Center of Gravity & Mass: Dist for 2m Rod

AI Thread Summary
The distance between the center of gravity and the center of mass of a uniform vertical rod of length 2 meters is zero, as both points coincide in a uniform gravitational field. The discussion highlights that while the mass distribution is uniform, the uniformity of the gravitational field is crucial for this conclusion. It is noted that in non-uniform gravitational fields, the center of mass and center of gravity can differ, with gravity exerting more force on lower parts of an object. The conversation also touches on the importance of understanding the context of the course material when considering these concepts. Overall, the relationship between center of mass and center of gravity is primarily dependent on the uniformity of the gravitational field.
Ellen Liu
Messages
1
Reaction score
0

Homework Statement


What is the distance between the center of gravity and the center of mass of a vertical rod of length 2 meters, and with a uniform mass distribution? (Center of gravity is defined like the center of mass with weight replacing mass in the formula).

Homework Equations


nothing

The Attempt at a Solution


the answer should be zero because it's uniform mass distributed.
Btw final tomorrow at 11:30
 
Physics news on Phys.org
Ellen Liu said:
the answer should be zero because it's uniform mass distributed.
Agree.
 
"Nothing" isn't very much of a relevant equation. The least you could come up with here is definitions of the center of mass and center of gravity in equation form. From there it's easy to see that the factor g that distinguishes the two can be brought outside a summation or integral in both numerator and denominator, so that it cancels and the two are shown to be equal.

The problem statement is weird: the mass distribution doesn't matter. The fact that the gravity field is uniform matters. And that isn't even stated. See hyperphysics

Good luck in the final :smile: !
 
Generally speaking, center of mass and center of gravity are not in the same location, even with uniform mass distributions (assuming a non-uniform gravity field). This is because gravity pulls harder on things that are closer to the attracting body. So for the case of a vertical rod, the "bottom" end of the rod will be pulled harder by the gravitational force. Obviously the CoM is the center of the rod. Think about this:
## df = \mu Mdm/r^2 ##
You know the total force acts on the CG:
## F = \mu Mm/r_{CG}^2 ##
Then, set up the integral in terms of dr, a length element of the rod:
## F = \int \mu Mdm/r_{CG}^2 = \int \mu Mmdr/(lr^2) ##
Evaluate the integral and compare to the known expression of the total force.
 
Last edited:
Agree with Sunny. So now we have to ask Ellen how the final went, and whether the context (level of the course) justifies taking the non-uniformity of the gravity field into consideration. To introduce this kind of detail necessitates knowledge of that field: the answer then depends on where you are wrt its sources (in deep space: COM still = center of gravity, or: COM exists, COG does not ?)
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top