Question

A big dumb bell is prepared by using a uniform rod of mass 40gm and length 20cm , two identical sphere of mass 20gm and radius 10cm reach at two ends of rod calculate moment of inertia of dumb bell when rotated about axes passing through it's centre and perpendicular to length

Answer 1

moment of inertia of the rod:

$$I_1=\frac 1{12}M_1L_1^2=1.33·10^{-4}\text{ kg m}^2.$$

The moment of inertia of two spheres:

$$I_2=2M_2\bigg(\frac25R^2+h^2\bigg)=1.76·10^{-3}\text{ kg m}^2.$$

Moment of inertia of the dumbell:

$$I=I_1+I_2=1.893·10^{-3}\text{ kg m}^2.$$