Question

a disc rotates with uniform angular velocity . in the centre of the disc a man with two blocks in his hands sits in the ventre he then streches his hands such that the two blocks are on the circumference . what is the new inertia?

Answer 1

     Let the mass of the disc be M.  Let the mass of the man be  m1.  Let the mass of each block be  m2.    Let the radius of the disc be R.  Le the angular velocity of the disc be  ω.

The moment of inertia of the disc, about the axis passing through its center and perpendicular to its surface.

    I = $I= \int\limits^R_0 {r^2*(\rho *2\pi r} \, dr)\\\\=\frac{M}{\pi R^2}*2\pi \int\limits^R_0 {r^3} \, dr\\\\=\frac{2M}{R^2}\frac{R^4}{4}\\\\=\frac{MR^2}{2}$

Moment of inertia of the man:    0  as he sits at the center.  So distance of the person from the axis of rotation is 0.

  Moment of inertia of the two blocks =  2 * m2 * R²

  The new inertia :  M R² /2 + 2  m2  R²
               =  ( M/2 + 2 m2)  R²