Answer:
I = 0.0025 kg.m²
Explanation:
Given that
m= 2 kg
Diameter ,d= 0.1 m
Radius ,[tex]R=\dfrac{d}{2}[/tex]
[tex]R=\dfrac{0.1}{2}[/tex]
R=0.05 m
The moment of inertia of the cylinder about it's axis same as the disc and it is given as
[tex]I=\dfrac{mR^2}{2}[/tex]
Now by putting the all values
[tex]I=\dfrac{2\times 0.05^2}{2}[/tex]
I = 0.0025 kg.m²
Therefore we can say that the moment of inertia of the cylinder will be 0.0025 kg.m².
