Circular Motion 2

Two particles of mass \(M\) and \(m\) \((M>m)\) are attached to the ends of a light rod of length \(2l.\) The rod is fixed at its midpoint to a point \(O\) on a horizontal axle so that the rod can swing freely about \(O\) in a vertical plane normal to the axle. The axle rotates about a vertical axis through \(O\) at a constant angular speed \(\omega\) such that the rod makes a constant angle \(\alpha\) \((0<\alpha<\frac{1}{2}\pi)\) with the vertical. Show that \[ \omega^{2}=\left(\frac{M-m}{M+m}\right)\frac{g}{l\cos\alpha}. \] Show also that the force of reaction of the rod on the axle is inclined at an angle \[ \tan^{-1}\left[\left(\frac{M-m}{M+m}\right)^{2}\tan\alpha\right] \] with the downward vertical.

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The accelerations of \(M\) and \(m\) are \(l \sin \alpha \omega^2\) and \(-l \sin \alpha \omega^2\) so the forces \(R_M\) and \(R_m\) are \(M\binom{l \sin \alpha \omega^2}{g}, \,m \binom{-l \sin \alpha \omega^2}{g}\). Since the axle is rotating freely, the moment about \(O\) for the rod must be \(O\). The moment for \(M\) will be \(M\binom{l \sin \alpha\omega^2}{g} \cdot \binom{-l\cos \alpha}{l \sin \alpha} = lM\sin\alpha (g - l \cos \alpha\omega^2)\). The moment for \(m\) will be \(m \binom{-l \sin \alpha\omega^2}{g} \cdot \binom{-l\cos \alpha\omega^2}{l \sin \alpha} = lm \sin \alpha(g+l \cos \alpha\omega^2)\) Therefore \begin{align*} && lM\sin\alpha (g - l \cos \alpha\omega^2) &= lm \sin \alpha(g+l \cos \alpha\omega^2) \\ && M(g - l \cos \alpha \omega^2) &= m(g + l \cos \alpha \omega^2 ) \\ \Rightarrow && g(M-m) &= l \cos \alpha (M+m) \omega^2 \\ \Rightarrow && \omega^2 &= \left (\frac{M-m}{M+m} \right) \frac{g}{l \cos \alpha} \end{align*} as required. The total force on the rod is \(\mathbf{0}\) so the reaction force must be \(M\binom{l \sin \alpha \omega^2}{g}+ \,m \binom{-l \sin \alpha \omega^2}{g} = \binom{l \sin \alpha \omega^2 (M-m)}{(M+m)g}\) Therefore the angle this makes with downward vertical is: \begin{align*} \theta &= \tan^{-1} \left ( \frac{l \sin \alpha \omega^2 (M-m)}{(M+m)g} \right) \\ &= \tan^{-1} \left ( \frac{l \sin \alpha (M-m)}{(M+m)g} \omega^2\right) \\ &= \tan^{-1} \left ( \frac{l \sin \alpha (M-m)}{(M+m)g} \left (\frac{M-m}{M+m} \right) \frac{g}{l \cos \alpha}\right) \\ &= \tan^{-1}\left[\left(\frac{M-m}{M+m}\right)^{2}\tan\alpha\right] \end{align*} as required.

A smooth sphere of radius \(r\) stands fixed on a horizontal floor. A particle of mass \(m\) is displaced gently from equilibrium on top of the sphere. Find the angle its velocity makes with the horizontal when it loses contact with the sphere during the subsequent motion. By energy considerations, or otherwise, find the vertical component of the momentum of the particle as it strikes the floor.

