Question

Calculate the moment of inertia of a uniform circular disc of mass 400g and radius 10 cm about
(a) diameter of the disc
(B) the axis , tangent to the
dist and parallel to its diameter (c) the axis through the centre of disc and perpendicular to its
plane
​

Answer 1

Answer :-

(a) 0.001 kg m²

(b) 0.005 kg m²

(c) 0.002 kg m²

Explanation :-

We have :-

→ Mass of the disc (M) = 400 g = 0.4 kg

→ Radius (R) = 10 cm = 0.1 m

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For number (a) :-

Moment of inertia about the diameter of the disc will be :-

= (½MR²)/2

= ¼MR²

= 0.25 × 0.4 × (0.1)²

= 0.1 × 0.01

= 0.001 kg m²

For number (b) :-

If we apply the parallel axis theorem, then moment of inertia of the disc in this case will be :-

= MR² + ¼MR²

= 5/4 × MR²

= 1.25 × 0.4 × (0.1)²

= 0.5 × 0.01

= 0.005 kg m²

For number (c) :-

Moment of inertia of the disc in this case :-

= ½MR²

= 0.5 × 0.4 × (0.1)²

= 0.2 × 0.01

= 0.002 kg m²

Answer 2

Given :-

Uniform circular disc of mass 400g and radius 10 cm about

To Find :-

(a) diameter of the disc

(B) the axis , tangent to the  dist and parallel to its diameter (c) the axis through the centre of disc and perpendicular to its  plane

​Solution :-

1] 1 kg = 1000 g

400 g = 400/1000 = 0.4 kg

1 m = 100 cm

10 cm = 10/100 = 0.1 m

Now

Diameter = 1/4 × MR²

Diameter = 1/4 × 0.4 × (0.1)²

Diameter = 1/4 × 0.4 × 0.01

Diameter = 1/4 × 0.004

Diameter = 0.001 kg m²

2] Apply the parallel axis theorem

MR² + ¼MR²

4 + 1/4 MR²

5/4MR²

5/4 × 0.4 × (0.1)²

5/4 × 0.4 × 0.01

5/4 × 0.04

0.05 kg m²

3]  1/2 MR²

1/2 × 0.4 × (0.1)²

0.2 × 0.01

0.002 kg m²

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