Question

Seven identical circular planar disks, each of mass m and radius r are welded symmetrically s shown. The moment of inertia of the arrangement about the axis normal to the plane and passing through the point p is:

Answer 1

you forgot to attach figure. see below , is it your perfect attachment , right ?

solution : we have to find moment of inertia about the axis normal to the plane through P

first of all, find moment of inertia about centre O .
e.g., $I_0=I_{cm}+Md^2$

where M = 6m [ if we assume mass of each disc is m, and ]
we also know, moment of inertia of disc about an axis passing through centre of mass and perpendicular to its plane , I = mr²/2

so, $I_0=7\times\frac{mr^2}{2}+6m\times(2r)^2$ [ here d = r + r = 2r]

$i_0=\frac{55mr^2}{2}$

so, moment of inertia about P , $I_P=I_0+md^2$ [ from parallel axis theorem]

so, $I_P=\frac{55mr^2}{2}+6m\times(2r)^2$

or, $I_P=\frac{181mr^2}{2}$

Answer 2

Answer:

tried to solve in easy steps..note that point A not focus on any sign I used because I did it for my help.

Explanation:

Find the moment of inertia of the disc (which is in the middle)along the axis perpendicular to its plane and passing through the centre.

Then use parallel axis theorem.find the moment of inertia of the remaining 6 discs about the centre of the middle disc and perpendicular to the plane.

Now you have the moment of inertia of seven discs about the axis passing through the centre and perpendicular to the plane of the middle disc.

Again use parallel axis theorem and shift the total inertia about the given point