∑ m i i r = . Remember that the position vectors ( r i ) are taken with respect to the CG. Now, in accordance with the reasoning given below Eq. ( .4a) in Sec.
. , if the sum is zero, the origin must be the centre of mass of the body. Thus, the centre of gravity of the body coincides with the centre of mass in uniform gravity or gravity- Fig. .
Determining the centre of gravity of a body of irregular shape. The centre of gravity G lies on the vertical AA through the point of suspension of the body A. u u free space. We note that this is true because the body being small, g does not vary from one point of the body to the other.
If the body is so extended that g varies from part to part of the body, then the centre of gravity and centre of mass will not coincide. Basically, the two are different concepts. The centre of mass has nothing to do with gravity. It depends only on the distribution of mass of the body.
In Sec. . we found out the position of the centre of mass of several regular, homogeneous objects. Obviously the method used there gives us also the centre of gravity of these bodies, if they are small enough.
Figure . illustrates another way of determining the CG of an irregular shaped body like a cardboard. If you suspend the body from some point like A, the vertical line through A passes through the CG. We mark the vertical AA .
We then suspend the body through other points like B and C. The intersection of the verticals gives the CG. Explain why the method works. Since the body is small enough, the method allows us to determine also its centre of mass.
Example . A metal bar cm long and . kg in mass supported on two knife- edges placed cm from each end. A .
kg load is suspended at cm from one end. Find the