Center of Gravity an Center of Mass 1 Introuction. Center of mass an center of gravity closely parallel each other: they both work the same way. Center of mass is the more important, but center of gravity is easier to grasp intuitively. Accoringly, I will evelop the basic ieas using center of gravity an then switch over to center of mass. 2 Torque an Center of Gravity. Say you take your younger sister 1 to the playgroun, an the two of you get on the seesaw. If, say, you weigh twice what she oes, then in orer for the seesaw to balance, she will have to sit twice as far from the fulcrum 2 as you o. This is because of the following facts. Because your weight an her weight both push the seesaw ownwars, each of you is exerting a twist, calle a torque, on the seesaw. Your torque an your sister s act in opposite irections. The magnitue of each torque is the prouct of the force (weight) times the istance to the fulcrum. The seesaw will balance the two torques cancel, which will happen if, an only if, (your weight) (your istance) = (your sister s weight) (your sister s istance). Calculating total torque. Say now that a number of chilren of various weights sit own at various positions on the seesaw. You can fin out whether the seesaw will tip to the left, tip to the right, or balance by carrying out a certain calculation. To set the stage, let us introuce an axis. Say the fulcrum of the seesaw is locate at x = a, an, for 1 i n, chil #i, who weighs w i pouns, is sitting at position x i. For 1 i n: this chil s torque exerte by chil #i is the prouct (1) τ i = w i (x i a); { } clockwise the torque pushes the seesaw accoring as counterclockwise 1 Probably not the same one who ha the fifth-grae lampshae project. 2 The place where the seesaw is supporte. { } τi > 0. τ i < 0 1
The total torque exerte by the chilren on the seesaw (with fulcrum at x = a) is then 3 (2) total torque about (x = a) = τ i = w i (x i a); the seesaw will balance at x = a the total torque about a equals zero. Center of gravity. Now suppose that the seesaw oes not balance, an you want to make it balance not by rearranging the chilren but instea by moving the fulcrum of the seesaw. 4 The center of gravity of this arrangement of chilren is the position a = x where you must place the fulcrum of the seesaw in orer to make the total torque on the seesaw equal to zero. (This is where the seesaw will balance.) It is important to know how to calculate x; but before iscussing how to o so, I will switch from weights to masses. 3 Moment an Center of Mass. If one were to ress the chilren in spacesuits an ferry them an the seesaw out between the stars, they woul weigh essentially nothing there woul be no gravity so it woul make no sense to talk about center of gravity. They woul still have mass, however, so that calculations similar to (1) an (2) coul still be mae. In this context, the quantity that correspons to torque is calle a moment about x = a: the moment of chil #i about x = a is given by (3) µ i = m i (x i a), where m i is the mass of chil #i; equation (2) is replace by (4) total moment about (x = a) = µ i = m i (x i a). In this context, The center of mass of the arrangement of chilren is the position x where you must place the fulcrum of the seesaw in orer to make the total moment equal to zero. The remainer of this hanout will focus on moments an centers of mass; but all of the ieas apply in the torque/center of gravity case as well. 3 Note that there are no absolute values in equation (1) or equation (2); allowing (x i a) to keep its sign permits opposite torques to cancel each other. 4 Ignore the weight of the seesaw itself. 2
4 Calculating the Center of Mass. The center of mass x must satisfy the equation (5) m i (x i x) = 0; the formula for x is obtaine by solving (5) for x. The solution to equation (5) is given by formula (6): (6) x = m i x i. m i I will not erive (6) from (5); rather, I will erive equation (11) on p.5 by solving equation (10) for x. The two erivations work the same way, but the erivation of (11) from (10) is easier to follow. Observe that the enominator of (6) is simply the total mass of the collection of objects; observe also that m i x i = m i (x i 0), so that putting a = 0 in (4) the numerator of (6) is the total moment about x = 0 (the y-axis). For this reason, equation (6) is often written x = M y mass, where M y stans for the moment about the y-axis. Do not let this notation confuse you: keep in min that M y is an x-irection quantity. 5 Center of Mass in R 2. Now consier a system of objects istribute through R 2 : say, for 1 i n, that object #i, with mass m i, is locate at (x i, y i ). Equation (4) can be interprete as the total moment of this system about the vertical line x = a the y-coorinates are irrelevant for this. Similarly, one can iscuss moments aroun horizontal lines: (7) total moment about (y = b) = m i (y i b) (the x-coorinates are irrelevant here.) The center of mass in the y irection (enote y) must satisfy the equation (8) m i (y i y) = 0; 3
the solution to this equation is given by (9) y = m i y i = m i M x mass. The point (x, y) is the two-imensional center of mass of the system of objects. In orer to calculate it, you nee to calculate the three quantities: M y ; M x ; an mass. You then must take ratios of them to fin x an y. Using Calculus III tools, it is not har to prove that the entire system can be balance on this point. (I will iscuss this statement further in class.) 6 Continuous Masses. None of the preceing iscussion requires Calculus; Calculus comes into the picture when a system of objects (treate as points) is replace by a two-imensional object having length an with. Let R be a region in the plane occupie by an object; assume the object is constructe out of a material of ensity ρ gm./cm. 2, so that for any piece of the object, mass of the piece = ρ (area of the piece). If the cross-section in the y irection is c(x) in length, then the slice of thickness x at a given x has area c(x)x an mass ρc(x)x; therefore, the moment of this slice about x = a is given by (x a)ρc(x)x. The total moment of the object about the line x = a is ρ(x a)c(x) x, so that x, the center of mass in the x-irection, must satisfy the equation (10) ρ(x x)c(x) x = 0. As promise, I will solve (10) for x. First, multiply out the integran an istribute the integral: ρxc(x) ρxc(x) x = 0 ρxc(x) ρxc(x) x = 0 4
Next, move the secon integral to the other sie of the equation an factor constants out of the integrals. ρ ρxc(x) x = xc(x) x = ρx ρxc(x) x c(x) x Finally, cancel the ρ s from both sies an ivie both sies by We have now solve (10) for x: (11) x = ( ) b xc(x) x = x c(x) x xc(x) x c(x) x xc(x) x = c(x) x = x. ρxc(x) x ρc(x) x = M y mass. c(x) x. In precisely the same way, the center of mass in the y irection, y, satisfies the equation (12) f e ρ(y y)c(y) y = 0; an (12) can be solve in precisely the same way as (10) to give: (13) y = f e yc(y) y = c(x) x f e ρyc(y) y ρc(x) x = M x mass. Just as in the non-calculus case: in orer to calculate the center of mass, you nee to calculate the three quantities M y M x an mass; you then must take ratios of them to fin x an y. 5
7 Calculating M x with a x Integral. Note that the integral for the mass in equation (13) was left as a x integral, even though solving (12) for y woul have resulte in a y integral. This is because the mass in (13) is the same as the mass in (11), so that the same integral can be use for the mass in both formulas. It turns out that in many cases, the numerator M x in (13) can also be calculate by a x integral. If the region R is boune above by the graph of y = f(x) an boune below by the graph of y = g(x) on x b, then M x can be calculate by the integral M x = In class, I will explain where this formula comes from. ρ [ (f(x) ) 2 ( ) ] 2 g(x) x. 2 6