Question

(C) Calculate the M. I. of disc about the transverse axis through centre of disc whose radius is 10 cm and mass 49.2 kg.​

Answer 1

Answer:

hope it's helpful

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

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.

$</p><p>R = 10 cm \\ Thickness, x = 5 mm = 0.5 cm \\ Density, ρ = 8 g {cm}^{ - 3} , \\ I = ? \\ Mass \: of \: the \: disc, \\ M = area \times thickness \times density \\ = π {R}^{2} \times ρ \\ or, \\ M = 22/7 \times 102 \times 0.58 \\ = 8800/7 gram$

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

$I = \frac{1}{2} M {R}^{2} \\ = \frac{1}{2} \times \frac{8800}{7} \times {10}^{2} \\ = 6.28 \times 104 g {cm}^{2}$