Do not round intermediate calculations however for display p

Do not round intermediate calculations; however, for display purposes, report intermediate steps rounded to four significant figures. Give your final answer(s) to three significant figures. The cam shown consists of a circular cylinder A with a circular hole to reduce its moment of inertia and a circular shaft BC about which it rotates. If the cam is made of cast iron with 7,200 kg/m^3 density, determine the mass moment of inertia about the z axis.

Solution

>> About Center Point, Iz = Moment of Inertia of cylinder rod = MR²/2

Now, Mass, M = Volume, V * Density

As Density = 7200 Kg-m³

and, Volume of big Cylinder = pi*R²*h = pi*44²*22 = 1.338*10^-4 m3

So, Mass, M = 1.338*10-4*7200 = 0.963 Kg

Now, Volume of cylindrical Hole, V2 =  pi*R\'²*h = pi*6²*22 = 2.488*10^-6 m³

So, Mass of Hole, M2 = V2*7200 = 0.0179 Kg

>> Volume of Shaft BC, V3 =  pi*R\"²*h\" = pi*3²*50 = 1.4137*10^-6 m³

So, Mass of Shaft BC, M3 = V3*7200 = 0.0102 Kg

>> So, Moment of Inertia of Cylinder With Hole = I1 - I2

I1 = Inertia of whole Cylinder about z-axis = MR²/2 + M*27² = 1.634*10^-3 Kg-m²

I2 = Inertia of Cylindrical Hole about z-axis = M2*R\'²/2 + M2*(27+ 22)² = 4.33*10^-5 Kg-m²

So, Inertia of cylinder = 1.59*10^-3 Kg-m²

As, I3 = Inertia of Shaft BC = M3*R\"²/2 = 4.59*10^-8 Kg-m²

So, Final Inertia = 1.59*10^-3 + 4.59*10^-8 = 1.59*10^-3 Kg-m²