Question

ascentric hole of radius r plan to is cut out of a uniform thin circular disc of radius R, resulting in the body shown the mass of this resulting body is m. the moment of inertia of this body about an axis p line in its plane and tangential it is​

Answer 1

Answer:

If the mass per unit area of the disc is m,

then mass of the portion remove from the disc is M

′

=π(R/2)

2

×m=(πR

2

/4)m=M/4

In figure centre of mass of M is at O and for M

′

is at O

′

But OO

′

=R/2.

After mass M

′

is removed the remaining portion can be taken as two masses M at O and −M

′

=−M/4 at O

′

we taking −M

′

because we are removing this mass.

distance of centre of gravity(P) of remaining part is :

X=

M+M

′

M×0−M

′

×R/2

X=

M−M/4

−M/4×R/2

X=−R/6