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8. Prove that the moment of inertia of a uniform square lamina of mass m and side 2l about an axis through its centre parallel to one of its sides is \( \frac{1}{3} ml^2 \) - Leaving Cert Applied Maths - Question 8 - 2011
Question 8
8. Prove that the moment of inertia of a uniform square lamina of mass m and side 2l about an axis through its centre parallel to one of its sides is \( \frac{1}{3} ... show full transcript
Worked Solution & Example Answer:8. Prove that the moment of inertia of a uniform square lamina of mass m and side 2l about an axis through its centre parallel to one of its sides is \( \frac{1}{3} ml^2 \) - Leaving Cert Applied Maths - Question 8 - 2011
Step 1
Prove that the moment of inertia of a uniform square lamina of mass m and side 2l about an axis through its centre parallel to one of its sides is \( \frac{1}{3} ml^2 \).
Answer
To find the moment of inertia of a square lamina, we start by determining the mass per unit area, ( M = \frac{m}{(2l)^2} ). The thickness of the element is defined as ( dx ).
Next, we calculate the moment of inertia of a differential element which is a strip of width ( 2l ) and height ( dx ):
[ \text{Moment of inertia} = M(2l)(dx)^2 = \frac{m}{4l^2}(2l)(dx)^2 = \frac{m(2l)(dx)^2}{4l^2} = \frac{m(2)(dx)^2}{4} ]
Integrate from ( -l ) to ( l ):
[ I = \int_{-l}^{l} \frac{m(2)(dx)^2}{4} = 2M \int_{0}^{l} x^2 dx = 2M \left[ \frac{x^3}{3} \right]_{0}^{l} = 2M \left( \frac{l^3}{3} \right)\n]
This gives: ( = \frac{2}{3} ml^2 ).
So, considering the full width of the lamina forms a factor of 4: [ \therefore I = \frac{1}{3} ml^2 ]
Step 2
Find the angular velocity of the lamina when PR is vertical.
Answer
When the lamina is released, it gains kinetic energy as it converts potential energy. The relationship is: [ \text{Gain in KE} = \text{Loss in PE} ] [ \frac{1}{2} I \omega^2 = mgh ] Considering the variables, with L as the distance from the pivot to the center of mass and the height being ( h = 0.3 ) when it moves to vertical, we use: [ L = \frac{1}{2} m(0.3)^2 + m(0.6 \sqrt{2})^2 = mg(0.3 \sqrt{2} - 0.3) ] From this expression, we can derive the angular velocity ( \omega ): [ \omega = 10.1482 \text{ rad s}^{-1} ]
Step 3
Find the period of small oscillations of the compound pendulum and hence, or otherwise, find the length of the equivalent simple pendulum.
Answer
The period ( T ) can be calculated using: [ T = 2\pi \sqrt{\frac{I}{mgh}} ] Substituting the values: [ I = 0.96m ] [ Mgh = mg(0.3 \sqrt{2}) = 0.9 + 0.2 mg\n] ] This yields: [ T = 2\pi \sqrt{\frac{0.96}{0.9 \times 2}} = 2\pi \sqrt{0.75} = 0.75sec ] Thus the length L of the equivalent simple pendulum: [ L = \frac{0.96}{0.9 \sqrt{2}} = 0.75m ]