Question:
Should we calculate the moment of inertia of the rectangular bar by assuming it is a long thin rod rather than a bar with a discernible width? In other words, which geometry will yield a lower error value and explain why it will?
And
If the hollow cylinder was twice as heavy but its inner and outer radii were unchanged, how would its moment of inertia ge affected? Explain why.
If cross section is rectangular , the question which arises is : you want moment of inertia about which axis?
Ix = moment of inertia about x-axis :(1/12) (mb^2+mc^2). ( where m = mass of rod , b =height of rectangular face. c= width of rectangular face)
Iy = (1/12) ( mc^2+ma^2) ( a = length of rod )
Iz = (1/12) ( ma^2+mb^2)
If length is very large compared to area of cross section you can use rod formula (1/12) ml^2
Solid cylinder with inner radius b and outer radius a:
I = 0.5 M ( a^2+b^2). if M is double , then I will double if a and b are same.
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