Assume the surface mass density to be 'S'
Area of elemental ring is dA=(2pi(xSin(A))dx)..hence its mass is Sda.
The moment of inertia cantribution is dI=dm(xSin(A))^2...
Integrate to get MR^2/2...
Points to note:
1)While taking a surface element..always take the length along the area.
2)The moment of inertia is same as that of a disc of same mass and radius...isn't it quite obvious..
3)For practice, try the COM of a hollow hemisphere..also find its MI.
yeah...i did that a long time ago using polar co-ordinates...COM was at R/2 above the base..
ReplyDeleteand if i am right with the COM calculation, then MOI about its COM parallel to the base should be [(5MR^2)/12]
ReplyDeletejust think why you used polar coordinates...to take the RdQ element along the surface..you had to the same for the hollow cone.
ReplyDeleteyeah..i had realised that..
ReplyDeletethis blog is too helpful...
also, could we have done this by normal approach..that is...without using the polar coordinates?
ya you could have done that..it would involve converting the dx (radial) into a length element alng. the surface..using trigo
ReplyDeletebut i don't recommend this approach!