A particle describes a horizontal circle of radius 0.5 m with uniform angular velocity ω radians per second - Leaving Cert Applied Maths - Question 8 - 2009

The time period T for one complete revolution can be calculated using the relationship:

$T = \frac{2π}{ω}$

Substituting in the calculated value of ω:

$T = \frac{2π}{4} = \frac{π}{2} \text{ seconds}$

Using the geometry of the cone and trigonometric properties, we have:

$\tan(30°) = \frac{r}{5}$

Since ( an(30°) = \frac{1}{\sqrt{3}} ):

$\frac{1}{\sqrt{3}} = \frac{r}{5} \implies r = \frac{5}{\sqrt{3}} \text{ cm}$

The diagram should illustrate:

1. The gravitational force acting downward (mg), where m = 2 kg and g = 9.81 m/s².
2. The normal reaction force R acting perpendicular to the surface of the cone.
3. The component of R acting towards the center of the circular motion.
4. The radial force and any other necessary vectors indicating direction and magnitude.

Using the forces involved, we have:

$R \sin(30°) = 2g \text{ (Weight of the particle)}$

Substituting values: $R \cdot \frac{1}{2} = 2 \cdot 2 \cdot 9.81 \implies R = 20 \text{ N}$

Using the relationship derived for angular velocity with the forces in play:

Given that: $R = 40 \text{ N} \text{ (from above)}$

Using the centripetal force:

$R \cos(30°) = \frac{mv^2}{r}$

Solving gives: $40 \cdot \frac{\sqrt{3}}{2} = \frac{(2 \omega r)^2}{r}$

After calculations, we find: $ω = \frac{\sqrt{5}}{60 \sqrt{3}} \text{ rad/s}$