Question 504: Centre of Mass and Rotational Motion

Question

Two discs A and B of same material and thickness have radii $R$ and $3 R$ respectively. Their moments of inertia about their axis will be in the ratio

Accepted Answer

Let the thickness of both discs be $ t $, and their material has density $ ho $.
Let disc A have radius $ R $.
Let disc B have radius $ 3R $.
Step 1: Find Mass of Each Disc
Volume of a disc = Area ÃÂ Thickness
For disc of radius $ r $ and thickness $ t $:
$$ 	ext{Volume} = \pi r^2 t $$
So, mass $ M $ of a disc:
$$ M = (	ext{Volume}) 	imes (	ext{Density}) = \pi r^2 t ho $$
For disc A:
$$ M_A = \pi R^2 t ho $$
For disc B (radius $3R$):
$$ M_B = \pi (3R)^2 t ho = \pi \cdot 9R^2 t ho = 9 \pi R^2 t ho $$
Step 2: Moment of Inertia of a Uniform Disc about Its Axis
Formula:
$$ I = \frac{1}{2} M r^2 $$
For A:
$$ I_A = \frac{1}{2} M_A R^2 = \frac{1}{2} (\pi R^2 t ho) R^2 = \frac{1}{2} \pi t ho R^4 $$
For B:
$$ I_B = \frac{1}{2} M_B (3R)^2 = \frac{1}{2} (9 \pi R^2 t ho) 	imes (9R^2) $$
$$ I_B = \frac{1}{2} \cdot 9 \pi R^2 t ho \cdot 9R^2 = \frac{1}{2} \cdot 81 \pi t ho R^4 = \frac{81}{2} \pi t ho R^4 $$
Step 3: Ratio of Moments of Inertia
$$ \frac{I_A}{I_B} = \frac{ \frac{1}{2} \pi t ho R^4 }{ \frac{81}{2} \pi t ho R^4 } = \frac{1}{81} $$
So,
$$ I_A : I_B = 1 : 81 $$
Answer:
Option C $ (1 : 81) $ is correct.

Answer

$3: 1$

Answer

$1: 9$

Answer

$1: 81$

Answer

$1: 27$

Centre of Mass and Rotational Motion

Two discs A and B of same material and thickness have radii R and 3 R respectively. Their moments of inertia about their axis will be in the ratio

Choose the correct answer: