Calculus: Analytic Geometry and Calculus, with Vectors

4.7 Mass, linear density, and moments 263

7 When M = 7, the graphs of the functions in the preceding problem are

different from thegraphs obtained when M = 0. What is the difference?

8 When a particle having mass m rotates in a circular path with angular

speed w radians per second at a constant distance r from an axis of rotation, its

speed is rw and its kinetic energy is 4mr2w2. With the aid of this information,

calculate the kinetic energy of a circular disk of radius a which has mass 8 per

unit area and which is rotating with speed w radians per second about an axis

through its center perpendicular to its plane. Hint: Base the solution on esti-

mates of the areaof a ring, the mass of the ring, and then the kinetic energy of

the ring. Ans.: KE _ jw6aaw2. The answer has the form KE _ yIw2, where I,

the polar moment of inertia of the disk about the axis used, is8a4.

9 The cone of Figure 4.51 has mass 3 per unit volume and is rotating w

radians per second about its axis. Find its kinetic energy. Hint: Use the answer

of Problem 8.

10 Figure 4.792 can make us wonder whether we are becoming wise enough

to determine the attractive force F upon a particle of mass in at P(x,y,z) that is

produced by a bar or rod concentrated upon an

interval a < x < b of the x axis of an x, y, z y

coordinate system. We suppose that the bar

has linear density 8(x) at the point (x,0,0) and

that 8(x) is integrable but not necessarily z a b `

constant over the interval a 5 x 5 b. The Figure 4.792

first task is to set up an integral for F. The

following solution of this problem should be read even by those who can solve

the problem without aid and assistance, because it fortifies our understanding

of the process by which integrals are set up. We make a partition Q of the inter-

val a 5 x < b, but we call the partition points to, t1, , t because the num-

ber x is the x coordinate of P. If the trick helps us, we can consider the x axis

to be simultaneously an x axis and a t axis. For each k = 1, 2, , it lettx

be chosen such that tk_1 5 tk 5 tk. We then use the number

(1) s(tk) ti

as an approximation to the mass of the part of the rod in the interval tk_I < t 5
tk. Supposing that this mass is concentrated at the point Pk(tk,0,0), we use the
number

(2) Gm

S(tk) Otk

IPPk12

as an approximation to the magnitude of the force OFk on the particle at P pro-
duced by the part of the rod in the interval tk_I 5 t < tk. Problem 19 of Prob-
lems 2.39 discusses this matter and shows how we derive the formula

(3) AFk = Gin

E(tk) Otk PPk

IPPkIs

by use of the fact that a nonzero vector is the product of its magnitude and a