If the temperature of a uniform rod is slightly increased by ?t, its moment of inertia I about a line parallel to itself will increase by what?
Asked by JAYACHANDRAN B
| 4th Feb, 2012,
09:49: AM
Expert Answer:
The moment of inertia of a rod about a perpendicular bisector is ML^2 / 12.
Change in M.I. is d(ML^2 / 12) = ML dL / 6
dL = aL dt
dI = aL dt ML / 6 = 2a(ML^2/12) dt = 2aIdt
The moment of inertia of a rod about a perpendicular bisector is ML^2 / 12.
Change in M.I. is d(ML^2 / 12) = ML dL / 6
dL = aL dt
dI = aL dt ML / 6 = 2a(ML^2/12) dt = 2aIdt
Answered by
| 4th Feb, 2012,
12:25: PM
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