Physics, asked by thakarev063, 6 months ago

From a rectangle having breadth 200 mm and depth 500 mm, a semicircular portion of diameter 200 mm is removed. Find M. I. of remainder about the centroidal X and Y axes.​

Answers

Answered by nikingowda70
0

Explanation:

The remaining area can be claculated by subtracting the area of semi circular portion from rectangular portion,

⟹remaining area = area of rectangle - area of semi circular portion,

now,let us define σ= mass per unit area = 2r2M,

where M= mass of rectangle of area 2r2

mass of semicircular portion = σ×πr2/2 =4Mπ

now centre of mass of remaining portion =m1−m2m1r1−m2r2,

where m1= mass of rectangle  and m2= mass of semi circular portion

r1 and r2 are position of centre of mass of rectangle and semi circular portion with respect to O

r1=r/2 and r2=3π4r

centre of mass of remaining portion=M−Mπ/4Mr/2−(Mπ/4)×(4r/3π)=3(4−π)2r

Answered by vishalboraste554
0

Answer:

0.51 *10^3mm^4

Explanation:

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