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Rim of a cast iron flywheel.

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The rim of a castiron flywheel is 0.5m wide and its inner and outer diameters are 1.8m and 2.6m respectively. Calculate a) the moment of inertia of the rim about its geometric axis and b) the radius of gyration k.

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Solution Summary

Moment of inertia of solid cylinder of radius R about it's axis: I = M*R^2/2 = (d*t*pi*R^2)*R^2/2.
Radius of gyration: I = M*k^2, where k = radius of gyration and M = mass = d*t*pi*(Ro^2 - Ri^2).

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Dear student, you have not mentioned the density of the cast iron therefore, let us assume the density of cast iron d = 7200 kg/m^3.

Thickness of the wheel t = 0.5 m
Outer diameter of the wheel Do = 2*Ro = 2.6 m => ...

Solution provided by:
  • BEng, Allahabad University, India
  • MSc , Pune University, India
  • PhD (IP), Pune University, India
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