Question 2: Based on the passage, how is moment of inertia related to kinetic energy?

Hint 1: Read the beginning of the subsection "Moments of Inertia" (p. 934 - 935).

934 Chapter 15 Multiple Integrals Moments of Inertia An object's first moments (Table 15.1) tell us about balance and about the torque the object experiences about different axes in a gravitational field. If the object is a rotating shaft, however, we are more likely to be interested in how much energy is stored in the Axis of marion shaft or about how much energy is generated by a shaft rotating at a particular angular velocity. This is where the second moment or moment of inertia comes in. Think of partitioning the shaft into small blocks of mass Am, and let a denote the FIGURE 15.40 To find an integral for distance from the th block's center of mass to the axis of rotation (Figure 15.40). If the the amount of energy stored in a rotating shaft rotates at a constant angular velocity of w = 20/de radians per second, the block's shaft, we first imagine the shaft to be parti- center of mass will trace its orbit at a linear speed of tioned into small blocks. Each block has its own kinetic energy. We add the contribu- de tions of the individual blocks to find the kinetic energy of the shaft. The block's kinetic energy will be approximately The kinetic energy of the shaft will be approximately Equ'n' Am. The integral approached by these sums as the shaft is partitioned into smaller and smaller blocks gives the shaft's kinetic energy: KEsean = / parr dm - bor / 1 am. (1) The factor 1 = r di is the moment of inertia of the shaft about its axis of rotation, and we see from Equation (1) that the shaft's kinetic energy isKE shaft The moment of inertia of a shaft resembles in some ways the inertial mass of a loco- motive. To start a locomotive with mass / moving at a linear velocity v, we need to pro- vide a kinetic energy of KE = (1/2)mv'. To stop the locomotive we have to remove this amount of energy. To start a shaft with moment of inertia / rotating at an angular velocity it, we need to provide a kinetic energy of KE = (1/2)/or. To stop the shaft we have to take this amount of energy back out. The shaft's moment of inertia is analogous to the locomotive's mass. What makes the locomotive hard to start or stop is its mass. What makes the shaft hard to start or stop is its moment of inertia. The moment of inertia depends not only on the mass of the shaft but also on its distribution. Mass that is farther away from the axis of rotation contributes more to the moment of inertia. We now derive a formula for the moment of inertia for a solid in space. If r(x, y. z) is the distance from the point (x, y, z) in D to a line L, then the moment of inertia of the mass Am, = 8(x, y, 2)AV about the line L. (as in Figure 15.40) is approximately 4X = "(x, 3, 4)Am; . The moment of inertia about L of the entire object is I = lim SAN = lim -6 dV. 11-+0015.6 Applications 935 If L is the x-axis, then ' = yo + 2' (Figure 15.41) and (/ (2 + 2) 8(x. y.2) dV. Similarly, if L is the y-axis or z-axis we have dy I, = / ( + 2) 8(x.y.2) dV and 1. = (x + y) 8(x, yz) dV. D + Table 15.2 summarizes the formulas for these moments of inertia (second moments because they invoke the squares of the distances). It shows the definition of the polar moment about the origin as well. FIGURE 15.41 Distances from dV to the coordinate planes and axes. TABLE 15.2 Moments of inertia (second moments) formulasTHREE-DIMENSIONAL SOLID About the x-axis: 1 = /(2 + 2)6ev About the y-axis: 1 = /(2 + 2)8av About the z-axis: 1. = (2 + >2 ) adv About a line L: out. y. 2) = distance from the point (x. y, 2) to Line & TWO-DIMENSIONAL PLATE About the x-axis: 6 = K(x, y) About the y-axis: 1, = / x 6 dA About a line L: rix, y) = distance from (x, y) In L About the origin 1 = /62+ )EdA = b+ ly (polar moment): EXAMPLE 3 Find 1,, I, , I for the rectangular solid of constant density 8 shown in Figure 15.42. `Center of block Solution The formula for 1, gives FIGURE 15.42 Finding Ly, I,, and I, for the block shown here. The origin lies at the We can avoid some of the work of integration by observing that (y' + 2 )8 is an even center of the block (Example 3). function of x, y, and z since 8 is constant. The rectangular solid consists of eight symmetric