What should be the surface of sphere pin??

Question

Hello. There is one steel rod with the of weight 0.6kg, length 150 mm & width 20mm. And i want to put a hemispherical shaped on top. So how big that hemispherical shape will be? How can i calculate the parameters of hemispherical design??

Answer 1

Well, if this is a rod that normally means it has a cylindrical shape so I am going to assume that when you say "width" you really mean "diameter". If that's the case, then the diameter of the hemisphere should just be the diameter of the rod: 20 mm. Now the volume of the rod is 3.14(150 mm)(20 mm)^2=188,400 mm^3 = 1.88X10^-4 m^3. You said the weight was 0.6 kg. This means the rod's material has a density of 0.6 kg divided by 1.88X10^-4 m^3=3.185X10^3 kg/m^3 = 3.185X10^-3 kg/cm^3. Since metric density is often stated in g/cm^3, the density converts to 3.185 g/cm^3. Based on this density, I could try to guess what the materials might be but this is as far as I can go. The hemisphere's material and the rod's material must be know before I can recommend a method of attaching the two pieces together.