Question 9.SP.2: (a) Determine the centroidal polar moment of inertia of a ci......

Vector Mechanics for Engineers Statics [1686415]

(a) Determine the centroidal polar moment of inertia of a circular area by direct integration. (b) Using the result of part (a), determine the moment of inertia of a circular area with respect to a diameter.

STRATEGY: Since the area is circular, you can evaluate part (a) by using an annular differential area. For part (b), you can use symmetry and Eq. (9.4) to solve for the moment of inertia with respect to a diameter.

$J_O=I_x+I_y$ (9.4)

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In this question, we are asked to find the polar moment of inertia and the moment of inertia with respect to a diameter for a circular area.

Step 1:

To find the polar moment of inertia, we consider an annular differential element of area, denoted as dA. Since all portions of the differential area are at the same distance from the origin, we can express the differential polar moment of inertia as dJ_O = u^2dA, where u represents the distance from the origin.

Step 2:

To find the total polar moment of inertia, we integrate the differential polar moment of inertia over the entire circular area. The differential area dA can be expressed as dA = 2πudu, where u ranges from 0 to r, the radius of the circular area. Therefore, the total polar moment of inertia, denoted as J_O, can be calculated as J_O = ∫dJ_O = ∫u^2(2πudu) = 2π∫u^3du.

Step 3:

Evaluating the integral, we find J_O = (π/2)r^4, where r is the radius of the circular area.

Step 4:

Next, we are asked to find the moment of inertia with respect to a diameter. Due to the symmetry of the circular area, the moment of inertia about the x-axis (I_x) is equal to the moment of inertia about the y-axis (I_y). Therefore, we can express the total polar moment of inertia as J_O = I_x + I_y = 2I_x. Step 5: Equating this expression

Step 5:

Equating this expression with the previously calculated value of J_O, we have (π/2)r^4 = 2I_x.

Step 6:

Solving for I_x, we find I_x = (π/4)r^4, which represents the moment of inertia with respect to a diameter.

In summary, the polar moment of inertia for a circular area is given by J_O = (π/2)r^4, and the moment of inertia with respect to a diameter is I_x = (π/4)r^4. These results can be obtained by considering the differential area and integrating over the circular region, taking advantage of the symmetry of the problem.

Final Answer

MODELING and ANALYSIS:

a. Polar Moment of Inertia. Choose an annular differential element of area to be dA (Fig. 1). Since all portions of the differential area are at the same distance from the origin, you have

$d J_O=u^2 d A \quad d A=2 \pi u d u$

$J_O=\int d J_O=\int_0^r u^2(2 \pi u d u)=2 \pi \int_0^r u^3 d u$

$J_O=\frac{\pi}{2} r^4$

b. Moment of Inertia with Respect to a Diameter. Because of the symmetry of the circular area, $I_x=I_y .$ . Then from Eq. (9.4), you have

$J_O=I_x+I_y=2 I_x \quad \frac{\pi}{2} r^4=2 I_x \quad I_{\text {diameter }}=I_x=\frac{\pi}{4} r^4$

REFLECT and THINK: Always look for ways to simplify a problem by the use of symmetry. This is especially true for situations involving circles or spheres.