Answer to Question #92087 in Classical Mechanics for Tapas

Question

Moment of inertia of a semicircular ring of mass πkg and radius 2cm about the axis passing through a point perpendicular to it's plane

Answer 1

Here the location of point is not given.

Assume the point is located at centre of mass which is located at position $d=\frac{2R}{\varPi}$ from horizontal axis

Then MI is given as $MI=MR^2$ =12.56 Kg-cm²

If the point is located at a point on horizontal axis,

Then MI is given as $MI=MR^2+Md^2$

Where,$d=2R/\varPi$

Thus

MI=3.14 x4+4/3.14

MI=12.56+1.27

MI=13.83 kg-cm²