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HELP : Centre of mass and Centre of gravity |
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| Jul19-06, 02:06 PM | #1 |
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HELP : Centre of mass and Centre of gravity
hello ,
could anyone explain to me the difference between a.) the centre of mass b.) the centre of gravity much aprreciated =) Cheers |
| Jul19-06, 03:33 PM | #2 |
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There isn't really any.
The terms are just used differently depending on context, but it's the same thing |
| Jul19-06, 04:14 PM | #3 |
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No, they are not the same.
The centre of gravity is the point (if it exists) so that if the net gravitational force acting upon the object is considered to act at that point (rather than diffusely distributed at the various mass points the object consists of), then the torque wrt. the C.M of the object is the same as the torque (wrt. C.M.) as calculated for the diffusely distributed gravitational force. Evidently, for a constant force of gravity, the C.M and the C.G coincide. |
| Jul19-06, 10:36 PM | #4 |
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HELP : Centre of mass and Centre of gravity
hmmm sorry but i'm stll a lil confuse , could elaborate slightly more =)
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| Jul19-06, 11:43 PM | #5 |
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Don't worry about the differences, they're essentially the same thing |
| Jul20-06, 05:54 AM | #6 |
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They are not the same thing.
As measured from the C.M of the object, where [itex]\vec{F}[/itex] is the net (grav.)force on the object, and [itex]\vec{\tau}[/itex] is the net (grav.) torque wrt. to the C.M, we have that that the position of C.G, [itex]\vec{r}_{C.G}[/itex] is given by the formula: [tex]\vec{r}_{C.G}=\frac{\vec{F}\times\vec{\tau}}{|\vec{F}|^{2}}[/tex] under the condition [itex]\vec{F}\cdot\vec{\tau}=0[/itex] It by no means follows that we have [itex]\vec{r}_{C.G}=\vec{0}[/itex] |
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