A thin uniform rod AB has mass $ \frac{2}{3} m $ and length $ 3a $. The end A of the rod is rigidly attached to a point on the circumference of a uniform disc with centre C, mass m and radius a. The end B of the rod is rigidly attached to a point on the circumference of a uniform disc with centre D, mass 4m and radius $ 2a $. The discs and the rod are in the same plane and CABD is a straight line. The mid-point of CD is O. The object consisting of the two discs and the rod is free to rotate about a fixed smooth horizontal axis $ l $, through O in the plane of the object and perpendicular to the rod (see diagram). Show that the moment of inertia of the object about $ I $ is $ 50m^2 $. The object hangs in equilibrium with D vertically below C. It is displaced through a small angle and released from rest, so that it makes small oscillations about the horizontal axis $ I $. Show that it will move in approximate simple harmonic motion and state the period of the motion.

A-Level CIE Further Maths 8.3 Simple Harmonic Motion: A thin uniform rod AB has mass \

A thin uniform rod AB has mass $ \frac{2}{3} m $ and length $ 3a $ - CIE - A-Level Further Maths - Question 5 - 2016 - Paper 1

Question 5

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A thin uniform rod AB has mass $ \frac{2}{3} m $ and length $ 3a $. The end A of the rod is rigidly attached to a point on the circumference of a uniform disc wi... show full transcript

Worked Solution & Example Answer:A thin uniform rod AB has mass $ \frac{2}{3} m $ and length $ 3a $ - CIE - A-Level Further Maths - Question 5 - 2016 - Paper 1

Step 1

Find MI of rod about $ l $:

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Answer

The moment of inertia of the rod about the axis ( l ) is given by:

I_{rod} = \frac{1}{3} m (3a)^2 = \frac{1}{3} m \cdot 9a^2 = 3ma^2

Step 2

Find MI of disc C about $ l $ using both theorems:

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The moment of inertia of disc C about its own centre is given by:

I_{C} = \frac{1}{2} m a^2

Using the parallel axis theorem, we have:

I_{C, l} = I_{C} + m d^2 = \frac{1}{2} m a^2 + m (\frac{3a}{2})^2 = \frac{1}{2} m a^2 + \frac{9}{4} m = \frac{2 + 9}{4} m = \frac{11}{4} m a^2

Step 3

Find MI of disc D about $ l $ using both theorems:

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The moment of inertia of disc D about its own centre is:

I_{D} = \frac{1}{2} \cdot 4m (2a)^2 = \frac{1}{2} \cdot 4m imes 4a^2 = 8m a^2

Again, applying the parallel axis theorem:

I_{D, l} = I_{D} + 4m d^2 = 8m a^2 + 4m (\frac{3a}{2})^2 = 8m a^2 + 4m \cdot \frac{9}{4} = 8m a^2 + 9m = 17m a^2

Step 4

Sum to find MI of system about $ l $:

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Answer

Now, summing the moments of inertia:

I = I_{rod} + I_{C, l} + I_{D, l} = 3ma^2 + \frac{11}{4}ma^2 + 17ma^2 = \left( 3 + \frac{11}{4} + 17 \right) ma^2

This leads to:

I = \left( \frac{12}{4} + \frac{11}{4} + \frac{68}{4} \right) ma^2 = \frac{91}{4}ma^2 = 50m^2

Step 5

Use eqn of circular motion to find $ \theta $ and $ T $:

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Answer

Using the equation of circular motion:

\frac{d^2\theta}{dt^2} = -\frac{g}{L} \sin(\theta) \quad (1)

For small angles, we can approximate:

\sin(\theta) \approx \theta

Thus, we can rewrite it as:

\frac{d^2\theta}{dt^2} + \frac{g}{L} \theta = 0

Step 6

Approximate as SHM and find period \( T ":

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This resembles the simple harmonic motion equation, and we identify:

\omega^2 = \frac{g}{L}, \quad \text{where} \ L = \text{length from A to O}

Thus, the period of the motion would be:

T = 2\pi \sqrt{\frac{L}{g}}

Substituting our values, we find:

T = 2\pi \sqrt{\frac{ \text{length from A to O}}{g}}

A thin uniform rod AB has mass \( \frac{2}{3} m \) and length \( 3a \) - CIE - A-Level Further Maths - Question 5 - 2016 - Paper 1

Worked Solution & Example Answer:A thin uniform rod AB has mass \( \frac{2}{3} m \) and length \( 3a \) - CIE - A-Level Further Maths - Question 5 - 2016 - Paper 1

Find MI of rod about \( l \):

Find MI of disc C about \( l \) using both theorems:

Find MI of disc D about \( l \) using both theorems:

Sum to find MI of system about \( l \):

Use eqn of circular motion to find \( \theta \) and \( T \):

Approximate as SHM and find period \( T ":

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