Question

Moment of Inertia (M.I.) of four bodies having same mass ‘ M ‘ and radius ‘2 R ‘ are as follows:

I 1 = M.I. of solid sphere about its diameter

I 2 = M.I. of solid cylinder about its axis

I 3 = M.I. of solid circular disc about its diameter.

I 4 = M.I. of thin circular ring about its diameter

If 2(I 2 + I 3) + I 4 = x⋅ I 1 then the value of x will be ______.

Answer 1

$2(\frac{1}{2}+\frac{1}{2})×M(2R)^2+\frac{1}{2}M(2R)^2=x\frac{2}{5}M(2R)^2$

$⇒1+\frac{1}{2}+\frac{1}{2}=x×\frac{2}{5}$

⇒ x = 5