}{(1-s) L \sin \omega t}\)
3. 

   + - ↺
   \begin{align*} \text{N2}(\rightarrow): && - F\cos \omega t + N \sin \omega t &= -m\omega^2 sL \cos \omega t\\ \text{N2}(\uparrow): && -mg + F\sin \omega t + N \cos \omega t &= -m\omega^2 (1-s) L \sin \omega t \\ \Rightarrow && \begin{pmatrix} \cos \omega t & -\sin \omega t \\ \sin \omega t & \cos \omega t \end{pmatrix} \begin{pmatrix} F \\ N \end{pmatrix} &= \begin{pmatrix} m\omega^2 s L \cos \omega t \\ mg - m\omega^2(1-s)L \sin \omega t \end{pmatrix} \\ \Rightarrow && \begin{pmatrix}F \\ N \end{pmatrix} &= \begin{pmatrix} \cos \omega t & \sin \omega t \\ -\sin \omega t & \cos \omega t \end{pmatrix} \begin{pmatrix} m\omega^2 s L \cos \omega t \\ mg - m\omega^2(1-s)L \sin \omega t \end{pmatrix} \\ \Rightarrow && N &= m \omega^2 s L (-\sin \omega t \cos \omega t) + mg \cos \omega t - m \omega^2 (1-s)L \sin \omega t \cos \omega t \\ &&&=mg \cos \omega t - m \omega^2 L \sin \omega t \cos \omega t \\ &&&= mg \cos \omega t \left (1 - \frac{L \omega^2}{g} \sin \omega t \right) \\ &&&= mg (1 - k \sin \omega t) \cos \omega t \\ \Rightarrow && F &= m \omega^2 s L \cos^2 \omega t + mg \sin \omega t - m \omega^2 (1-s) L \sin ^2 \omega t \\ &&&= m \omega^2 s L + mg \sin \omega t - m \omega^2 L \sin^2 \omega t \\ &&&= mg \frac{\omega^2 L}{g} s + mg(1-\frac{\omega^2 L}{g} \sin \omega t)\sin \omega t \\ &&&= mg sk + mg(1-k \sin \omega t) \cos \omega t \tan \omega t \\ &&&= mgsk + N \tan \omega t \end{align*}
4. The particle will not slip if \(F < \tan \alpha N\). When \(t = 0\), \(N = mg, F = mgsk\), but clearly \(sk < \tan \alpha \Rightarrow mgsk = F < \tan \alpha mg = \tan \alpha N\). The particle will slip when: \(F > \tan \alpha N\), but we have \(F = mgsk + N \tan \omega t\). Clearly when \(\omega t = \alpha\) we have reached a point where \(F > \tan \alpha N\). Therefore we must slip before we reach this point, ie at a point where the plank makes an angle of less than \(\alpha\) to the horizontal. Notice also that \(N\) changes sign when \(1-k \sin \omega t = 0\), however, to do this \(N\) must become very small, smaller than \(mgsk\), therefore we must slip before this point too. Since we slip before either condition occurs, we must be in a position when \(N\) is positive AND the plank still makes a shallow angle.