Question

Determine the moment of inertia of the half-ring of mass m about its diametral axis a-a and about axis b-b through the midpoint of the arc normal to the plane of the ring. The radius of the circular cross section is small compared with r. Use the values m = 7.0 kg and r = 300 mm. m Answers: laa kg.m2 Ibb = kg.m2

Answer 1

Solution:

If you found this solution useful then please upvote

Thank you

Given that,

b o a G b a 0

For a full Ring,

у X

given, mass of the half-Ring = m = 7.0 kg Radius of the half- Ring = =r=0.3 m For a full ring, Тоо Too = 1 mg ² Ixx = Iyy 2 (from perpendicular Axis theorem Ioo = Ida & Tyy From symmetry Insa Iyy) This for a half Ring Ioo = t - (MP) = my? (: m= m) $ (mo) I un = 4 mm (*: M12 = m) Iuq = 2 * 7 * (0.3² = 0.315 kg-m² Iga 10 2 Tag = 0.315 kg-m 2 10o = I GG tm Let a be Centroid, distance between 0-0 and G-G Axis = 20/T From purullel Axis theorem (I = I + m2) Ibb = IGGtm (7_24/7) (217) Ibb-Too (r- upon (:cf-b = (4-6) (4+b)) Ibb my² + mr my (144) : Ibb = 2 mm* (- 6) = 2*7*(0-35* ("- 3*) (1-0.63) = mr = 2x7X0.3% 21 Ibb = 0.466 kg-m Thank you.