Two rough planes inclined to the horizontal at angles \(\alpha\) intersect in a horizontal line, forming a V. A thin rod, whose centre of gravity divides it into segments of lengths \(l_1\) and \(l_2\), rests in equilibrium with its extremities on the planes, the rod being perpendicular to their line of intersection and making an angle \(\theta\) with the horizontal. The angle \(\theta\) is measured positive when the portion \(l_2\) is higher than the portion \(l_1\). Show that if \(\lambda < \alpha\), where \(\lambda\) is the angle of friction, the angle \(\theta\) must lie between the values given by the equations \[ \tan\theta = (\sin^2\alpha-\sin^2\lambda)^{-1/2}\left[\frac{l_1-l_2}{l_1+l_2}\sin\alpha\cos\alpha \pm \sin\lambda\cos\lambda\right], \] but that if \(\lambda \ge \alpha\) equilibrium is possible in any position.
A given line \(L\) is perpendicular to a given force \(P\) and to the axis of a given couple \(G\). Show that the system consisting of \(P\) and \(G\) may in general be resolved into a force along \(L\) and another force, and give a construction for the point of intersection of the second force and the plane through \(P\) perpendicular to \(L\). Hence or otherwise show that given any line \(L\) and a system of forces which does not reduce to a single force or a couple, the system may be resolved uniquely into a non-zero force along \(L\) and a second force, unless \(L\) is parallel to the central axis or the moment of the system about \(L\) is zero, in which cases the resolution is impossible. Show further that if the system reduces to a single force or a couple, the resolution is impossible unless \(L\) is coplanar with the force or parallel to the plane of the couple, in which cases the resolution is possible in an infinity of ways.
A particle is projected from a point \(O\) at an angle \(\phi\) with the horizontal in a medium which causes a resistance proportional to the velocity. It meets the horizontal plane through \(O\) in a point \(P\), and at \(P\) the direction of motion of the particle makes an angle \(\omega\) with the horizontal. Prove that when \(\phi\) is such that \(OP\) is a maximum for a given velocity of projection, the values of \(\phi\) and \(\omega\) are connected by the relation \[ \phi+\omega = \frac{1}{2}\pi. \] Deduce that for this trajectory \(\phi < \frac{1}{4}\pi\).
A uniform rod \(AB\), of mass \(m\) and length \(a\), is free to turn about a fixed point \(A\). The end \(B\) is connected by an elastic string, of natural length \(l_0\) and modulus \(\lambda\), to a point \(C\) distant \(d\) from \(A\) and vertically above it. Show that the steady motions, in which the rod rotates with angular velocity \(\omega\) about a vertical axis and makes a fixed angle \(\alpha\) with the vertical, are stable. Show that to any given value of \(\alpha\) there is in general one and only one value of \(\omega\), but that if \[ \lambda = \frac{mgl_0(a^2+d^2)^{3/2}}{2d(a^2+d^2)^{3/2}-l_0^3} \] the steady motion \(\alpha=\pi/2\) is possible with any value of \(\omega\); and show that in this case the period of a small oscillation about the steady state is \[ 2\pi\left[\omega^2 + \frac{3\lambda d^2}{m(a^2+d^2)^2}\right]^{-1/2}. \]
State the principle of virtual work as applied to impulses. Four heavy uniform rods, smoothly jointed together at their ends, rest on a smooth horizontal table in the form of a square \(ABCD\), and the joints \(A, C\) are connected by a light inextensible strut \(AC\). A blow of impulse \(I\) is applied at \(A\), in the direction and sense of \(AB\). Show that the impulsive thrust in the strut is \(I/\sqrt{2}\).
If \(\mathbf{A, B, C}\) are three linearly independent vectors, show that necessary and sufficient conditions for a vector \(\mathbf{P}\) to be equal to a vector \(\mathbf{Q}\) are \[ \mathbf{P.A=Q.A, P.B=Q.B, P.C=Q.C}. \] A point \(O\) of a rigid body is fixed. If \(\mathbf{v_1, v_2}\) are vectors representing the velocities of any two points \(K_1, K_2\), show that \[ \mathbf{v_1.r_2+v_2.r_1=0}, \] where \(\mathbf{r_1, r_2}\) are the vectors \(\mathbf{OK_1, OK_2}\). Show further that, provided \(\mathbf{v_1}\) and \(\mathbf{v_2}\) are not parallel, the equations \[ \mathbf{\Omega\wedge r_1 = v_1, \quad \Omega.v_2 = 0} \] define a unique vector \(\mathbf{\Omega}\) which has the property that \[ \mathbf{\Omega\wedge r_2 = v_2}, \] and that if \(\mathbf{v_r, r_r}\) refer to any other point \(K_r\) of the rigid body then \[ \mathbf{\Omega\wedge r_r=v_r}. \] [The notation \(\mathbf{X.Y}\) denotes a scalar product, \(\mathbf{X\wedge Y}\) a vector product.]
