distance from the axis, by k. Find the mass of the body (Art. 11), and denote it by M.

We have, from Equation (102),

Substituting for M, 1, and k, the values already found, and the value of K will be the moment of inertia, with respect to the assumed axis. Subtract from this the value of Mk2, and the remainder will be the moment of inertia with respect to a parallel axis through the centre of gravity.

The moment of inertia of a homogeneous body of regular figure, is most readily found by means of the calculus. A few examples of the application of the calculus to finding the moment of inertia of bodies are subjoined.

Application of the Calculus to determine the Moment of Inertia.

131. To render Formula (104) suitable to the application of the calculus, we have simply to change the sign of summation, Σ, to that of integration, S, and to replace m by dM, and r by x. This gives,

Example 1. To find the moment of inertia of a rod or bar of uniform thickness with respect to an axis through its centre of gravity and perpendicular to the length of the rod.

D

Let AB represent the rod, G its centre of gravity, and E any element contained by planes at right angles to the length of the rod and infinitely near each other. Denote the mass of the rod by M, its length, by 27, the distance GE, by x, and the thickness of the element E,

G

E

IC

Fig. 110.

by dx. Then will the mass of the element E be equal to

M

27

dx. Substituting this for dM, in Equation (106), and integrating between the limits 7 and +, we have,

For any parallel axis whose distance from G is d, we shall have,

These two formulas are entirely independent of the breadth of the filament in the direction of the axis DC. They will, therefore, hold good when the filament AB is replaced by the rectangle KF. In this case, M becomes the mass of the rectangle, 27 the length of the rectangle, and d the distance of the centre of gravity of the rectangle from the axis parallel to one of its ends.

Example 2. To find the moment of inertia of a thin circular plate about one of its diameters.

Let ACB represent the plate, AB the axis, and C'D' any element parallel to AB. Denote

the radius OC, by r, the distance OE, by x, the breadth of the element EF, by da, and its length DC, by 2y. If we denote the entire mass of the plate, by M, the mass of the element

CD will be equal to M

2ydx

; or,

D

O EF

C

Substituting in Equation (106), we have,

2√ p2 M

- x2

dx.

B

Fig. 111.

Integrating by the aid of Formulas A and B (Integral Calculus), and taking the integral between the limits - r, and x = + r, we find,

x=

and for a parallel axis at a distance from AB equal to d,

Example 3. To find the moment of inertia of a circular plate with respect to an axis through its centre perpendicular to the face of the plate.

Let the dimensions and mass of the plate be the same as before. Let KL be an elemetary ring whose radius is x, and whose breadth dx. Then will the mass of the elementary ring be equal 2πxdx 2 Mxdx

to M

or dM =

προ

A

K

X

B

Fig. 112.

Substituting this in Equation (106), and taking the integral between the limits = 0, and x = r, we have,

For a parallel axis at a distance d from the primitive

Example 4. To find the moment of inertia of a circular ring, such as may be generated by revolving a rectangle about a line parallel to one of its sides, taken with respect to an axis through the centre of gravity and perpendicular to the face of the ring. This case differs but little from the preceding. Denote the inner radius by r, the outer radius by r', and the mass of the ring by M. If we take, as before,

an elementary ring whose radius is

Fig. 113.

2, and whose breadth is dx, we shall have for its mass,

Substituting in Equation (106), and integrating between the limits r, and r', we have,

For a parallel axis at a distance from the primitive axis equal to d, we have,

If in these values of K and K' we make r = 0, we shall deduce the results of the last example.

Example 5. To find the moment of inertia of a right cylinder with respect to an axis through the centre of gravity and perpendicular to the axis of the cylinder.

Let AB represent the cylinder, CD the axis through

A

E

B

its centre of gravity, and E an element of the cylinder between two planes perpendicular to the axis, and distant from each other, by da. Denote the length of the cylinder by 27, the area of its cross section by «r2, being the radius of the base; the distance of the section E from the centre of gravity, by x, and the mass of the cylinder, by M.

Fig. 114.

The mass of the element E is equal to M

moment of inertia with respect to its diameter parallel to

Integrating this expression between the limits x = and xl, we have,

For an axis parallel to the primitive one, and at a distance from it equal to d,

132. The centre of gyration of a body with respect to an axis, is a point at which, if the entire mass be concentrated, its moment of inertia will remain unchanged. The distance from this point to the axis is called the radius of gyration.