A particle describes a horizontal circle of radius 0.5 metres with uniform angular velocity 3 radians per second - Leaving Cert Applied Maths - Question 8 - 2017

The centripetal acceleration can be calculated using the formula: a = \frac{v^2}{r} Substituting the values we have: a = \frac{(1.5)^2}{0.5} = 4.5 \text{ m/s}^2.

The horizontal force can be found using Newton's second law: F = m a where the mass m = 2 kg and a = 4.5 m/s².

Calculating: F = 2 \times 4.5 = 9 \text{ N}.

The total distance covered in nine revolutions is given by: Distance = 2 \pi r \times 9 Substituting r = 0.5: Distance = 2 \pi (0.5) \times 9 = 9\pi

Now, using the relationship: time = \frac{\text{distance}}{\text{speed}}:

t = \frac{9\pi}{1.5} = 6\pi \text{ seconds}.

Given: \tan \alpha = \frac{r}{42} From the provided information: \tan \alpha = \frac{20}{21} Setting the two equations equal gives: \frac{20}{21} = \frac{r}{42}

Cross multiplying yields: 20 \times 42 = 21r\n840 = 21r\nr = \frac{840}{21} = 40 \text{ cm}.

In the diagram, the forces acting on the particle include:

1. Weight (mg) acting downwards.
2. Normal reaction force (R) acting perpendicular to the surface of the cone.
3. Centripetal force acting towards the center of the circular motion.

Using the equilibrium of forces in the vertical direction: R \cos \alpha = mg\nSubstituting values and knowing $m = 2$ kg and $g = 10$ m/s²: R \cos \alpha = 20 Now we have: R = \frac{20}{\cos \alpha} = R = \frac{20\cdot21}{20} = 29 \text{ N}.

Using the formula for centripetal force: R = \frac{mv^2}{r} Substituting R = 29 N, m = 2 kg, and r = 0.4: 29 = \frac{2v^2}{0.4}

Solving for v: v^2 = \frac{29 \cdot 0.4}{2} = 5.8\nv = \sqrt{5.8} \approx 2.05 \text{ m/s}.