Question

The moment of inertia of a uniform rod of mass 0.50 kg and length 1 m is 0.10 kg-m² about a line perpendicular to the rod. Find the distance of this line from the middle point of the rod?

Answer 1

Given in the question,
Length of the Rod - 1 m
Mass of the Rod - 0.5 g

Moment of Inertia at perpendicular bisector

$I = \frac{ml^2}{12}$

=0.50 × 1² /12
=1/ 24 kg.m

At the distance r the moment of inertia , rod middle point l' = 0.10 kg-m²

$I' = I + mr^2$ 
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.


Hope it Helps :-)

Answer 2

Length of the rod = 1 m, mass of the rod = 0.5 kg

Let at a distance d from the center the rod is moving

Applying parallel axis theorem :

The moment of inertial about that point

(mL2 / 12) + md2 = 0.10

(0.5 × 12)/12 + 0.5 × d2 = 0.10

d2 = 0.2 – 0.082 = 0.118

d = 0.342 m from the centre.