Question 857974
With a rim thickness of 2 cm, and an average radius of 5 cm,
the radius of the outside of flywheel must be 5cm + 2cm/2= 6cm,
and the inner edge of the rim must be 5cm - 2cm/2= 4cm from the axle.
The rim is a 2-cm tall 6-cm diameter cylinder with a 4-cm circulat hole in the middle.
The volume of the 2-cm tall 6-cm diameter cylinder is
{{{pi*(6cm)^2(2cm)=pi*36*2}}}{{{cm^3=72pi}}}{{{cm^3}}} .
The volume of the 4-cm hole is
{{{pi*(4cm)^2(2cm)=pi*16*2}}}{{{cm^3=32pi}}}{{{cm^3}}} .
The volume of the rim is the difference. In cubic centimeters, it is
{{{72pi-32pi=40pi}}}
Using {{{pi=3.1412}}} , the volume of the rim of the flywheel is
{{{125.648}}}{{{cm^3}}} .