8. (a) A particle describes a horizontal circle of radius $r$ metres with uniform angular velocity $\omega$ radians per second - Leaving Cert Applied Maths - Question 8 - 2010

Substituting the value of $r$ into equation (1):

$6 = 3\omega$ Solving for $\omega$ gives:

$\omega = \frac{6}{3} = 2 \text{ radians per second}$

For the conical pendulum, we use the relationship involving the angle $\alpha$ and the component of the forces:

Using the given value:

$\tan \alpha = \frac{4}{3} \implies \sin \alpha = \frac{4}{5}, \cos \alpha = \frac{3}{5}$

Considering vertical and horizontal components of the tension $T$, we have:

1. Vertical component: $T \cos \alpha = mg \implies T \cdot \frac{3}{5} = 3g\implies T = 5g$

Substituting $g \approx 10$ ms$^{-2}$ results in:

$T = 50 \text{ N}$

To find $r$, we can again examine the circular motion: $T \sin \alpha = m\frac{v^2}{r}$ So: $50 \cdot \frac{4}{5} = 3\frac{v^2}{r} \implies 40 = 3 \cdot \frac{v^2}{r}$ Substituting for $v = \omega r = 0.8 \omega$ yields:

$r = 0.8 \text{ m}$

From our earlier calculation, we established that: $T = 50 \text{ N}$

We know: $T \sin \alpha = m \frac{v^2}{r}$ Substituting what we have: $50 \cdot \frac{4}{5} = 3\frac{(0.8\omega)^2}{0.8}$ This simplifies to: $40 = 3 \cdot 0.8 \omega^2 \implies \omega^2 = \frac{40}{2.4}$ Solving gives: $\omega \approx 4.08 \text{ radians per second}$