8. (a) A particle describes a horizontal circle of radius 3 m with uniform angular velocity $ ho$ radians per second - Leaving Cert Applied Maths - Question 8 - 2013

To find the time taken for one complete revolution, we use the relationship between linear speed and angular velocity.

The angular velocity $ho$ can be calculated as:

$v = r \cdot \omega$

Rearranging gives:

$\omega = \frac{v}{r} = \frac{6}{3} = 2 \, \text{radians/second}$

Next, the time period $T$ for one revolution is given by:

$T = \frac{2\pi}{\omega} = \frac{2\pi}{2} = \pi \, \text{s}$

To find the value of $r$, we can use the tangent function from trigonometry:

$\tan(45^\circ) = \frac{r}{4}$

Since $\tan(45^\circ) = 1$:

$1 = \frac{r}{4}$

Thus, we find:

$r = 4 \, \text{cm}$

To show the forces acting on the particle, draw a free-body diagram that includes:

• Gravitational force $2g$ acting downwards.
• Normal reaction force $R$ acting perpendicular to the surface of the cone at the position of the particle. The angle of this force will be adjusted according to the angle of the cone.

From the free-body diagram, we can apply the following relationship based on vertical forces:

$R \sin(45^\circ) = 2g$

Substituting $g \approx 10 \, \text{m/s}^2$ leads to:

$R \cdot \frac{1}{\sqrt{2}} = 20 \implies R = 20 \sqrt{2} = 28.3 \, \text{N}$

Using the centripetal motion formula: $R \cos(45^\circ) = \frac{mv^2}{r}$ Substituting $R$ and rearranging gives:

$20 \cdot \frac{1}{\sqrt{2}} = \frac{2v^2}{0.04}$ This leads to:

$v^2 = 0.4$ Taking the square root, we find: $v \approx 0.63 \, \text{m/s}$