Counterintuitive Result Regarding COM Of A Hemispherical Shell

  • Thread starter GPhab
  • Start date
  • Tags
    Shell
In summary, the conversation discusses how to amuse oneself by rotating a semicircular wire about its center of mass (COM) and creating a new wire at each position. The COMs of all the wires still coincide at the same point, but the COM of a hemispherical shell is different due to the accumulation of mass at the center of the hemisphere. The conversation also briefly touches on calculating the COM of a hemisphere, which involves integrating the sum of semicircular wires.
  • #1
GPhab
25
0
Follow the below four steps to amuse yourself

1) Take a semicircular wire. You know that its COM is at [tex](0,2R/\pi)[/tex]. 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
 
Physics news on Phys.org
  • #2
because the wire isn't shaped like a wedge.
 
  • #3
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.
 
  • #4
Jeff got it right. To what extent did it amuse you?
 
  • #5
Speaking of which, can anyone show me how to calculate the COM of a hemisphere (integral way)?
 
  • #6
GPhab said:
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 :confused:
 

1. What is the meaning of a "counterintuitive result"?

A counterintuitive result is a conclusion or finding that goes against common sense or expectations. It may seem illogical or surprising based on prior knowledge or beliefs.

2. Can you explain the concept of "COM" in relation to a hemispherical shell?

COM stands for "center of mass", which is the point at which all the mass of an object is considered to be concentrated. In a hemispherical shell, the COM is located at the geometric center of the object.

3. What is the typical expectation for the COM of a hemispherical shell?

Many people would expect the COM of a hemispherical shell to be located at the center of the flat base, since this is where the majority of the mass is concentrated. This expectation is based on the symmetry of the shape.

4. What is the counterintuitive result regarding the COM of a hemispherical shell?

The counterintuitive result is that the COM of a hemispherical shell is actually located above the flat base, closer to the curved surface. This means that the majority of the mass is distributed above the center point, rather than being evenly distributed throughout the object.

5. What is the significance of this counterintuitive result?

This result challenges our understanding of how mass is distributed within a hemispherical shell, and highlights the importance of considering the shape and geometry of an object in relation to its COM. It also has practical applications in fields such as engineering and physics, where the location of the COM is important for stability and balance calculations.

Similar threads

C.O.M and Inertia of a sphere; top half twice as dense as lower half
    • Calculus and Beyond Homework Help
Replies
1
Views
2K

Hot Threads

Recent Insights

Back
Top