Question

Calculate the moment of inertia of a thin rod rotating about an axis through its center perpendicular to its long dimension. Do the same for rotation about an axis through one of the ends (perpendicular to the length again). Does this confirm the parallel axis theorem? Remember ?? = ? ??C????.

Answer 1

M = total mass of the rod

L = total length

M/L = linear mass density

dm = mass of small element of length "dx" at small distance "x" = (M/L) dx

small moment of inertia of small length is given as

dI = dm x²

dI = (M/L) x² dx

Total moment of inertia is given as

I = (M/L) x² dx

I = ML²/12

M = total mass of the rod

L = total length

M/L = linear mass density

dm = mass of small element of length "dx" at small distance "x" = (M/L) dx

small moment of inertia of small length is given as

dI = dm x²

dI = (M/L) x² dx

Total moment of inertia is given as

I = (M/L) x² dx

I = ML²/3

yes this confirms parallel axis theorem