Moments and Centers of Mass

Moments and Centers of Mass In our initial treatment of objects under constant acceleration we treated the object as though it were a single point. We ignored any possible effects due to an irregular shape of the object, such as rotation. This does not seem like it would be an especially useful approximation, except for the fact that we can treat many physical objects as though all of their mass was concentrated at a single point, called the center of mass of an object. We can describe the overall motion of an object by treating like a point at its center of mass, and later consider ancillary effects such as rotation around the center of mass. For a rigid body (a body that can only negligibly be deformed), the center of mass is a fixed point. Let us consider a one-dimensional bar, which we are trying to balance on a fulcrum. Let us ignore the mass of the bar, and only consider mass contributed by a number of point masses put at various places along the bar. Each of these point masses will exert a torque in conjunction with the force of gravity, which will try to rotate the bar. Each of these torques depends on the distance of the mass on the bar from the fulcrum, so if we place the fulcrum at just the right place, the bar will balance. This specific point is the center of mass of the bar, which makes sense because if all of the mass of the point masses were put at exactly this point (above the fulcrum), then clearly the bar would balance, as there would be no torque acting on it. The torque is given by the force of each mass acting downward on the bar (m · g) times the signed distance from the flucrum (because a mass will rotate the lever in a different direction depending on whether it is to the left or right of the fulcrum). If we let each of these masses reside at a point xi , with mass mi , we find the total torque is given by X xi mi g i Now when we move the fulcrum, the signed distance will become xi − x̄, where x̄ is the new location of the fulcrum, the center of mass, where the system balances. Since the system balances, the total torque will be zero, so X (xi − x̄)mi g = 0 i which finally yields that P mi xi x̄ = P mi The numerator in this equation is called the system moment about the origin, and the denominator is simply the total mass of the system. Once again, x̄ is the system’s center of mass. In the above illustration we are using an idealization, by claiming that the mass is all located at a number of discrete points. More realistically, the mass of the system will be continuously distributed along the entire length of the bar. In order to represent this, we consider the quantity δ(x), which is the density of the bar. This is a linear density, insofar that by multiplying by a length, the resulting quantity is mass. If we imagine a very small strip of our total bar, then its mass will be given by ∆mi ≈ δ(xi )∆xi By summing up all of these contributions (over an interval a to b) in the limit that each length approaches zero, we find the exact moment and center of mass of the bar, given by Rb xδ(x)dx x̄ = Ra b δ(x)dx a where the numerator is the system moment, and then denominator is the total mass of the bar. Example 1 A 10-m has a density that increases from left to right, given by δ(x) = 1+(x/10), where the left end of the rod begins at x = 0. Find the rod’s center of mass. Solution We begin by finding the rod’s moment about the origin. Since our rod is 10m long, we will integrate from 0 to 10. Z 10 Z 10  h x2 x3 10 x 100 250 xδ(x)dx = x 1+ = 50 + dx = + = 10 2 30 0 3 3 0 0 The rod’s mass is given by Z 10 Z δ(x) = 0 10  1+ 0 h x2 x dx = x + 10 20 10 = 10 + 5 = 15 0 Thus, the center of mass is given by x̄ = 250/3 50 = ≈ 5.56 15 9 Example 2 Consider a 100-cm vertical bar composed of a substance that has settled over time, so it has higher density near the bottom of the bar. Let z denote height, and δ(z) = e−0.01z in grams per centimeter. Find the mass of the bar, and its center of mass. Solution The mass is given by Z 100 100 1 e−0.01z dz = e−0.01z = −100(e−1 − 1) ≈ 63.21 −0.01 0 0 Here we cannot find the center of mass of the object because we would need to solve the integral Z ze−0.01z dz which unfortunately requires the method of integration by parts. However, we can verify that −(100z + 10000)e−0.01z is a solution through differentiation. We find that  d − (100z + 10000)e−0.01z = −100e−0.01z + ze−0.01z + 100e−0.01z = ze−0.01z dz Thus, the system’s moment is given by Z 100 ze−0.01z dz = −(100z + 10000)e−0.01z 100 =− 0 0 20000 + 10000 ≈ 2642.41 e and so 2642.41 = 41.8 63.21 which is slightly below the middle of the bar, because it has higher density at the bottom. x̄ = Example 3 Show that the center of mass of a straight, thin strip or rod of constant density lies halfway between its two ends. Solution Suppose that a rod extends from a to b, so that we need to show x̄ = a+b 2 which is halfway between its two ends. In this situation the density is constant, so δ(x) = δ, which is simply some constant. Now we find that Z b δ x2 b xδdx = δ = (b2 − a2 ) 2 a 2 a and Z b b = δ(b − a) δdx = δx a a Finally, x̄ = δ 2 (b 2 − a2 ) (a + b)(b − a) a+b = = δ(b − a) 2(b − a) 2 When we have an object with constant density, the center of mass corresponds with the object’s centroid, which is a geometric feature of the object. Thus, in such cases, the actual constant does not matter at all, only the fact that the density is constant.