The mass of the cylinder is 25 kg and radius of cylinder is 5 m. Find the moment of inertia of a solid cylinder about a tangent to the cylinder.

Explanation

The moment of inertia of a cylinder about its own axis (YY')
`= (1)/(2)MR^(2)`
Applying the theorem of parallel axes, the moment of inertia of the cylinder about tangent (XX') is
`I_(t)=I+MR^(2)=(1)/(2)MR^(2)+MR^(2)rArr I_(t)=(3)/(2)MR^(2)`
Here, `M=25 kg, R=5 m :. I_(t)=(3)/(2)(25)(5)^()=937.5 "kg-m"^(2)`