C.O.M. of Different Bodies and Tabulation of C.O.M.

C.O.M OF DIFFERENT BODIES AND TABULATION OF C.O.M Let's consider a semicircular disc with uniform mass M and radius R as depicted in the figure. Due to the symmetrical nature of the disc about the y-axis, the center of mass will be positioned along the y-axis. Let the coordinates of the center of mass be (0,y cmm ) Let's examine a thin circular element with thickness dr and mass dm located at a radius r from the center, as illustrated in the figure. (0,y mal ) Now, the area of the element is, Ignoring (dr) ^ 2 as it is very small, we get, Mass of the element is Therefore, the coordinates of the centre of mass are (0. (4R)/(3pi) ). y com = 1 M * int 0 ^ R ( 2r pi * 2M R^ 2 * r* dr) y com = 4 pi R^ 2 * int 0 ^ R r^ 2 dr= 4 pi R^ 2 *[ r^ 3 3 ] 0 ^ R dm = M/A * dA = (2M)/(pi * R ^ 2) * pi*rdr = (2Mrdr)/(R ^ 2) y com = 4R 3 pi dA = 1/2 * pi[(r + dr) ^ 2 - r ^ 2] dA = 1/2 * pi\{r ^ 2 + (dr) ^ 2 + (2rdr) - r ^ 2\} dA = pi*rdr Centre of Mass of Hollow Hemisphere Let's consider a hollow hemisphere with uniform mass M and radius R as depicted in the figure. Because the hemisphere exhibits symmetry about the y-axis, the center of mass will be positioned along the y-axis. The center of mass is located at coordinates 0,ycom. The hollow hemisphere can be conceptualized as comprising numerous thin rings arranged vertically, with their radii decreasing from the base to the top. Let's examine a slender ring with mass dm positioned at an angle 6 from the center, as depicted in the figure. Now, the area of the element Circumference Thickness dA = 2pi * R ^ 2 * cos theta*d*theta dA = 2pi(R * cos theta)(Rd*theta) Surface density of the hemisphere is, Mass of the element is, Now, the-coordinate of the centre of mass is given by. y coun = 1/M * integrate y dm Here, y = Centre of mass of the elemental ring = Rsin 0 sin 20dexos20 2sin cos sin 20] Therefore, the coordinates of the centre of mass are (0, R/2) dm = sigma*dA = M/(2pi * R ^ 2) * 2pi * R ^ 2 * cos theta * d*theta = M * cos theta*d*theta sigma = M/(2pi * R ^ 2) y com = 1 M * int 0 ^ 4 2 (Rsin theta* M* cos theta d theta) y com = R 2 R (0,y i = 0 ) R * cos theta Rd*theta Rsine R Centre of Mass of a Solid Hemisphere Let's contemplate a solid hemisphere with uniform mass M and radius R as depicted in the figure. Due to its symmetry about the y-axis, the center of mass of the hemisphere will be positioned along the y-axis. The center of mass is situated at coordinates 0, y com . The solid hemisphere can be conceptualized as comprising multiple thin discs arranged in a stacked manner, where the radius decreases as one moves from the base to the top. R * cos theta Risin R Let's examine a slender disc with mass dm positioned at an angle 0 from the center, as depicted in the figure. Now, the volume of the element Base area Height dV = pi * (R * cos theta) ^ 2 * (R * cos theta*d*theta) Volume density of the hemisphere is, rho = M/((2pi * R ^ 3)/3) = (3M)/(2pi * R ^ 3) Mass of the element is, Now the y coordinate y com = 1 M int ydm Here, Centre of mass of the elemental disc = Rsin 0 y = Let From properties of definite integration we have, Adding equation (2) and equation (3), we get, Put in equation (1) to get, Therefore, the coordinates of the centre of mass are. y cun = 1 M * int 0 ^ pi 2 (R sin theta* 3M 2 * cos^ 3 theta d theta) y con = 3R 2 * int 0 ^ pi 2 sin theta cos^ 3 theta d theta dm = rho*dV = (3M)/(2pi * R ^ 3) * pi * R ^ 3 * cos^3 theta * d*theta = (3M)/2 * cos^3 theta * d*theta 1 = integrate sin theta * cos^3 theta dtheta from 0 to pi/3 int 0 ^ pi 2 sin theta cos^ 3 theta d theta= int 0 ^ pi 2 sin(pi/2 - theta) * cos^3 ((pi/2 - theta) * d*theta) integrate f(x) dx from a to b = integrate f(a + b - x) dx from a to b y com = 3R s 2l = integrate sin theta * cos theta * (sin^2 theta + cos^2 theta) dtheta from 0 to pi/2 = integrate sin theta * cos theta dtheta from 0 to pi/2 (0, (3R)/n) 1 = integrate cos theta * sin^3 theta dtheta from 0 to pi/2 dV = pi * R ^ 3 * cos^3 theta I = 1/4 1 = 1/2 * integrate sin theta * cos theta dtheta from 0 to pi/2 = 1/4 * integrate sin 2theta dtheta from 0 to pi/2 = 1 4 [-cos 2 theta] 0 ^ pi 2 R * d*theta * cos theta R de B C ...(1) ... (2) ... (3) Table for COM of Certain Shapes Geometry Figure Coordinates of COM Rectangular lamina (d/2, 1/2) (b/2, 1/2) Triangular lamina (a/2, a/(2sqrt(3))) Semi-circular ring (0, (2theta)/s) (0, (2R)/n) Semicircular disc Delta * (4pi)/(3pi) ) (0, (4R)/(3pi)) Hollow hemisphere \langleplus/minus mu/2\rangle (0, R/2) Solid hemisphere (0; (2pi)/n) (0, (3R)/8) Hollow cone K, K/2 )= (0, H/3) Solid cone ( 0 , H bullet ) (0, H/4)