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8. (a) Prove that the moment of inertia of a uniform rod, of mass m and length 2l about an axis through its centre, perpendicular to the rod, is \( \frac{1}{3}ml^2 \) - Leaving Cert Applied Maths - Question 8 - 2019
Question 8
8. (a) Prove that the moment of inertia of a uniform rod, of mass m and length 2l about an axis through its centre, perpendicular to the rod, is \( \frac{1}{3}ml^2 \... show full transcript
Worked Solution & Example Answer:8. (a) Prove that the moment of inertia of a uniform rod, of mass m and length 2l about an axis through its centre, perpendicular to the rod, is \( \frac{1}{3}ml^2 \) - Leaving Cert Applied Maths - Question 8 - 2019
Step 1
Prove that the moment of inertia of a uniform rod, of mass m and length 2l about an axis through its centre, perpendicular to the rod
Answer
To prove that the moment of inertia ( I ) of a uniform rod about an axis through its center, perpendicular to the rod, we can start by defining the mass per unit length as:
[ M = \frac{m}{2l} ]
Next, we consider a differential element of the rod:
- Let ( dx ) be a small segment around the center.
- The moment of inertia of this element about the center is given by:
[ dI = M \cdot dx \cdot x^2 ]
Thus, the total moment of inertia of the rod can be determined by integrating over its length:
[ I = \int_{-l}^{l} dI = M \int_{-l}^{l} x^2 dx ]
Calculating the integral yields:
[ I = \frac{M}{3} [x^3]_{-l}^{l} = \frac{M}{3} (l^3 - (-l)^3) = \frac{2Ml^3}{3} ]
Inserting ( M = \frac{m}{2l} ) gives:
[ I = \frac{1}{3} ml^2 ]
Step 2
Find (i) the moment of inertia of the compound body about the axis of rotation
Answer
To determine the moment of inertia ( I_a ) of the compound body about the axis of rotation located at point A, we consider the contributions from both the rod and the disc:
- The moment of inertia of the rod about its own center is:
[ I_{rod} = \frac{1}{3} m(0.5)^2 + m(0.5)(1.1^2) ]
Calculating:
[ = \frac{1}{3} m(0.25) + m(0.5)(1.21) ] [ = \frac{m}{12} + 1.21m = 1.548m ]
Thus, the total moment of inertia is:
[ I_a = 1.548m ]
Step 3
Find (ii) the period of small oscillations correct to two decimal places
Answer
Given that the oscillation occurs about a horizontal axis:
Using the formula for the period of a physical pendulum, where ( mgh = mgx ):
[ Mg h = mg(0.5) + mg(1.1) ] [ = 1.6mg ]
Thus, the period ( T ) is given by:
[ T = 2\pi \sqrt{\frac{I}{mg}} ]
Substituting the values gives:
[ T = 2\pi \sqrt{\frac{929m}{600\times 1.6g}} ] [ = 1.97s ]
Step 4
Find (iii) the length of the equivalent simple pendulum
Answer
To find the length ( L ) of the equivalent simple pendulum, we relate it as follows:
[ 2\pi \sqrt{\frac{L}{g}} = T ]
Thus,
[ L = \left(\frac{T}{2\pi}\right)^2 g ]
Calculating gives:
[ L = \left(\frac{1.97}{2\pi}\right)^2 g ] [ = 0.97m ]