Where is the Center of Mass in a Solid Hemisphere?

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
5 replies · 5K views
Westin
Messages
87
Reaction score
0

Homework Statement


Consider a solid hemisphere of uniform density with radius R. Where is the center of mass?

z=0
0
char3C.png
z
char3C.png
R
char3D.png
2
z=R
char3D.png
2
R
char3D.png
2
char3C.png
z
char3C.png
r
z=R

Image is provided.

Homework Equations

None

The Attempt at a Solution



Answer A and E do not seem logical. I thought it was answer C from my eyes. Center of mass is the average of the masses factored by their distances from a reference point. I didn't think the answer could range like B and D do.[/B]
 

Attachments

  • Screen Shot 2015-03-30 at 3.33.15 PM.png
    Screen Shot 2015-03-30 at 3.33.15 PM.png
    66.5 KB · Views: 2,744
Last edited by a moderator:
Physics news on Phys.org
Westin said:

Homework Equations

None
Really? No equation for the center of mass?

Westin said:
Answer A and E do not seem logical. I thought it was answer C from my eyes. Center of mass is the average of the masses factored by their distances from a reference point. I didn't think the answer could range like B and D do.
Do you understand that the range is there to keep you from having to calculate the exact value? It doesn't mean that the center of mass can be anywhere within that range.
 
If the center of mass were precisely at R/2, there will be more mass below than above that point. Hence, it must be somewhere between R/2 and...
 
NTW said:
If the center of mass were precisely at R/2, there will be more mass below than above that point.
I don't understand what you mean.
 
DrClaude said:
I don't understand what you mean.
To make a mental experiment: If I imagine a given point on the Z axis, precisely at R/2, and also imagine the hemisphere as formed by a very large, but finite number of particles, the number of particles with z-coordinates lower than R/2 will be larger than the number of particles with z-coordinates higher than R/2. Hence, in order to reach a 50% partition in the values of the z-coordinates, that point must be placed somewhere between 0 and R/2.
 
NTW said:
If I imagine a given point on the Z axis, precisely at R/2
Ok, but that's not the same as saying "if the center of mass were precisely at R/2."

Also, please to not give direct answers in the homework forums. The poster has to do the work.