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
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
0
Answer:
0.51 *10^3mm^4
Explanation:
hxcvoczjcjsxjfsckchvkfsbjdh
Similar questions
Math,
2 months ago
Physics,
6 months ago
Business Studies,
11 months ago
English,
11 months ago
Physics,
11 months ago