The wheels of a wagon can be approximated as the combination of a thin outer hoop, of radius ????h=0.209 m and mass 4.32 kg , an
1 answer:
Answer:
r= 98.3 mm
Explanation:
For rim
R= 0.209 m
M= 4.32 kg
For rods
m= 7.37 kg
L= 2 R= 2 x 0.209 = 0.418 m
The Total moment of inertia of the wagon
I=MR²+2 x 1/12 m L²
Now by putting the values
I=0.413 kg.m²
For disk:
t= 0.0462 m
Density ρ = 5990 kg/m³
Lets take r is the radius of disk
So the mass of the disc
m'=ρ πr² t
The moment of inertia of disc
I'=1/2 m'r²
I'=1/2 x r² x ρ πr² t
Given that
I = I'
1/2 x r² x ρ πr² t = 0.413 kg.m²
1/2 x r³ x ρ π t = 0.413
r³ x ρ π t = 0.826
r³=0.00095
r=0.0983 m
r= 98.3 mm
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