The object is now detached from the spring and attached to the end of a string of fixed length 11 cm - Leaving Cert Physics - Question vi, vii, viii, ix - 2022

To find the angular velocity (ω), we use the formula for angular velocity in terms of the period (T):

$\omega = \frac{2\pi}{T}$

Given that the period T is 0.5 s:

$\omega = \frac{2\pi}{0.5} = 4\pi \approx 12.57 \text{ rad s}^{-1}$

Using the linear velocity formula derived earlier, with the radius r = 11 cm = 0.11 m and the calculated angular velocity ω:

$v = r \cdot \omega = 0.11 \cdot 12.57 \approx 1.38 \text{ m s}^{-1}$

The minimum tension (T_min) in the string can be calculated using the centripetal force formula:

$T_{min} = m \cdot g + \frac{m v^2}{r}$

Where:

• m is the mass of the object,
• g is the acceleration due to gravity (9.8 m/s²),
• v is the linear velocity (1.38 m/s),
• r is the radius (0.11 m).

Assuming an object mass of 0.2 kg:

1. Calculate gravitational force: $m \cdot g = 0.2 \cdot 9.8 = 1.96 \text{ N}$
2. Calculate centripetal force: $\frac{m v^2}{r} = \frac{0.2 \cdot (1.38)^2}{0.11} \approx 0.34 \text{ N}$
3. Therefore, total minimum tension:

$T_{min} = 1.96 + 0.34 = 2.30 \text{ N}$

The diagram should include:

• Weight (mg) acting downwards, labelled.
• Tension (T) acting upwards along the string, labelled.

Ensure that the directions of the forces are correct: gravitational force acts downwards and tension acts upwards along the string.