Question

For a thin plate shaped like a quarter circle, the polar moment of inertia at the center of the circle is given by

• A. \( \pi R^4 / 4 \)
• B. \( \pi R^4 / 8 \)
• C. \( \pi R^4 / 16 \)
• D. \( \pi R^4 / 2 \)

Hint:

Area Moments of Inertia. For a quarter circle about axes passing through the center of the full circle (corner of the quarter circle): \(I_x = I_y = \pi R^4 / 16\). Polar moment about that corner: \(J_O = I_x + I_y = \pi R^4 / 8\).

Answer 1

The Correct Option is B

This question asks for the polar moment of inertia of the *area* of a quarter circle about the center of the full circle (which is one corner of the quarter circle).
The polar moment of inertia (\(J_O\)) about an origin O is related to the area moments of inertia about the x and y axes passing through O by the Perpendicular Axis Theorem for areas: \(J_O = I_x + I_y\).
For a quarter circle of radius R in the first quadrant (bounded by x=0, y=0, and \(x^2+y^2=R^2\)): The area moment of inertia about the x-axis is \(I_x = \int_A y^2 dA\).
The area moment of inertia about the y-axis is \(I_y = \int_A x^2 dA\).
Due to symmetry of the quarter circle with respect to the line y=x in this quadrant, \(I_x = I_y\).
The standard formula for the area moment of inertia of a quarter circle about one of its straight edges (e.
g.
, the x-axis or y-axis passing through the center of the full circle) is \(I_x = I_y = \frac{\pi R^4}{16}\).
Using the Perpendicular Axis Theorem, the polar moment of inertia about the origin O (center of the full circle) is: $$ J_O = I_x + I_y = \frac{\pi R^4}{16} + \frac{\pi R^4}{16} = \frac{2\pi R^4}{16} = \frac{\pi R^4}{8} $$ (Note: The polar moment of inertia for a *full* circle about its center is \(J = \frac{\pi R^4}{2}\).
For a *semi-circle* about the center of the diameter, \(J = \frac{\pi R^4}{4}\).
For a quarter circle about the center, \(J = \frac{\pi R^4}{8}\)).