Center of mass of two connected different density blocks

In summary, the center of mass for a composite slab with dimensions d1 = 11.6 cm, d2 = 2.85 cm, and d3 = 13.4 cm is located at (x = 2.9 cm, y = 6.7 cm, z = 13.4 cm).
  • #1
J-dizzal
394
6

Homework Statement


The figure shows a composite slab with dimensions d1 = 11.6 cm, d2 = 2.85 cm, and d3 = 13.4 cm. Half the slab consists of aluminum (density = 2.70 g/cm3) and half consists of iron (density = 7.85 g/cm3). What are (a) the x coordinate, (b) the y coordinate, and(c) the z coordinate of the slab's center of mass?

http://edugen.wileyplus.com/edugen/courses/crs7165/art/qb/qu/c09/fig09_40.gif

Homework Equations


20150707_161138_zpsvokykdw2.jpg


The Attempt at a Solution


I thought i applied the formulas correctly
20150707_161132_zpsa7yycxn5.jpg
[/B]
 
Physics news on Phys.org
  • #2
Its easier without calculus. Compute the mass and center of the two separate pieces, then compute the net CM.
 
  • #3
Dr. Courtney said:
Its easier without calculus. Compute the mass and center of the two separate pieces, then compute the net CM.
I don't see why the com in the vertical (x-axis) is not 13.4/2 = 6.7cm.
 
  • #4
I would approach this by getting the COM for each of the metal blocks. You then look at the different mass density ratio's of the two blocks. In the case above iron to aluminium is pretty much a 3:1 ratio. You then take the block as a whole and get the COM if both sides were the same density. Then use the COM for both blocks and adjust it with respect to the COM of the separate Fe and Al blocks in the 3:1 ratio.

For example the x direction, the COM of Fe is halfway through the block at 5.8cm, and the same for Al. Since Fe is 3 times denser than Al move the COM for both blocks closer to the COM of the Fe in a 3:1 ratio. So the overall COM in the x direction is 2.9cm from the COM of the Fe block.

Repeat that train of thought for y and z and then get the point that is closest to all three.

Sorry if that answer seems convoluted.
 
  • #5
J-dizzal said:
I don't see why the com in the vertical (x-axis) is not 13.4/2 = 6.7cm.
It should be. The answer says it's not?
 
  • #6
Nathanael said:
It should be.
yea, I've done it by symmetry and using the xcom formula and each time 6.7cm
 
  • #7
Nathanael said:
It should be. The answer says it's not?
20150707_172625_zpsy7gbuml4.jpg
 
  • #8
Perhaps it's because it lies on the negative side of the axis. Try -6.7 cm
 
  • Like
Likes J-dizzal
  • #9
Nathanael said:
Perhaps it's because it lies on the negative side of the axis. Try -6.7 cm
yep that's why thanks sir.
 

1. What is the definition of center of mass?

The center of mass is the point at which the entire mass of an object can be considered to be concentrated. It is the point where the object would balance if it were suspended at that point.

2. How is the center of mass of two connected blocks calculated?

The center of mass of two connected blocks can be calculated by finding the weighted average of the positions of the individual center of masses. This takes into account the mass and distance of each block from the axis of rotation.

3. What factors affect the center of mass of two connected blocks?

The center of mass is affected by the mass and position of each block. The closer a block is to the axis of rotation, the more it will contribute to the overall center of mass.

4. How does the density of each block impact the center of mass?

The density of each block impacts the center of mass by affecting the mass of each block. A block with a higher density will have more mass, and therefore, will have a greater influence on the overall center of mass.

5. Can the center of mass of two connected blocks be outside of the physical object?

Yes, the center of mass can be outside of the physical object if the distribution of mass is uneven. This can occur if one block is significantly larger or denser than the other, causing the center of mass to shift towards that block.

Similar threads

  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
9
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
11K
  • Introductory Physics Homework Help
Replies
13
Views
8K
  • Introductory Physics Homework Help
Replies
2
Views
7K
  • Introductory Physics Homework Help
Replies
9
Views
6K
  • Introductory Physics Homework Help
Replies
11
Views
1K
  • Introductory Physics Homework Help
Replies
17
Views
4K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
3K
Back
Top