Obtain Euler's equations for the motion of a rigid body about a fixed point in the form \[ A\dot\omega_1 - (B-C)\omega_2\omega_3 = L, \] and two similar equations. Show that \((B-C)\omega_2\omega_3\) is equal to the sum of the moments round the axis \(O\xi\) of the centrifugal forces of the separate element of the body considered as arising from their motion of rotation about the instantaneous axis. Show further that the moment of the same centrifugal forces about the axis of resultant angular momentum is zero.
A particle of mass \(m\) at the point \((x,y)\) is acted on by a force whose rectangular components are \(X,Y\). It is found experimentally that the equations of motion of the particle in Cartesian co-ordinates are \[ m\frac{d}{dt}(\beta\dot x) = X, \quad m\frac{d}{dt}(\beta\dot y)=Y, \] where \[ \beta = \left(1-\frac{\dot x^2+\dot y^2}{c^2}\right)^{-1/2} \] and \(c\) is a constant. Show that the equations of motion in polar co-ordinates can be written in the form \[ mc^2\frac{d\beta}{dt} = P\dot r + Qr\dot\theta, \quad m\frac{d}{dt}(r^2\dot\theta\beta)=Qr, \] where \(P,Q\) are the components of force along and perpendicular to the radius vector respectively. Show that if \(P=-m\mu r^{-2}, Q=0\), the polar equation of the orbits is of the form \[ \frac{1}{r} = A+B\cos(\gamma\theta-C), \] where \[ \gamma = \left(1-\frac{\mu^2}{c^2h^2}\right)^{1/2}, \] \(h\) is the constant value of \(r^2\dot\theta\beta\) and \(A,B,C\) are constants depending on the initial conditions.
Obtain the conditions which must be satisfied by the electric intensity and the electric displacement at the interface between two dielectrics. What modifications are necessary if there is a surface charge located at the interface? The distance between the plates \(A_1, A_2\) of a parallel plate condenser is \(a_1+a_2\), and the space between them is entirely filled with two slabs of dielectric \(S_1\) and \(S_2\), of thicknesses \(a_1\) and \(a_2\), whose sides are parallel to the faces. The dielectric constants are \(K_1\) and \(K_2\). The slabs are slightly conducting, and have specific resistances \(r_1\) and \(r_2\). At the instant \(t=0\), the plate \(A_1\) (in contact with \(S_1\)) is connected to the positive pole of a battery of electromotive force \(V\), and the plate \(A_2\) is simultaneously connected to the negative pole. Prove that a charge accumulates at the interface between \(S_1\) and \(S_2\), and that at time \(t\) the surface density at the interface is \[ \frac{V}{4\pi}\frac{K_2r_2-K_1r_1}{a_1r_1+a_2r_2}(1-e^{-\alpha t}), \] where \[ \alpha = \frac{4\pi(a_1r_1+a_2r_2)}{r_1r_2(a_1K_2+a_2K_1)}. \] (The internal resistance of the battery and connecting wires and the effects of electro-magnetic induction are to be neglected.)
The magnetic vector-potential \(\mathbf{U}\) in a magnetic field \(\mathbf{H}\) is defined to be any vector function satisfying the relation \[ \mathbf{H}=\text{curl }\mathbf{U}. \] Show that the field at a point distant \(r\) from a doublet at \((x,y,z)\) of strength represented by the vector \(\boldsymbol{\mu}\) may be derived from a vector-potential given by \[ \mathbf{U} = \boldsymbol{\mu}\wedge\text{grad}\left(\frac{1}{r}\right), \] and hence that the vector-potential due to a normally magnetised magnetic shell of uniform strength \(\phi\) may be taken to be \[ \mathbf{U} = \phi\oint\frac{1}{r}d\mathbf{s}, \] where \(d\mathbf{s}\) is an element of arc of the boundary curve of the shell and the integral is taken round the boundary curve. Deduce that the mutual potential energy of two currents of intensities \(i,i'\) in closed circuits may be expressed in the form \[ -ii'\iint\frac{1}{r}d\mathbf{s}.d\mathbf{s}', \] the integrals being taken round the circuits.