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8. (a) Prove that the moment of inertia of a uniform circular disc, of mass m and radius r, about an axis through its centre perpendicular to its plane is \( \frac{1}{2} mr^{2} \) - Leaving Cert Applied Maths - Question 8 - 2013
Question 8
8. (a) Prove that the moment of inertia of a uniform circular disc, of mass m and radius r, about an axis through its centre perpendicular to its plane is \( \frac{1... show full transcript
Worked Solution & Example Answer:8. (a) Prove that the moment of inertia of a uniform circular disc, of mass m and radius r, about an axis through its centre perpendicular to its plane is \( \frac{1}{2} mr^{2} \) - Leaving Cert Applied Maths - Question 8 - 2013
Step 1
Prove that the moment of inertia of a uniform circular disc, of mass m and radius r, about an axis through its centre perpendicular to its plane is \( \frac{1}{2} mr^{2} \)
Answer
To find the moment of inertia (I) of a uniform circular disc, we consider a small element of the disc.
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Define mass per unit area: [ M = \frac{m}{\pi r^2} \ ]
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The mass of the small ring of radius (x) and thickness (dx) is: [ dm = M \cdot 2\pi x , dx = \frac{m}{\pi r^2} \cdot 2\pi x , dx = \frac{2m}{r^2} x , dx ]
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The moment of inertia of the ring about the central axis is given by: [ dI = x^{2} , dm = x^{2} \left( \frac{2m}{r^2} x , dx \right) = \frac{2m}{r^2} x^3 , dx ]
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To find the total moment of inertia, integrate from (0) to (r): [ I = \int_{0}^{r} dI = \int_{0}^{r} \frac{2m}{r^2} x^3 , dx = \frac{2m}{r^2} \cdot \left[ \frac{x^4}{4} \right]_{0}^{r} = \frac{2m}{r^2} \cdot \frac{r^4}{4} = \frac{mr^{2}}{2} ]
Thus, we conclude that the moment of inertia of a uniform circular disc is: [ I = \frac{1}{2} mr^{2} ]
Step 2
Find (i) the period of small oscillations of the compound pendulum
Answer
The period of small oscillations (T) can be derived using:
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The total gravitational potential energy (Mgh) where (h) is the height of the center of mass: [ Mgh = 8mg h + mg h + mg r + mg (2r) = 12mgr ]
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The moment of inertia (I) of the compound body: [ I = \frac{1}{6} (8m)(r^2) + (m)(\sqrt{2}r)^2 + m(r^2) + m(2r)^2 = 20mr^{2} ]
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The formula for the period is: [ T = 2\pi \sqrt{\frac{I}{Mgh}} = 2\pi \sqrt{\frac{20mr^{2}}{12mgr}} = 2\pi \sqrt{\frac{20}{12g}} \sqrt{r} = 2\pi \sqrt{\frac{5r}{3g}} ]
Step 3
Find (ii) the length of the equivalent simple pendulum
Answer
The length (L) of the equivalent simple pendulum can be calculated as follows:
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Using the formula for the period of a simple pendulum: [ T = 2\pi \sqrt{\frac{L}{g}} ]
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Setting it equal to the derived period for the compound pendulum: [ 2\pi \sqrt{\frac{L}{g}} = 2\pi \sqrt{\frac{5r}{3g}} ]
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Solving for (L): [ L = \frac{5r}{3} ]