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Chapter 5: Q9P (page 268)

Let the solid in Problem 7 have density $= c o s 胃$ .
Show that then $I_{z} = \frac{3}{10} M a^{2} s i n^{2} 伪$ .

Short Answer

Expert verified

The required value of $\frac{I_{z}}{M} i s \frac{3}{10} a^{2} \sin^{2} 伪$ .

Step by step solution

01

Given information

The density of the solid is $\cos 胃 .$

02

Concept of spherical coordinates

The spherical coordinates $(r, 胃, 蠒)$ is related to Cartesian coordinates(x,y,z) by:

$r = \sqrt{x^{2} + y^{2} + z^{2} 胃} = \tan^{- 1} (\frac{y}{x}) 蠒 = \cos^{- 1} (\frac{z}{r})$

The cylindrical coordinates $(r, 胃, 蠒)$ is related to Cartesian coordinates(x,y,z) by:

$r = \sqrt{x^{2} + y^{2} 胃} = t a n^{- 1} (\frac{y}{x}) z = z$

03

Evaluate the mass of the solid

Calculate the mass of the solid as follows:

$M = 鈭� 蚁 d V = {鈭�}_{0}^{a} r^{2} d r {鈭�}_{0}^{a} \cos 胃 d 胃 {鈭�}_{0}^{2 蟿蟿} d 蠒 = \frac{2 蟿蟿}{3} a^{3} {鈭�}_{0}^{a} \sin 胃 d 胃 (\cos 胃) = \frac{蟿蟿 a^{3}}{3} \sin 伪$

The moment of inertia is as follows:

$I_{z} = 鈭� 蚁 (x^{2} + y^{2}) d V = 蚁 (r^{2} \sin^{2} 胃) d V = {鈭�}_{0}^{2 蟿蟿} d 蠒 {鈭�}_{0}^{伪} \cos 胃 \sin^{3} 胃 d 胃 {鈭�}_{0}^{a} r^{4} d r = - \frac{2 蟿蟿 a^{5}}{5} {鈭�}_{0}^{伪} \sin^{3} 胃 d 胃 (\sin 胃)$

04

Use the above values to prove .

Write integral as follows:

$I_{Z} = - \frac{蟿蟿 a^{5}}{10} {(\sin^{4} 胃)}_{0}^{伪} = \frac{蟿蟿 a^{5}}{10} \sin^{4} 伪$

Substitute $\frac{蟿蟿 a^{5}}{10} \sin^{4} 伪$ for $I_{z}$ and $\frac{蟿蟿 a^{5}}{3} \sin 伪$ for M in $\frac{I_{z}}{M}$ .

$\frac{I_{z}}{M} = \frac{3}{10} a^{2} \sin^{2} 伪$

Therefore, the required value of $I_{z}$ is $\frac{3}{10} M a^{2} \sin^{2} 伪$ .

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Most popular questions from this chapter

For the solid bounded above by the sphere $x^{2} + y^{2} + z^{2} = 4$ and below by a horizontal plane through (0, 0, 1), find

(a) the volume (see Problem 6 and Problem 3.12);

(b) the z coordinate of the centroid (use cylindrical coordinates).

Above the triangle with vertices (0,2),(1,1) and (2,2) , and under the surface z = xy.

In the problems of this section, set up and evaluate the integrals by hand and check your results by computer ${鈭�}_{y = 0}^{2} {鈭�}_{x = 2 y}^{4} d x d y$

In the problems of this section, set up and evaluate the integrals by hand and check your results by computer.

${鈭�}_{x - 0}^{4} {鈭�}_{y - 0}^{x / 2} y - dy - dx$

Above the square with vertices at, (0,0), (2,0),(0,2) and (2,2) and under the plane z = 8-x+y.

See all solutions

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