Answer:
I' = 2.34 kg.m².
Explanation:
Given ,R= 40 cm = 0.4 m
m = 28 kg
d= 6 cm = 0.06 m
The mass moment of inertia of the cylinder about center given as
[tex]I=\dfrac{1}{2}mR^2[/tex]
By using Parallel axis theorem
[tex]I'=\dfrac{1}{2}mR^2+md^2[/tex]
Now by putting the values
[tex]I'=\dfrac{1}{2}mR^2+md^2[/tex]
[tex]I'=\dfrac{1}{2}\times 28\times 0.4^2+28\times 0.06^2[/tex]
I'=2.34 kg.m²
There the answer will be 2.34 kg.m².
