A particle describes a horizontal circle of radius 2 m with uniform angular velocity $\omega$ radians per second - Leaving Cert Applied Maths - Question 8 - 2015

The speed $v$ of the particle can be calculated using the formula:

$v = r\omega$

where $r=2$ m is the radius. Substituting the value of $\omega$:

$v = 2 \times \frac{\pi}{3} = \frac{2\pi}{3} \approx 2.1 \text{ m s}^{-1}$

The centripetal acceleration $a$ can be calculated as:

$a = \frac{v^2}{r} = \frac{\left(\frac{2\pi}{3}\right)^2}{2} = \frac{4\pi^2}{18} \approx 2.2 \text{ m s}^{-2}$

To find the tension $T$ in the string, we can use the centripetal force equation:

$T \sin \alpha = \frac{mv^2}{r}$

Substituting known values, where $m = 2 \text{ kg}$, $v = 1.2 \text{ m s}^{-1}$, and $r = 0.25 \text{ m}$:

$T \left(\frac{4}{5}\right) = \frac{2(1.2)^2}{0.25}$

Solving gives:

$T = 14.4 \text{ N}$

The vertical forces acting on the particle can be represented by:

$T \cos \alpha + R = 20$

Substituting the values:

$14.4 \left(\frac{3}{5}\right) + R = 20$

Calculating gives:

$8.64 + R = 20 \implies R = 20 - 8.64 = 11.36 \text{ N}$