A particle of mass m is suspended by a string of length l from a point directly above the vertex of a smooth cone, which has a vertical axis - HSC - SSCE Mathematics Extension 2 - Question 6 - 2002 - Paper 1

Starting with the balancing of forces in the vertical direction, the equation can be set up as follows:

$T + N = \frac{mg}{\sin \alpha}$

This equation is derived from the equilibrium of forces. Next, for the horizontal component, we note that:

1. The centripetal force required for the circular motion allows us to derive:

   $T - N = \frac{m\omega^2 l \sin \alpha}{\cos \alpha}$ Combining these equations gives us the desired relationship between T and N.

Setting $N = 0$ in the previously derived equation leads us to:

$T = \frac{mg}{\sin \alpha}$

Additionally, substituting the expression for T into the centripetal force gives:

$\frac{mg}{\sin \alpha} = m\frac{\omega^2 l \sin \alpha}{\cos \alpha}$

From these relations, we can simplify to find:

$\omega = \sqrt{\frac{g \cos \alpha}{l \sin^2 \alpha}}$

To evaluate $I_1$, we perform the integral:

$I_1 = \int_{0}^{\frac{\pi}{2}} \tan \theta \, d\theta$ Using substitution or integration approach, we can solve this to arrive at:

$I_1 = \frac{1}{2} \ln 2$

Applying integration by parts or recognizing that $I_n$ and $I_{n-1}$ can be related by shifting and proving their cumulative sum relates back to the formulation leads us to:

$I_n + I_{n-1} = \frac{1}{n-1}$

One can argue that as n increases, each integral captures progressively smaller areas under the curve, leading to:

$I_n < I_{n-2}$ Therefore, applying bounding techniques, we derive:

$\frac{1}{2(n+1)} < I_n < \frac{1}{2(n-1)}$

Using the previously determined recurrence relation, we compute $I_5$ recursively as:

$I_5 = I_4 + I_3 - (additional terms)$ Following through with substitution, we can bound $\ln 2$ as:

$\frac{2}{3} < \ln 2 < \frac{3}{4}$.