Arun
Last Activity: 6 Years ago
Given in the question,
Length of the Rod - 1 m
Mass of the Rod - 0.5 g
Moment of Inertia at perpendicular bisector
I = ml^2 / 12
=0.50 × 1² /12
=1/ 24 kg.m
At the distance r the moment of inertia , rod middle point l' = 0.10 kg-m²
I’ = I + m r^2
0.10 = 1/ 24 +0.50 r²
1+0.50× 24 r² =0.10× 24
12 r² =2.4 -1
r² =1.4/12
r²=0.467/4
r= 0.68/2
r= 0.34 m.
Hence the distance from the middle point is equal to 0.34 m.