What is the moment of inertia of a hollow cylinder?
My textbook says: 0.5M(R2^2-R1^2) and wikipedia says: 0.5M(R2^2+R1^2)
where R1=inner radius and R2=outer radius and M=mass of hollow cylinder

Question

What is the moment of inertia of a hollow cylinder?
My textbook says: 0.5M(R2^2-R1^2) and wikipedia says: 0.5M(R2^2+R1^2) 
where R1=inner radius and R2=outer radius and M=mass of hollow cylinder

anonymous · Answer

about axis through the centre of the hollow part

surry99 · Answer

I = 0.5M(R2^2+R1^2)

surry99 · Answer

what textbook are you using

anonymous · Answer

Physics for scientists and engineers

anonymous · Answer

In that case, to calculate the moment of inertia of:

anonymous · Answer

|dw:1416773127677:dw|

anonymous · Answer

smaller radius circle shows hollow feature

anonymous · Answer

can i say that the moment of inertia is: 0.5MR1^2+(0.5MR2^2+Md^2) where R1=outer radius, R1= radius of smaller circle and d=distance between two COM's.

surry99 · Answer

can you send me the actual drawing please to look at?

surry99 · Answer

I will be back on line if a few hours.

anonymous · Answer

ok

anonymous · Answer

attachment

anonymous · Answer

is this right?

surry99 · Answer

I will check tonight...gotta run now.

surry99 · Answer

yes they appear to be correct.

anonymous · Answer

why can i get the same answer for (b) when i use parallel axis theorem on (a)?