A uniform solid sphere with centre C, radius 2a and mass 3M, is pivoted about a smooth horizontal axis and hangs at rest. The point O on the axis is vertically above C and OC = a. A particle P of mass M is attached to the sphere at its lowest point (see diagram). Show that the moment of inertia of the system about the axis through O is $rac{3}{4}Ma^2$. The system is released from rest with OP making a small angle $\alpha$ with the downward vertical. Find i. the period of small oscillations. ii. the time from release until OP makes an angle $rac{1}{2}\alpha$ with the downward vertical for the first time.

Photo AI

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

Question 5

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.png

A uniform solid sphere with centre C, radius 2a and mass 3M, is pivoted about a smooth horizontal axis and hangs at rest. The point O on the axis is vertically above... show full transcript

Worked Solution & Example Answer: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

Step 1

i. the period of small oscillations.

96%

114 rated

Answer

To find the period of small oscillations, we start with the principle of angular motion and use the equation of motion for the swinging system:

The moment of inertia of the sphere about its diameter is given as: $I_c = \frac{2}{5} \cdot 3M (2a)^2 = \frac{24Ma^2}{5}$
Next, we find the moment of inertia about the axis through O, which is: $I_O = I_c + Mh^2$ where $h = 2a$ (the vertical distance from the center of the sphere to point O). Thus, $I_O = \frac{24Ma^2}{5} + 3M(2a)^2 = \frac{24Ma^2}{5} + 12Ma^2 = \frac{84Ma^2}{5}$
The equation of motion for small oscillations based on SHM yields: $I \frac{d^2\theta}{dt^2} = -Mg \sin\theta$ For small angles, $\sin\theta$ can be approximated with $\theta$ : $I \frac{d^2\theta}{dt^2} = -Mg\theta$
Thus, we can express the angular frequency $\omega$ as: $\omega^2 = \frac{Mg}{I}$ Substituting our expression for $I_O$ into this, we get: $\omega^2 = \frac{Mg}{\frac{84Ma^2}{5}} = \frac{5g}{84a^2}$
The period $T$ can then be calculated as: $T = 2\pi \sqrt{\frac{I}{Mg}} = 2\pi \sqrt{\frac{84Ma^2}{5Mg}} = 2\pi \sqrt{\frac{84a^2}{5g}}$

Step 2

ii. the time from release until OP makes an angle $rac{1}{2}\alpha$ with the downward vertical for the first time.

99%

104 rated

Answer

To calculate the time from release until OP makes an angle $rac{1}{2}\alpha$ with the downward vertical:

From the earlier analysis, the angular position as a function of time can be written as: $\theta(t) = \alpha \cos(\omega t)$
We need to find the time $t$ when: $\theta(t) = \frac{1}{2}\alpha$ Setting the equations equal gives: $\frac{1}{2}\alpha = \alpha \cos(\omega t)$ This simplifies to: $\frac{1}{2} = \cos(\omega t)$
The solution to this is: $\omega t = \frac{\pi}{3} \Rightarrow t = \frac{1}{\omega} \cdot \frac{\pi}{3}$
Recalling that: $\omega = \sqrt{\frac{5g}{84a^2}}$ The time can then be expressed as: $t = \frac{1}{\sqrt{\frac{5g}{84a^2}}} \cdot \frac{\pi}{3} = \frac{\pi \sqrt{84a^2}}{3 \sqrt{5g}}$ . Thus, the final expression for the time is: $t = \frac{\pi a \sqrt{84}}{3 \sqrt{5g}}$ .

97% of Students

Report Improved Results

98% of Students

Recommend to friends

100,000+

Students Supported

1 Million+

Questions answered

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

Worked Solution & Example Answer: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

i. the period of small oscillations.

ii. the time from release until OP makes an angle $rac{1}{2}\alpha$ with the downward vertical for the first time.

Join the A-Level students using SimpleStudy...

Other A-Level Further Maths topics to explore

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

Worked Solution & Example Answer: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

i. the period of small oscillations.

ii. the time from release until OP makes an angle $ rac{1}{2}\alpha$ with the downward vertical for the first time.

Join the A-Level students using SimpleStudy...

Other A-Level Further Maths topics to explore

ii. the time from release until OP makes an angle $rac{1}{2}\alpha$ with the downward vertical for the first time.