Find the angular momentum of a disc about the axis shown in figure in the following situations.

Explanation

Case a: In this case the disc is rotating about fixed axis which is parallel to the `z`-axis.
`Hence vecL=I_(z)(hatk)=((MR^(2))/2)omega(hatk)`
direction can be determined using right hand rule.
Case b: In this case also the disc rotates about fixed axis. We can write angular momentum for this situation.
`vecL=I_(z)omega(hatk)`
Uisnng parallel axis we can find the value of `I_(2)=(MR^(2))/2+Md^(2)` Hence `vecL=((MR^(2))/2+Md^(2))omega(hatk)`
Case c: In this case the disc is rotating about an axis parallel to the `z`-axis and centre of disc is also translating.
Hence, angular momentum,
`vecL=I_(cm)vecomega+vecrxxmvecv`
`=(MR^(2))/2omega(-hatk)+Mvd(-hatk)`
`=((MR^(2))/2omega+Mvd)(-hatk)`
Case d: In the case also the the disc is rotating about an axis passing through it centre and parallel to `y`-axis and centre of mass of disc is translating with velocity `v` parallel to `x`-axis
`vecL=I_(CM)omega(hatj)+Mvd(hatk)`
Here `I_(CM)=(MR^(2))/4`
Hence , `vecL=(MR^(2))/4omega(hatj)+Mvd(hatk)`