Two thin discs each of mass M and radius R are placed at either and of a rod of mass m length l and radius r. Moment of inertia of the system about an axis passing through the centre of rod and perpendicular to its length is

Options

A. `(mL^(2))/(12)+(1)/(2)MR^(2)+(1)/(2)ML^(2)`
B. `(ML^(2))/(12)+(1)/(2)mR^(2)+(1)/(2)mL^(2)`
C. `(1)/(2)mL^(2)+(mR^(2))/(2)+(ML^(2))/(12)`
D. `(mL^(2))/(12)+MR^(2)+(1)/(2)ML^(2)`

Answer

A

Explanation

Moment of inertia of rod about the given axis `=(mL^(2))/(12).`
Moment of inertia of each disc about its diameter `=(MR^(2))/(4)`
Using theorem of parallel axes, moment of inertia of each disc about the given axis `=(MR^(2))/(4)+M((L)/(2))^(2)`
`=(MR^(2))/(4)+(ML^(2))/(4)`
`therefore` For theorem of parallel axes, moment of inertia about the given axis is
`l=(mL^(2))/(12)+((MR^(2))/(4)+(ML^(2))/(4))=(mL^(2))/(12)+(MR^(2))/(2)+(ML^(2))/(2)`