moment of inertia about the x axis of the semicircular segment is:

Question

moment of inertia about the x axis of the semicircular segment  is:

Quizplus · Accepted Answer

The moment of inertia of a semicircular segment about its own axis of symmetry (diameter) is given by:

$$I = \frac{\pi}{8} R^4 - \frac{8}{15}R^4 = \frac{\pi}{120} R^4$$

where R is the radius of the semicircle.

To find the moment of inertia about the x-axis, we need to use the parallel axis theorem. The distance between the x-axis and the centroid of the segment is:

$$d = \frac{4R}{3\pi}$$

The moment of inertia about the x-axis is then:

$$I_x = I + Ad^2$$

where A is the area of the segment.

The area of a semicircular segment is:

$$A = \frac{1}{2}\pi R^2 - \frac{1}{2}R^2 = \frac{\pi - 2}{4}R^2$$

Substituting the expressions for I, A, and d, we get:

$$I_x = \frac{\pi}{120} R^4 + \left(\frac{\pi - 2}{4}ight)\left(\frac{4R}{3\pi}ight)^2 = \frac{13\pi - 24}{36\pi}R^4$$

Plugging in R = 6 in., we get:

$$I_x = \frac{13\pi - 24}{36\pi}(6	ext{ in.})^4 \approx 408	ext{ in.}^4$$

Therefore, the best choice is C.

Moment of Inertia About the X Axis of the Semicircular 4^ { 4 }4

Moment of Inertia About the X Axis of the Semicircular $^ { 4 }$