8. A particle describes a horizontal circle of radius 1.5 metres with uniform angular velocity $\omega$ radians per second - Leaving Cert Applied Maths - Question 8 - 2016

The time period $T$ for one complete revolution is given by:

$T = \frac{2\pi}{\omega}$

Substituting the value of $\omega$:

$T = \frac{2\pi}{2} = \pi \text{ seconds}$

The centripetal force $F$ can be calculated using the formula:

$F = m \frac{v^2}{r}$

Given the mass $m = 2$ kg, velocity $v = 3$ m/s, and radius $r = 1.5$ m:

$F = 2 \frac{3^2}{1.5} = 2 \cdot \frac{9}{1.5} = 12 \text{ N}$

To find the radius $r$, we apply the Pythagorean theorem in the triangle formed:

$r^2 + 8^2 = 17^2$

Solving for $r$ gives:

$r = \sqrt{17^2 - 8^2} = \sqrt{289 - 64} = \sqrt{225} = 15 \text{ cm}$

Using the equation for the force components, we find:

$R \sin \alpha = 10$

And from the triangle:

$R \cdot \frac{8}{17} = 10\implies R = 21.25 ext{ N}$

Using the centripetal force relation:

$R \cos \alpha = m r \omega^2$

Substituting the known values:

$21.25 \cdot \frac{15}{8} = 1 \cdot 15 \cdot \omega^2$

Solving for $\omega$:

$\omega = \sqrt{\frac{21.25 \cdot 15/8}{15}} = 11.18 \text{ radians per second}$