8. (a) A particle describes a horizontal circle of radius 2 metres with constant angular velocity $ heta$ radians per second - Leaving Cert Applied Maths - Question 8 - 2008

The centripetal force can be calculated using the formula:

F_c = m rac{v^2}{r}

Where:

• $m = 3$ kg (mass of the particle)
• $v = 5$ m/s (speed)
• $r = 2$ m (radius)

Substituting the values:

F_c = 3 rac{5^2}{2} = 3 imes 12.5 = 37.5 ext{ N}

To find the radius $r$, we can use the geometry of the situation. We have a height of 2 cm and a diameter of 10 cm. Thus, the radius of the bowl is:

ext{radius of the bowl} = rac{10 ext{ cm}}{2} = 5 ext{ cm}

Using the Pythagorean theorem:

r = ext{hypotenuse} = rac{ ext{diameter}}{2} = ext{radius of bowl}

Calculating:

r = ext{hypotenuse} = rac{ ext{diameter}^2 - ext{height}^2}{ ext{height}}

This simplifies the calculation giving:

$r = ext{hypotenuse} = ext{sqrt}(5^2 - 2^2) = ext{sqrt}(25 - 4) = ext{sqrt}(21) ext{ cm}$

The forces acting on the particle in this scenario include:

1. Gravitational Force ($F_g = mg = 2g$ downwards)
2. Normal Reaction Force ($R$ perpendicular to the surface of the bowl)

In a diagram, represent:

• The particle at the center of the circle,
• The gravitational force pointing down,
• The normal force acting radially outward from the surface.

This can be visualized as vectors from the particle indicating the direction of each force.

Using the vertical component of forces:

From equilibrium, we have:

$R ext{sin}( heta) = 2g$

Where $heta = ext{angle of the radius to the horizontal}$. Since we want to find R using the components:

R ext{cos}( heta) = rac{100}{3} ext{ N}

Substituting the values back into these equations yields:

ightarrow R = 100/3$$

Given the radius $r$ found previously and the force:

Using the centripetal force equation:

$F_c = m r heta^2$

You can find the angular velocity:

rac{F_c}{R} = 2 heta^2 ightarrow heta = ext{sqrt}rac{100}{3}$$ Thus, the angular velocity is given as: $$ heta = rac{10}{3} ext{ rad/s}$$