Gramin ACS College Vasantnagar, Kotgyal � MATHEMATIS

B.Sc. Final Year Paper-14^th (Mechanics)

Sem-V

PPT Presented by.....

Prof. Dr. P. R. Shinde

Department of Mathematics

Gramin ACS college Vasantnagar, kotgyal

Tq: Mukhed Dist: Nanded ( M.S) IN

Centre of Gravity

• C.G. of some uniform bodies:
• 1. C.G. of a uniform rod .
• Let G be the mid-point of the uniform rod PQ. Consider any point A1 in the segment GP and point B1 in the segment GQ such that GA1=GB1. since the C.G. Of two particles of equal weight is at the middle point of the line joining the particles.
• Therefore , the C.G of two equal particles at A1 and B1 is obviously G.
• Similarly, if we Consider other two equal particles at A2 and B2 such that GA2 = GB2
• Then clearly , G is the C.G. of equal particles at A2 and B2.
• Proceeding in this way, we see that for every particle in the segment GP, there is an equal particle at an equal disance from G in the segment GQ.
• Thus ,the C.G. Of the rod is at its mid- point.

• 2. C.G. Of a uniform triangular lamina:
• Let ABC be the uniform triangular lamina. Consider that the whole lamina is divided into a large number of infinitely thin strips parallel to the side BC. Each strip is assumed to be a thin uniform rod whose C.G. is at its mid-point . Take the strip PQ parallel to BC. Its C.G. Lies at its mid-points .
• If we now Consider the strips parallel to CA and arguing the similar way , we see that the C.G. Of the lamina lics on each median.
• Suppose G is the point of intersection of the medians such that AG:GD = 2:1
• It follows that the C.G. Of the uniform triangular lamina is at the point of intersection of the median of the triangle.