A small particle hanging on the end of a light inextensible string 2 m long is projected horizontally from the point C - Leaving Cert Applied Maths - Question 6 - 2016

Using the relationship of forces at the point of slackness:

The centripetal force needed is directed towards the center and is equal to the tension in the string. At the moment it goes slack, we set up the equations of motion.

Considering the vertical components:

$v^2 = g(4 - 4 \cos \alpha)$

Solving gives:

$7^2 = g(4 - 4 \cos \alpha)$

Substituting $g = 9.8:
$49 = 9.8(4 - 4 \cos \alpha)$

Simplifying results in:

$\cos \alpha = \frac{1}{6}$

Therefore:

$\alpha = \arccos \left(\frac{1}{6}\right) \approx 80.41^\circ.$

For the particle P, the forces acting on it can be expressed using Hooke's law. The extension of the string when it is taut is given by:

$T = 2g$

The oscillation will occur under the restoring force provided by the spring, which leads to a simple harmonic motion (S.H.M.). We can show this by considering the acceleration:

$a = -\frac{T}{m}$

Substituting the tension gives:

$a = -\frac{98 \times \delta}{2}$

This matches the form of S.H.M. with

• $\,\omega^2 = 49$.

Thus, the particle exhibits simple harmonic motion.

In this scenario, the potential energy transitions to kinetic energy as the string becomes slack:

$v^2 = g(4)(0.2)$

Solving gives:

$v = \sqrt{4 \times 9.8 \times 0.2} \approx 5.88 \, m/s.$

The time taken is calculated by considering the motion of the particle from the lowest point to the highest.

Using the equation for time:

$t = \frac{T}{4} + t_1 + t_2$

Where $x = 4 \sin \alpha$, substituting the values leads to:

$t_{1} = 0.0748 \, s \quad\text{and}\quad t_{2} = 0.2469 \, s$

Finally, adding these gives:

$t = 0.5461 \, s.$