Answer :
Answer:
1.75 kg.m²
Explanation:
moment of inertia (I) = moment of inertia of ring + 6(moment of inertia of the spoke about the axis of rotation)
[tex]I =[/tex] [tex]I_{rim}+6*I_{spoke}(axis of rotation)[/tex]
[tex]I =[/tex] [tex](M_{rim}*(\frac{R}{2})^2 )+ (6*\frac{1}{3}M_{spoke}(\frac{R}{2})^2 )[/tex]
Given that:
diameter wagon wheel = 1
radius = [tex]\frac{1}{2}[/tex]
mass of the rim = 4 kg
number of spokes = 6
mass of spoke = 1.5 kg
substituting our values into the above equation, we have:
[tex]I =[/tex] [tex](4*(\frac{1}{2})^2 )+ (6*\frac{1}{3}1.5(\frac{1}{2})^2 )[/tex]
[tex]I =[/tex] [tex](4* \frac{1}{4})+(2*1.5*\frac{1}{4})[/tex]
[tex]I =[/tex] 1 + 0.75
[tex]I =[/tex] 1.75 kg m²
∴ the moment of inertia of the wagon wheel for rotation about its axis = 1.75 kg.m²