8. (a) A smooth particle of mass 2 kg performs uniform horizontal circular motion on the inside surface of a smooth hollow sphere of radius 1.7 m - Leaving Cert Applied Maths - Question 8 - 2021

In the diagram, we would illustrate the following forces acting on the particle:

1. The gravitational force downward: $mg$ where $m$ is the mass of the particle (2 kg) and $g$ is the acceleration due to gravity (approximately 9.81 m/s²).
2. The normal force $R$ exerted by the surface of the sphere, acting perpendicular to the surface.
3. The component of the gravitational force acting towards the center of the sphere.

We can resolve the forces in the vertical direction:

Using the relation: $R \cos{\alpha} = mg \,\text{and}\, R \sin{\alpha} = 20$

Given that: $R = \frac{20}{\sin{\alpha}}$

Using trigonometric relations for $\, \alpha$: $R \sin{\alpha} = 20 \Rightarrow R = \frac{20}{\frac{15}{17}} = 22.7 \, N$

To find the speed of the particle, we can use the relation for centripetal acceleration:

$T = \frac{mv^2}{r}$

Here, $T$ is the tension acting on the particle. Using the previous values:

$T = 22.7 \, N$

This leads to:

$v = \sqrt{\frac{68 \times 9.81}{r}} = 2.07 \, m/s$

The force diagram should include:

1. The tension $T$ in the string acting upwards along the direction of the string.
2. The weight of the object, acting downward as $mg$ where $m$ is the mass (7 kg) and $g$ is the gravitational acceleration.
3. The centripetal force required for circular motion.

Using the equation for the tension in the string at uniform circular motion:

$T \sin{\alpha} = 70$

Substituting to find tension: $T = 87.5 \, N$

Then we can find the angular velocity as:

$T \cos{\alpha} = \frac{7 \times v^2}{3}$

Substituting values, we can find: $v = 22.5 \, m/s$