A rod of length L is composed of a uniform Length L/2 of wood whose mass is m1 and a uniform length-L/1 of brass whose mass is m2. Find the moment of inertia I of the rod about an axis perpendicular to the rod and through its centre.
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The moment of inertia I of the rod about an axis perpendicular to the rod and through its centre M = (mw + mb)L^2 / 12
Explanation:
- Moment of inertia
M1 = m(w) × (L/2)^2 / 3
= m(w) =L^2 / 12
- Moment of inertia
M2 = mb × (L/2)^2/ 3
= mb×L^2 / 12
- Moment of inertia
M = M1 + M2
= mw L^2 / 12 + mb × L^2 / 12
= (mw + mb)L^2 / 12
Hence the moment of inertia I of the rod about an axis perpendicular to the rod and through its centre M = (mw + mb)L^2 / 12
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