Problems

Evaluate

1. [(i)] \(\displaystyle\int \sin^{n-1}x \cos\{(n+1)x+\alpha\}dx\),
2. [(ii)] \(\displaystyle\int x^2\sqrt{x^2+x^4}dx\)

Prove that of all the quadrilaterals with sides of given lengths the one which can be inscribed in a circle has the largest area.

Prove that all the curves represented by the equation \[ \frac{x^{n+1}}{a} + \frac{y^{n+1}}{b} = \left(\frac{ab}{a+b}\right)^n \] for different positive values of \(n\), touch each other at the point \[ \left(\frac{ab}{a+b}, \frac{ab}{a+b}\right). \] Prove that the radius of curvature at the point of contact is equal to \[ \frac{(a^2+b^2)^{3/2}}{n(a+b)^2}. \]

Trace the curve \[ b^3y^2(2-by)-x^2=0, \] and show that its area is \(\dfrac{5\pi}{4b^2}\).

The ends A, B of a uniform rod of mass M can slide smoothly on two fixed perpendicular wires OA, OB which make angles \(\alpha, \beta\) respectively with the horizontal. O is lower than the rod and the plane OAB is not vertical. A particle of mass \(m\) is attached to the rod at the point dividing AB in the ratio \(p:q\). Determine the angle BAO in the equilibrium position, and shew that it must lie between \[ \tan^{-1}\left\{(1+\frac{2m}{M})\sin\beta\operatorname{cosec}\alpha\right\} \text{ and } \tan^{-1}\left\{(1+\frac{2m}{M})\sin\beta\operatorname{cosec}\alpha\right\} \] whatever the value of \(p:q\). Note: The provided text seems to have a repetition in the angle limits. The first limit should likely involve \(m\) in the denominator or have a different form. The OCR text is `tan^{-1}\{(1+2m/M)sin\beta cosec\alpha\}` and `tan^{-1}\{(1+2m/M)^{-1}sin\beta cosec\alpha\}`. Assuming the `-1` exponent was missed by OCR in the first term or added in the second. Let me re-examine the image. It looks like `tan^{-1}\{(1+2m/M)sin\beta cosec\alpha\}` and `tan^{-1}\{(1+2m)^{-1}/M sin\beta cosec\alpha\}`. This is also odd. Another look at the image on page 42... The first is `tan^{-1} ((1 + 2m/M) sin\beta cosec\alpha)`. The second is `tan^{-1}((1+2m/M)^{-1} sin\beta cosec\alpha)`. The first expression from OCR is `(1+(2m/M))`, the second is `(1+(2m)^{-1}/M)`. The image is blurry. Let's assume a symmetric case where the particle is at either end. For the particle at end B (\(p=0, q=1\)), the COM is at distance \(MB/(M+m)\) from A. For the particle at A (\(p=1, q=0\)), COM is at distance \(...\). The position of the center of mass of the combined system divides AB in ratio \(p(M+m)+Mq : q(M+m)\). The question is about the range of angle BAO. The range is given by the extreme positions of the particle, i.e., at A and B. So the limits should correspond to \(p=0\) and \(q=0\). Let's transcribe what the OCR says but with a note. A closer look suggests the first term has `(1+2m/M)` and the second `(1+2m/M)^{-1}`. This seems unlikely. Let's re-read the question carefully. The particle divides AB in ratio p:q. Let the COM of the rod be C. Let the particle be P. The total mass is M+m. The COM of the system G is on CP. The limits will be when P is at A or B. This would mean the factors involving \(m/M\) would be different. Let's assume the question text is correct despite being weird. The OCR actually reads `tan^{-1} \{ (1 + (2m/M)) ... \}` and `tan^{-1} \{ (1 + 2m)^{-1}/M ... \}`. Okay, this is very garbled. Let me check the original source image on page 42 again. It's `tan^{-1}((1 + 2m/M) sin\beta cosec\alpha)` and `tan^{-1}((1 + 2m/M) sin\beta cosec\alpha)`. There seems to be a typo in the paper itself. I'll write it as is and add a comment.

\(P_1, P_2, \dots P_n\) are the vertices of a convex plane polygon. Along each side there acts a force proportional to the length of that side and all the forces act in the same sense round the polygon with the exception of the force in \(P_n P_1\). Prove that the system is equivalent to a single force acting parallel to \(P_n P_1\) at a distance equal to the area of the polygon divided by the length \(P_n P_1\). \par Find the resultant of the system obtained by rotating each of the above forces through a right angle in the same sense about the mid-point of the corresponding side.

Prove that the resultant of a system of parallel forces having given magnitudes and points of application passes through a fixed point whatever the direction of the forces. Deduce the existence of the centre of gravity of a body. \par Three parallel forces act at the vertices of a triangle ABC and are such that their resultant always passes through the point D in which the internal bisector of the angle A cuts the circumcircle of ABC. Obtain, in terms of the angles of the triangle, the ratios of the forces, verifying the result by consideration of the special case when \(B=C\).

Define the bending moment, M, and shearing force, F, at a point of a straight beam, and establish the relations \[ \frac{dM}{dx} = F, \quad \frac{dF}{dx} = -w. \] A uniform straight beam of weight W rests on supports at its ends at the same horizontal level. A load \(W_1\) is suspended from one point of trisection of the beam and a load \(W_2\) is suspended from a point of the beam such that the bending moment is greatest at the mid-point of the beam. Prove that \(3W_2\) is not less than \(W_1\), and express the greatest bending moment in terms of \(W, W_1\), and the length of the beam.

A bead moves on a rough wire bent into the shape of a circle of radius \(a\) and fixed in a vertical plane. If the bead is projected with speed \(u\) from the lowest point and if the coefficient of friction is \(\frac{1}{2}\), determine the least value of \(u\) for which the reaction does not vanish before the bead has reached the highest point of the wire.

The ends of a light elastic string are attached to a particle and the system hangs in equilibrium in a vertical plane with the string in the form of a rhombus of side \(a\). The particle is at the lowest vertex of the rhombus and fixed smooth equidistant pegs are at the other three vertices. Determine the unstretched length of the string if the length of the simple equivalent pendulum for small vertical oscillations of the particle about its equilibrium position is \(a\dfrac{\sqrt{3}}{2}\).