Define (i) velocity, (ii) angular velocity - Leaving Cert Physics - Question 6 - 2006

For a particle in uniform circular motion:

• Linear velocity $v$ is related to angular velocity $\omega$ by the relationship:

$$v = r \omega$$

where $r$ is the radius of the circular path. Thus, we can express angular displacement $\theta$ in terms of $v$ and $r$ as:

$$\theta = \frac{s}{r}$$

where $s$ is the arc length that the particle travels along the circular path. This leads to the conclusion that:

$$v = r \frac{d\theta}{dt}$$

The velocity $v$ of the ball can be calculated using the formula:

$$v = r \omega$$

Here, $r = 0.7$ m and $\omega = 10$ rad/s. Therefore,

$$v = (0.7 \text{ m})(10 \text{ rad/s}) = 7.0 \text{ m/s}$$

To find the time period $T$ for one complete revolution:

$$T = \frac{2\pi}{\omega} = \frac{2\pi}{10} = \frac{2\pi}{10} \approx 0.63 \text{ s}$$

At position A, the forces acting on the ball are:

1. Weight (W) acting downwards, equal to the gravitational force $mg$.
2. Tension (T) in the string acting upwards, which provides the necessary centripetal force to keep the ball in circular motion.

(Diagram should illustrate these forces with arrows indicating direction.)

When the ball is released from position A, its initial height is 130 cm above the ground. The maximum height can be calculated as follows:

Using conservation of energy, we know:

$$v_{u}^2 = 2g h$$

Where:

• $v_{u}$ is the initial velocity (the same as the ball's upward velocity, which is 7.0 m/s).
• $g$ is gravity (9.8 m/s$^2$).
• $h$ is the height gained. Solving this gives:

$$130 ext{ cm} + h = \frac{(7.0)^2}{2 \cdot 9.8}$$

The time taken for the ball to fall after being released can be calculated using:

$$t = \sqrt{\frac{2h}{g}}$$

Where h is the height of the ball from the ground and g = 9.8 m/s$^2$. The calculations lead to:

• $h = 1.30$ m Thus,

$$t = \sqrt{\frac{2 \cdot 1.30}{9.8}} = 0.51 \text{ s}$$