A particle describes a horizontal circle of radius $r$ m with uniform angular velocity $\omega$ radians per second - Leaving Cert Applied Maths - Question 8 - 2007

Using the value of $r$ from the previous part, we can find the angular speed $\omega$:

From the first formula $v = r \omega$, substituting in:

$2 = 1 \cdot \omega$

So,

$\omega = 2\text{ rad/s}$.

We will determine the reaction force using the given angle $\alpha$ where $\tan \alpha = \frac{3}{4}$.

To find $R$, the reaction force, we consider the forces acting on the particle. The vertical component of the forces must balance:

$15 \cos \alpha + R = 20$

Using $\tan \alpha = \frac{3}{4}$, we find:

• $\sin \alpha = \frac{3}{5}$
• $\cos \alpha = \frac{4}{5}$

Substituting these values:

$15 \cdot \frac{4}{5} + R = 20$

Calculating:

$12 + R = 20 \implies R = 20 - 12 = 8 \text{ N}$

Using the equilibrium of forces, we apply:

$15 \sin \alpha = m r \omega^2$

Here, we know $m = 2$ kg and $r = 0.5$ m, so:

Substituting values:

$15 \cdot \frac{3}{5} = 2 \cdot 0.5 \omega^2$

Calculating gives:

$9 = 1 \cdot \omega^2$

Thus,

$\omega^2 = 9 \implies \omega = 3 \text{ rad/s}$.