Solveeit Logo

Question

Physics Question on System of Particles & Rotational Motion

For spheres each of mass MM and radius RR are placed with their centres on the four comers A,B,CA, B, C and DD of a square of side bb. The spheres AA and BB are hollow and CC and DD are solids. The moment of inertia of the system about side ADAD of square is

A

83MR2+2Mb2\frac{8}{3}MR^{2}+2Mb^{2}

B

85MR2+2Mb2\frac{8}{5}MR^{2}+2Mb^{2}

C

3215MR2+2Mb2\frac{32}{15}MR^{2}+2Mb^{2}

D

32MR2+4Mb232MR^{2}+4Mb^{2}

Answer

3215MR2+2Mb2\frac{32}{15}MR^{2}+2Mb^{2}

Explanation

Solution

Moment of inertia of a hollow sphere of radius RR about the diameter passing through DD is
IA=23MR2(i)I_{A}=\frac{2}{3} M R^{2} \ldots(i)
Moment of inertia of sol id sphere about diameter
IB=25MR2(ii)I_{B}=\frac{2}{5} M R^{2} \ldots(i i)
\therefore Moment of inertia of whole system about side
AD=IA+ID+IB+ICA D=I_{A}+I_{D}+I_{B}+I_{C}
=23MR2+25MR2+(Mb2+23MR2)+(Mb2+25MR2)=\frac{2}{3} M R^{2}+\frac{2}{5} M R^{2}+\left(M b^{2}+\frac{2}{3} M R^{2}\right)+\left(M b^{2}+\frac{2}{5} M R^{2}\right)
=3215MR2+2Mb2=\frac{32}{15} M R^{2}+2 M b^{2}