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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 \( \frac{1}{3} m ℓ^2 \) - Leaving Cert Applied Maths - Question 8 - 2009
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} m ℓ^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 \( \frac{1}{3} m ℓ^2 \) - Leaving Cert Applied Maths - Question 8 - 2009
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 is \( \frac{1}{3} m ℓ^2 \)
Answer
To prove the moment of inertia of a uniform rod, we start by considering a small element of length ( dx ) in the rod, which has a mass given by:
[ dm = M \cdot dx \quad (M = \frac{m}{L}) ]
The moment of inertia of this small element about the axis is:
[ dI = dm \cdot (\frac{L}{2})^2 = M \cdot dx \cdot (\frac{L}{2})^2 ]
To find the moment of inertia of the entire rod, we integrate from 0 to L:
[ I = \int_0^L M \cdot dx \cdot (\frac{L}{2})^2 = M \cdot \frac{L^3}{12} ]
Re-arranging in terms of mass, we get the final result:
[ I = \frac{1}{3} m ℓ^2 ]
Step 2
Find the moment of inertia of the frame abc about an axis through a perpendicular to the plane of the triangle.
Answer
To find the moment of inertia of the equilateral triangle frame abc:
- Each rod has a length of 2ℓ, and there are three rods, so:
- The moment of inertia for one rod about its own centroid is ( \frac{1}{3} m (2ℓ)^2 = \frac{4}{3} m ℓ^2 )
- For three rods:
- The distance from the centroid of the triangle to the axis through point a is ( d = \frac{\sqrt{3}}{3} imes 2ℓ = \frac{2ℓ\sqrt{3}}{3} )
- Using the parallel axis theorem:
- Total moment of inertia:
[ I = 3 \left( \frac{1}{3} m (2ℓ)^2 + m \left(\frac{2ℓ\sqrt{3}}{3}\right)^2 \right) = 3 \left( \frac{4}{3} m ℓ^2 + m \cdot \frac{4ℓ^2}{3} \right) = 6mℓ^2 ]
Step 3
Find the maximum angular velocity of the triangle in the subsequent motion.
Answer
Using energy conservation:
[ \text{Gain in KE} = \text{Loss in PE} ]
Potential energy lost when the triangle falls: [ PE = mgh = mg \cdot \frac{h}{3} = mg \cdot \frac{2ℓ}{3} ]
Kinetic energy gain: [ KE = \frac{1}{2} I \omega^2 ]
Setting them equal: [ mg \cdot \frac{2ℓ}{3} = \frac{1}{2} (6mℓ^2) \omega^2 ]
Solving for ( \omega ): [ \omega^2 = \frac{g}{\sqrt{3}} ] and hence, [ \omega = \sqrt{\frac{g}{\sqrt{3}}} ]