Vikas TU
Last Activity: 7 Years ago
Give us a chance to consider a round shell of mass M and range R. Give us a chance to take a distance across XOX' of the circular shell about which snapshot of dormancy of the shell is to be resolved.
Presently, let us isolate the round shell into substantial no. of basic rings by a plane, opposite to the width XOX'. Give us a chance to take any of these rudimentary rings having range "y" and thickness 'dx'. On the off chance that dθ be the point made by the thickness dx of the ring at the focal point of the shell, then we can compose
From figure,
dx=Rdθandy=Rsinθdx=Rdθandy=Rsinθ
Mass per unit surface territory of the shell= M4πR2M4πR2
Surface territory of the basic ring=2πydx2πydx
Mass of the ring, m=M4πR2×2πydxM4πR2×2πydx
m=msinθdθ2m=msinθdθ2
Presently, snapshot of idleness of the basic ring is given by,
dI=mass×(radius)2(∵foraringI=MR2)dI=mass×(radius)2(∵foraringI=MR2)
=(Msinθdθ2)y2=(Msinθdθ2)y2
=MR2sin3θdθ2=MR2sin3θdθ2
Since, these rings are shifting from θ= 0 at X to θ= π at X', the snapshot of dormancy of the shell is given by,
I=∫dI=∫dI
=∫π0MR2sin3θdθ2=∫0πMR2sin3θdθ2
=MR2/2[−π 0∫dcosθ+π0∫ cos2Ѳdcosθ]=MR2/2[−0 π ∫dcosθ+∫0π cos2ѳdcosθ]
I=MR2/2×4/3I=MR2/2×4/3
I=2MR2/3
