Question

the moment of inertia of a thin rod of mass M and length L about an Axis passing through its centre and normal to its length is giving by ML²/ 12 using theorem of parallel axis.find its moment of inertia about an Axis passing through On end and normal to its length if L=1 M and M = 0.2 Kg

Answer 1

${ml }^{2} + m {(l \div 2)}^{2}$

${ml}^{2} \div 12 + {ml }^{2} \div 4 \\ = {ml}^{2} \div 3 \\ = 0.2 {(1)}^{2} \div 3 \\ = 1 \div 15 \\ = 0.066$