   Chapter 15, Problem 10P

Chapter
Section
Textbook Problem

(a) A lamina has constant density ρ and takes the shape of a disk with center the origin and radius R. Use Newton’s Law of Gravitation (see Section 13.4) to show that the magnitude of the force of attraction that the lamina exerts on a body with mass m located at the point (0, 0, d) on the positive z-axis is F = 2 π G m ρ d ( 1 d − 1 R 2 + d 2 ) [Hint: Divide the disk as in Figure 15.3.4 and first compute the vertical component of the force exerted by the polar subrectangle Rij.](b) Show that the magnitude of the force of attraction of a lamina with density ρ that occupies an entire plane on an object with mass m located at a distance d from the plane is F = 2 π G m ρ Notice that this expression does not depend on d.

(a)

To determine

To show: The magnitude of the force of attraction is 2πGmρd(1d1R2+d2).

Explanation

Given:

The lamina has the constant density and takes the shape of the disk with center at origin and radius R.

Formula used:

The Newton’s law of gravitation is given by, F=GMmr2u.

If g(x) is the function of x and h(y) is the function of y then,

abcdg(x)h(y)dydx=abg(x)dxcdh(y)dy (1)

Calculation:

Since the given region is the disk centered at origin and radius R, use of polar coordinates is a wise choice. So separate the given region into small sub regions by split the r into r1,r2,,rn and θ into θ1,θ2,,θm. Denote each sub region by Rij where i denotes the value of r (ri*) and j denotes the position of θ (θj*). Therefore, the value of ΔAi is given by ΔAi=ri*ΔrΔθ. From the given conditions, it is observed that the mass of the region Rij is

M=ρA(Rij)=ρΔAi

The distance of Rij from the point m is denoted by sij and it is given by, sij(ri*)2+d2.

Substitute the value of M and u and rewrite the above mentioned formula as follows.

F=GmρΔAisij2

The total force of the lamina present only in the direction of z-axis as mentioned in the problem. Thus, consider only the components in the vertical direction. So, the force value becomes, F=GmρΔAisij2sinα where this α denotes the angle between the origin, ri* and m. Hence, sinα=dSij. Substitute the value of  in the equation of the force of attraction.

F=GmρΔAisij2(dsij)=GmρdΔAisij3.

Thus, the total force of attraction can be mentioned by the Riemann sum for double integrals as given below

(b)

To determine

To show: The magnitude of the force attraction of the given lamina that occupies an entire plane is given by, F=2πGmρ.

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