A particle describes a horizontal circle of radius 2 metres with uniform angular velocity \( \omega \) radians per second - Leaving Cert Applied Maths - Question 8 - 2014

The speed ( v ) of the particle can be calculated using the relationship:

$v = r \omega$

Substituting the given radius (2 m) and the calculated ( \omega ):

$v = 2 \times 5 = 10 \text{ m s}^{-1}$

The radial acceleration ( a ) can be calculated using the formula:

$a = r \omega^2$

Thus, substituting the values we get:

$a = 2 \times (5)^2 = 50 \text{ m s}^{-2}$

For the conical pendulum, using the sine function:

$\sin(30°) = \frac{r}{1}$

Solving gives:

$r = 0.5 \text{ m}$

Using the vertical component of the tension:

$T \cos(30°) = 20$

From this:

$T = \frac{20}{\cos(30°)} = 20 \times \frac{2}{\sqrt{3}} = 23.09 \text{ N}$

From the geometry, we balance the forces:

$T \cos(60°) = m r \omega^2$

Substituting the known values, we find:

$23.09 \times 0.5 = 2 \times 0.5\omega^2$

Solving this gives:

$\omega^2 = \frac{23.09 \times 0.5}{2 \times 0.5} = 11.545 \Rightarrow \omega = \sqrt{11.545} \approx 3.4 \text{ rad s}^{-1}$