More generally, for any symmetry of a body, its center of mass will be a fixed point of that symmetry.
2007-01-05 19:11:27
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answer #1
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answered by The Last Paladin 4
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yes the center of the mass can lie outside the body also as in case of a ring which is hollow from the middle and its center of gravity lies in the air in the middle of it.
2007-01-06 05:00:10
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answer #2
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answered by Daksh Jain 1
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Yes, any object that curves back on itself can have its center of mass outside its body. Think of a hook or a ring.
2007-01-06 03:12:23
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answer #3
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answered by hznfrst 6
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of course s the center of mass of the object can lie outside. for example take a returning boomberang.
2007-01-07 07:25:59
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answer #4
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answered by sona 1
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yes
cosider a ring ,a hose shoe . even think of a stone tied to a thread and whirled arround in a circle.then the centre of mass of the system lies at the pt of rotation .
also the CM of eath moon system lies closer to earth!
2007-01-06 05:30:47
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answer #5
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answered by khandavillimahesh k 2
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yes definately...centre of mass lies outside the body in case of a ring...etc etc..
2007-01-06 03:25:04
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answer #6
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answered by Neelu 2
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Do you mean Centre of Gravity...???? If so... it does lie outside the body sometimes...
2007-01-06 03:11:35
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answer #7
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answered by Rrrahul 2
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yes, think pole-vaulter... during their ascent, their centre, physically is actually UNDER the bar, while their body curves over the top. Cool stuff.
2007-01-06 02:53:59
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answer #8
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answered by Anonymous
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yes centerof mas lie outside of body in case hollow objects it llie outside of the object example in ringgs , hollow cylinder,hollow hemisphere
2007-01-06 03:57:02
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answer #9
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answered by vande_matram27011987 2
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http://hyperphysics.phy-astr.gsu.edu/hbase/cm.html
http://en.wikipedia.org/wiki/Center_of_mass
The center of mass \mathbf{R} of a system of particles is defined as the average of their positions \mathbf{r}_i, weighted by their masses mi:
\mathbf{R} = \frac 1M \sum m_i \mathbf{r}_i
where M is the total mass of the system, equal to the sum of the particle masses.
For a continuous distribution with mass density \rho(\mathbf{r}), the sum becomes an integral:
\mathbf R =\frac 1M \int \mathbf{r} \; dm = \frac 1M \int\rho(\mathbf{r})\, \mathbf{r} \ dV =\frac{\int\rho(\mathbf{r})\, \mathbf{r} \ dV}{\int\rho(\mathbf{r})\ dV}
If an object has uniform density then its center of mass is the same as the centroid of its shape.
http://www.phys.psu.edu/~demo/Scott/energy/center_of_mass.html
http://dev.physicslab.org/Chapter.aspx?cid=22
2007-01-06 03:01:57
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answer #10
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answered by Anonymous
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