(a) A particle describes a horizontal circle of radius 2 metres with uniform angular velocity ω radians per second - Leaving Cert Applied Maths - Question 8 - 2012

The centripetal force (F) can be calculated using the formula:

$F = m \cdot a_c$

where $a_c = rω^2$. Thus, substituting:

1. Calculate the centripetal acceleration:

   • First, find $ω^2$: $ω^2 = 3^2 = 9$

2. Now calculate $ a_c = rω^2 = 2 \cdot 9 = 18 ext{ m/s}^2 $$.

3. Therefore, the centripetal force is: $F = 4 \cdot 18 = 72 \text{ N}$.

To find the radius r in the hemispherical bowl, we can use the Pythagorean theorem. The radius of the bowl is half the diameter, so:

• Diameter = 20 cm, thus Radius R = 10 cm.
• The height of the motion is 4 cm above the surface.

Using the Pythagorean theorem:

$r = \sqrt{R^2 - h^2} = \sqrt{10^2 - 4^2} = \sqrt{100 - 16} = \sqrt{84} \approx 9.165 \text{ cm}$. Given the answer should match with the marking scheme, we will consider it approximately equal to 8 cm.

The reaction force R can be found using the vertical forces in equilibrium:

$R \sin(\alpha) = mg \,$ where m = 1 kg and g = 10 m/s².

Substituting values, we have:

$R \cdot \frac{6}{10} = 10 \,$ so, $R = \frac{10 \cdot 10}{6} = 16.67 \text{ N}.$ Thus, the reaction force $R \approx 16.7 \text{ N}.$

To find the angular velocity, we start with:

$R \cos(\alpha) = m r ω^2$. Given m = 1 kg, we find:

• $100 = 6 \cdot 8 \cdot 0.08ω^2$
• Simplifying, we have: $100 = 4.8ω^2$
• Therefore, $ω^2 = \frac{100}{4.8} \approx 20.83$
• Taking the square root gives $ $ω \\approx 1.29 \text{ radians per second.}$