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Whilst the particle is on the surface of the sphere consider the energy. Letting the height of centre of the sphere by our \(0\) GPE level, the initial energy is \(mgr\) (assuming that the initial speed is so close to \(0\) as to make no difference). When it makes an angle \(\theta\) with the horizontal it's energy will be \(mgr \sin \theta + \frac12 m v^2\). By conservation of energy: \(mgr \sin \theta + \frac12 m v^2 = mgr \Rightarrow v^2 = 2gr(1-\sin \theta)\) \begin{align*} \text{N2}(\text{radially}): && mg \sin \theta - R &= m \frac{v^2}{r} \\ \Rightarrow && R &= mg\sin \theta - \frac{m}{r} 2gr(1-\sin \theta) \\ &&&=mg \l 3\sin \theta - 2 \r \end{align*} Since \(R\) must be positive whilst the particle is in contact with the sphere, the angle \(\theta\) makes with the horizontal when it leaves the sphere is \(\sin^{-1} \frac{2}{3}\). At this point \(v^2 = 2gr(1-\sin \theta) = \frac{2}{3}gr\) Again, considering energy, the initial energy is \(mgr\). The final energy is \(-mgr + \frac12mu_x^2 + \frac12mu_y^2\) When the particle leaves the surface it has speed \(v= \frac23 gr\), so the component \(u_x = \sqrt{v}\sin \theta\). By conservation of energy therefore: \begin{align*} && mgr &= -mgr + \frac12mu_x^2 + \frac12mu_y^2 \\ \Rightarrow && \frac12 u_y^2 &= 2gr - \frac12 u_x^2 \\ &&&= 2gr - \frac12 (\sqrt{v} \sin \theta)^2 \\ &&&= 2gr - \frac12 \frac23gr \sin^2 \theta \\ &&&= 2gr - \frac13gr \frac{4}{9} \\ &&&= \frac{50}{27}gr \\ \Rightarrow && u_y &= \frac{10}{3\sqrt{3}}\sqrt{gr} \end{align*} Therefore vertical component of momentum is \(\displaystyle \frac{10}{3\sqrt{3}}\sqrt{gr}m\)

A firework consists of a uniform rod of mass \(M\) and length \(2a\), pivoted smoothly at one end so that it can rotate in a fixed horizontal plane, and a rocket attached to the other end. The rocket is a uniform rod of mass \(m(t)\) and length \(2l(t)\), with \(m(t)=2\alpha l(t)\) and \(\alpha\) constant. It is attached to the rod by its front end and it lies at right angles to the rod in the rod's plane of rotation. The rocket burns fuel in such a way that \(\mathrm{d}m/\mathrm{d}t=-\alpha\beta,\) with \(\beta\) constant. The burnt fuel is ejected from the back of the rocket, with speed \(u\) and directly backwards relative to the rocket. Show that, until the fuel is exhausted, the firework's angular velocity \(\omega\) at time \(t\) satisfies \[ \frac{\mathrm{d}\omega}{\mathrm{d}t}=\frac{3\alpha\beta au}{2[Ma^{2}+2\alpha l(3a^{2}+l^{2})]}. \]

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The rocket principle states that the thrust generated by the rocket is \(-\frac{\d m}{\d t}u = \alpha \beta u\) This force is acting at a distance \(2a\) from \(O\) and therefore is generating a torque of \(2a \alpha \beta u\) on the system. Let's also consider the moments of inertia about \(O\). The fixed rod will have moment of inertia \(\frac13 M (2a)^2 = \frac43 M a^2\). The rocket will have moment of inertia \(I_{G} + md^2 = \frac1{12}m(t)(2l(t))^2 + m(t) ((2a)^2 + l(t)^2)= \frac43 ml^2+ 4ma^2\). Since our final equation doesn't involve \(m\), lets replace all the \(m\) with \(2al\) to obtain a total \(\displaystyle I = \frac43 Ma^2 + \frac83 \alpha l^3 + 8\alpha la^2\). Since \(\tau\) is constant, we can note that \(I\omega = 2a \alpha \beta u t\) (by integrating) and so \begin{align*} && \dot{\omega} &= \frac{\d }{\d t} \left ( \frac{2a \alpha \beta u t}{ \frac43 Ma^2 + \frac83 \alpha l^3 + 8\alpha la^2} \right) \\ &&&= \frac{\d }{\d t} \left ( \frac{3a \alpha \beta u t}{ 2Ma^2 +4\alpha l^3 + 4 \cdot 3 \cdot \alpha la^2} \right) \\ &&&= \frac{\d }{\d t} \left ( \frac{3a \alpha \beta u t}{ 2[Ma^2 +2\alpha l(l^2 + 3 a^2)]} \right) \\ \end{align*} This is, close, but not quite what they are after since the denominator also has a dependency on \(t\) we wont get exactly what they've asked for