6. (a) A particle of mass m kg lies on the top of a smooth sphere of radius 2 m - Leaving Cert Applied Maths - Question 6 - 2010

At point Q, the total energy is conserved. The speed at point Q can be found using energy principles.

At B, we’ve established the total energy as:

$E = PE_B + KE_B = 0 + \frac{1}{2} mv^2$

At point Q, the height h is 0:

$E = \frac{1}{2} mv_Q^2 + mg(0)$

Setting the total energy at point B equal to that at point Q:

$\frac{1}{2} m (2\sqrt{g})^2 = \frac{1}{2} mv_Q^2$

Thus:

$2mg = \frac{1}{2} mv_Q^2$

Canceling m yields:

$4g = v_Q^2$

Therefore:

$v_Q = \sqrt{4g} = 2\sqrt{g}$

Again substituting g = 9.81 m/s²:

$$v_Q = \sqrt{4 \cdot 9.81} \approx 6.26 , \text{m/s}.$

In simple harmonic motion, the maximum speed (v_max) can be calculated using the formula:

$v_{max} = \omega A$

Where:

• A is the amplitude (0.75 m)
• (\omega) is the angular frequency calculated as:

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

Given the period (T) is 4 s, we have:

$\omega = \frac{2\pi}{4} = \frac{\pi}{2} \, \text{rad/s}$

Now substituting:

$$v_{max} = \frac{\pi}{2} \cdot 0.75 = \frac{3\pi}{8} \approx 1.178 \text{ m/s}.$

To find this time, we know maximum speed occurs at the equilibrium position.

Half maximum speed is given by:

$v = \frac{1}{2} v_{max} = \frac{3\pi}{16}.$

Using the equation:

$v^2 = \omega^2(A^2 - x^2)$

Setting the parameters:

• A = 0.75 m
• (\omega = \frac{\pi}{2})

We thus have:

$\left(\frac{3\pi}{16}\right)^2 = \left(\frac{\pi}{2}\right)^2 ((0.75)^2 - x^2)$

From which we find:

$\frac{9\pi^2}{256} = \frac{\pi^2}{4} (0.5625 - x^2)$

This leads to: