A toy motor car consists of a body of mass \(4m\) and four road wheels, each of mass \(m\), radius \(a\), and moment of inertia \(\frac{1}{2}ma^2\). A second motor car is similar except for the addition of a flywheel of mass \(m\) and moment of inertia \(\frac{1}{2}ma^2\) geared to the road wheels in the ratio \(\sqrt{(23)}:1\). Show that, for given initial velocity, the second motor car will travel twice as far as the first up an inclined rough plane before coming to rest.
A thin uniform rod of mass \(m\) and length \(2a\) can turn freely about one end which is fixed, and a circular disk of mass \(12m\) and radius \(\frac{1}{4}a\) can be clamped to the rod so that its centre is on the rod. Show that, for oscillations in which the plane of the disk remains vertical, the length of the equivalent simple pendulum lies between \(2a\) and \(\frac{8}{3}a\).
A rocket in rectilinear motion is propelled by ejecting all the products of combustion of the fuel from the tail at a constant rate and at a constant velocity relative to the rocket. Show that, for a given initial total mass \(M\), the final kinetic energy of the rocket is greatest when the initial mass of fuel is \((1-e^{-2})M\).
The ends \(A\), \(B\) of a light rod \(AB\) are joined by light inextensible strings \(AO\), \(BO\) to a fixed point \(O\), and \(AO\) and \(BO\) are equal in length and perpendicular to each other. If weights \(W_1\) and \(W_2\) are now suspended from \(A\) and \(B\), find the angle to the horizontal that the rod will take up in equilibrium.
A uniform chain of length \(b\) and weight \(w\) per unit length has one end free to slide on a smooth vertical wire and passes over a smooth peg at distance \(a\) from the wire, the whole system being in a vertical plane. To the other end of the chain is attached a weight \(wm\). Show that for equilibrium to be possible \(n + b/a\) must be not less than \(e\).
A loop of light inextensible string \(OABCO\) passes in a vertical plane over a horizontal circular cylinder, and a weight \(W\) is attached at \(O\). The portions \(OA\), \(OC\) are straight and perpendicular to each other and tangential to the cylinder at \(A\) and \(C\), and the weight hangs in equilibrium. A gradually increasing couple is applied to the cylinder and slipping is found to occur when it attains the value \((5 \sqrt{2}/13) W \cdot OA\). Find the coefficient of friction between the string and cylinder.
The angle of elevation of a point \(P\) from an origin \(O\) is \(\theta\), and a particle is projected under gravity from \(O\) with given speed \(V\) to pass through \(P\). Show that in general there are two possible trajectories, and that if \(z_1\) and \(z_2\) are the two angles of projection \(z_1 + z_2 = 90^\circ + \theta\). Prove also that if \(T_1\) and \(T_2\) are the corresponding times required to reach \(P\), then \(gT_1T_2\) depends only on the distance \(OP\).
Solution: Suppose we are projected with speed \(V\) at angle \(\alpha\), then: \(u_x = V \cos \alpha, u_y = V\sin \alpha - g t\) \(x= V \cos \alpha t, y= V \sin \alpha t - \frac12 g t^2 \Rightarrow t = \frac{x}{V \cos \alpha}\) So: \(y = x\tan \alpha - \frac12 \frac{g}{V^2} \sec^2 \alpha \cdot x^2\) Suppose are at the point \(P\) at \((k, k \tan \theta)\), then we must have: \begin{align*} && k \tan \theta &= k \tan \alpha - \frac12 \frac{g}{V^2} \sec^2 \alpha k^2 \\ && 0 &= -k \tan \theta + k \tan \alpha -\frac12 \frac{gk^2}{V^2} (1 + \tan^2 \alpha) \\ && &= k \tan \theta + \frac12 \frac{gk^2}{V^2} - k \tan \alpha + \frac12 \frac{g k^2}{V^2} \tan^2 \alpha \end{align*} This is a quadratic in \(\tan \alpha\) which will (in general) have two solutions for \(\alpha\). Notice that if \(\tan z_1\) and \(\tan z_2\) are the two solutions then note that: \begin{align*} \tan z_1 + \tan z_2 &= \frac{k}{\frac12 \frac{g k^2}{V^2}} \\ &= \frac{2V^2}{gk} \\ \\ \tan z_1 \tan z_2 &= \frac{k \tan \theta}{\frac12 \frac{g k^2}{V^2}} + 1 \\ &= \frac{2V^2 \tan \theta}{gk} + 1 \end{align*} and so \begin{align*} && \tan (z_1 + z_2) &= \frac{\tan z_1 + \tan z_2}{1 - \tan z_1 \tan z_2} \\ &&&= \frac{\frac{2V^2}{gk} }{1 - \left ( \frac{2V^2 \tan \theta}{gk} + 1\right)} \\ &&&= -\frac{\frac{2V^2}{gk}}{\frac{2V^2 \tan \theta}{gk}} \\ &&&= - \cot \theta \\ &&&= \tan (\theta + 90^\circ) \end{align*} Therefore \(z_1 + z_2 = \theta + 90^\circ\) by considering physical options for the projectile. \begin{align*} && \cos \alpha &= \frac{k}{Vt} \\ && \sin \alpha &= \frac{k \tan \theta + \frac12 g t^2}{Vt} \\ \Rightarrow && 1 &= \frac{k^2}{V^2} \frac{1}{t^2} + \frac{k^2 \tan^2 \theta}{V^2 t^2} + gk \tan \theta + \frac14 \frac{g^2}{V^2} t^2 \\ \Rightarrow && 0 &= \frac{k^2}{V^2} (1 + \tan^2 \theta) + (gk \tan \theta -1)t^2 + \frac14 \frac{g^2}{V^2} t^4 \end{align*} This is a quadratic in \(t^2\), with the product of the roots being \(\frac{\frac{k^2}{V^2} (1 + \tan^2 \theta) }{ \frac14 \frac{g^2}{V^2}} = \frac{4}{g^2} (k^2 + k^2 \tan^2 \theta^2)\), therefore \(gT_1T_2 = g \sqrt{ \frac{4}{g^2} (k^2 + k^2 \tan^2 \theta^2)} = 2 \cdot OP\)
The ends \(P\), \(Q\) of a thin straight rod are constrained to move on two straight lines \(OX\), \(OY\) respectively that are perpendicular to each other. If the velocity of \(P\) is constant, prove that the acceleration of any point on the rod is at right angles to \(OX\), and find how it varies for different points of the rod.
A train of mass \(M\) is pulled by its engine against a constant resistance \(R\). The engine works at constant power equal to \(H\) units of work per second. Find the time taken for the velocity to be increased from \(v_0\) to \(v_1\) feet per second.
A smooth wedge of mass \(M\) and inclination \(\alpha\) (\(< 90^\circ\)) has one face in contact with a horizontal plane. A particle of mass \(2M\) is placed on the inclined face and allowed to slide down. Show that the horizontal acceleration of the wedge is \[g\sin 2\alpha/(2 - \cos 2\alpha),\] and find the force exerted on the table during the motion.