A uniform solid sphere with centre C, radius $2a$ and mass $3M$, is pivoted about a smooth horizontal axis and hangs at rest - CIE - A-Level Further Maths - Question 5 - 2011 - Paper 1

To find the period of small oscillations, we will use the simple harmonic motion (SHM) principles.

1. Setting Up the Equation of Motion: The equation of motion for small oscillations can be expressed as: $I \frac{d^2\theta}{dt^2} = -Mg \sin(\theta)$ For small angles, we use the approximation $\sin(\theta) \approx \theta$: $I \frac{d^2\theta}{dt^2} = -Mg\theta$

2. Substituting the Moment of Inertia: Substitute the moment of inertia $I$ we found earlier: $\frac{39}{5}Ma^2 \frac{d^2\theta}{dt^2} = -Mg\theta$ Simplifying gives: $\frac{d^2\theta}{dt^2} + \frac{5g}{39a} \theta = 0$

3. Finding the Period (T): The general form of SHM gives us: $T = 2\pi \sqrt{\frac{I}{MgL}}$ Plugging in the values, where $L = 39/5a$, the period becomes: $T \approx 2\pi \sqrt{\frac{39}{5g}}$

To find the time $t$ until OP makes an angle $\frac{1}{2}\theta$:

1. Use the SHM Formula: The angle $\theta$ as a function of time can be approximated in SHM: $\theta = \alpha \cos(\sqrt{\frac{g}{L}}t)$ Where $L$ is the effective length of oscillation.

2. Finding Time for Half the Angle: Set $\theta = \frac{1}{2}\alpha$: $\frac{1}{2}\alpha = \alpha \cos(\sqrt{\frac{g}{L}}t)$ Simplifying gives: $\frac{1}{2} = \cos(\sqrt{\frac{g}{L}}t)$

3. Solving for t: The solution to this equation yields: $\sqrt{\frac{g}{L}}t = \frac{\pi}{3}$ Which leads to: $t = \frac{\pi}{3}\sqrt{\frac{L}{g}}$ Substituting $L$ provides the required time.