Counterintuitive Result Regarding COM Of A Hemispherical Shell

[GPhab]

Follow the below four steps to amuse yourself

1) Take a semicircular wire. You know that its COM is at $$(0,2R/\pi)$$. Now pass an axis through its COM and perpendicular to the line joining its ends.

2)Rotate this "half-lollipop" about the axis fixed till it comes to another position. The COM obviously didn't undergo any displacement.

3)Do this is an umpteen number of times and imagine as if a new wire is created for each position. The COMs of all of these wires coincide and should be at the coordinate mentioned in step 1 (Including a Z-co-ordinate, which can be taken as zero)

4)But the COM of a hemispherical shell is R/2 above the centre!

COM-centre of mass

 

[granpa]

because the wire isn't shaped like a wedge.

 

[rcgldr]

If the wires have finite thickness, then there's an accumulation of mass at the center of the hemisphere where multiple wires would cross. The density at the center (top) of the hemisphere would be higher than at the edges, which differs from a shell of finite thickness.

 

[GPhab]

Jeff got it right. To what extent did it amuse you?

 

[ibc]

Speaking of which, can anyone show me how to calculate the COM of a hemisphere (integral way)?

 

[matematikawan]

Jeff got it right. To what extent did it amuse you?

We assume that the hemispherical shell is being built up from the sum of semicircular wire. Each element of the semicircular wire has COM at the point A. But the cummulative sum of the elements has COM at B. That puzzling