What is the triangle’s moment of inertia about an axis perpendicular to the screen and passing through point O? Point G? The image depicts a triangular frame made from wire, which has a mass per unit length of 0.4 kg/m. The triangle's sides are each of length \( s = 89 \) mm. The triangle is suspended from a point \( O \) at its top vertex, and point \( G \) represents the centroid of the triangle.### Diagram Details:- **Points**: The triangle is labeled with points \( A \), \( B \), and \( O \).- **Angles**: The internal angles at points \( A \) and \( B \) are each \( 60^\circ \).- **Dimensions**: - The horizontal distance along the base \( AB \) is labeled as \( s \). - The vertical distance from the base to the vertex \( O \) is labeled as \( \overline{y} \).### Additional Details:- The triangle is equilateral, with all angles being \( 60^\circ \) degrees.- The size of the suspension ring at \( O \) is considered negligible in calculations.

Given data: s=89 mmm=0.4 kg/m Need to determine the moment of inertia about point O and G. Calculate the mass of each side of the triangle. m=0.4×0.089 =0.0356 kg Calculate the moment of inertia about O. IO=IOA+IOB+IABAbout point O =mL23+mL23+mL212+mLsin602 =2mL23+mL212+mL2×34 =18mL212 =18×0.0356×0.089212 =4.23×10-4 kg.m2 Calculate the distance OD. sin60=ODOAOD=89sin60OD=77.08 mm Calculate the distance GD. GD=13OD =13×77.08 mm =25.7 mm Calculate the moment of inertia about the point G. IG=mL212+m×GD2×3 =30.0356×0.089212+0.0356×0.02572IG=32.35×10-5+2.35×10-5IG=1.4×10-4 kg.m2

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Engineering

Mechanical Engineering

What is the triangle’s moment of inertia about an axis perpendicular to the screen and passing through point O? Point G?

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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