A particle describes a horizontal circle of radius 1.5 metres with uniform speed 6 m s⁻¹ - Leaving Cert Applied Maths - Question 8 - 2018

The centripetal acceleration ( a ) of the particle can be calculated using:

a = rω^2$$ Substituting the known values:

a = 1.5 imes (4)^2 = 24 , ext{m s}^{-2}$$

Centripetal force ( F ) is given by the equation:

F = m r ω^2$$ Using the mass (1.2 kg), radius (1.5 m), and previously calculated angular velocity (4 rad s⁻¹):

F = 1.2 imes 1.5 imes (4)^2 = 28.8 , ext{N}$$

The time taken for one revolution is given by:

T = \frac{2π}{ω} = \frac{2π}{4} = \frac{π}{2} \, ext{s}$$ For ten revolutions:

10T = 10 \times \frac{π}{2} = 5π , ext{s}$$

To find radius r, we use the sine function:

$\sin(30°) = \frac{r}{1}$

Thus:

r = 0.5 \, ext{m}$$

The forces acting on the particle include:

• Gravitational force downwards (mg)
• Tension in the string at an angle of 30° to the vertical The diagram would show the tension force angled above and the gravitational force acting downwards.

Considering vertical forces:

$T \, \sin(60°) = mg$

Where m = 1.4 kg and g = 14 N:

$T = \frac{14}{\sin(60°)} = \frac{14}{\frac{\sqrt{3}}{2}} = \frac{28}{\sqrt{3}} \approx 16.2 \, ext{N}$

Using the equation for angular velocity with the radius:

$T \, \cos(30°) = m r ω^2$

Substituting the values and solving for ω:

$16.2 \, \times 0.5 = 1.4 imes 0.5 imes ω^2$

Solving gives:

$ω ≈ 3.4 \, ext{rad s}^{-1}$