Question

the moment of inertia of a circular disc of radius 10 cm. thickness 5 mm and uniform density 8 g cm-3 about an axis perpendicular to the plane and passing through the centre of the disc​

Answer 1

Explanation:

R = 10 cm

Thickness, x = 5 mm = 0.5 cm

Density, ρ = 8 g cm-3, I ?

Mass of the disc, M = area x thickness x density = πR2 x ρ or,

M = 22/7 x 102 x 0.58 = 8800/7 gram

Moment of inertia of the disc about a transverse axis through the centre,

I = 1/2 MR2 = 1/2 x 8800/7 x (10)2 = 6.28 x 104 g cm2.

Answer 2

Given:

A circular disc of radius 10 cm. thickness 5 mm and uniform density 8 g/cm³ is provided.

To find:

Moment of Inertia across geometric axis ?

Calculation:

• First, we find out the mass of the disc.

$Mass = Volume \times density$

$\implies \: M =( \pi {r}^{2} \times d) \times \rho$

$\implies \: M =( \pi {10}^{2} \times 0.5) \times 8$

$\implies \: M =50\pi\times 8$

$\implies \: M =400\pi \: gram$

$\implies \: M \approx 1.25\:kg$

Now, Moment of Inertia is :

$MI = \dfrac{ M \times {r}^{2} }{2}$

$\implies MI = \dfrac{ 1.25 \times {(0.1)}^{2} }{2}$

$\implies MI = 0.0625 \: kg {m}^{2}$

So, value of MI is 0.0625 kg m².