Question

A triangle of height r& base ar is removed from a semicircuar lamina of radius r. The distance of centroid of remaining area from base is,​

Answer 1

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 =

2r

2

M

,

where M= mass of rectangle of area 2r

2

mass of semicircular portion = σ×πr

2

/2 =

4

Mπ

now centre of mass of remaining portion =

m

1

−m

2

m

1

r

1

−m

2

r

2

,

where m

1

= mass of rectangle and m

2

= mass of semi circular portion

r

1

and r

2

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

r

1

=r/2 and r

2

=

3π

4r

centre of mass of remaining portion=

M−Mπ/4

Mr/2−(Mπ/4)×(4r/3π)

=

3(4−π)

2r