Question

A circular disc A of radius r is made from an iron plate of thickness t and another circular disc B of radius 4r is made from an iron plate of thickness t/4. The relation between the moments of inertia IA and IB is
(a) IA > IB
(b) IA = IB
(c) IA < IB
(d) depends on the actual values of t and r.

Answer 1

Answer:

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Answer 2

(c) IA < IB

The relation between the moments of inertia IA and IB is IA < IB when a circular disc A with radius with thickness 't' and another circular plate with radius '4r' and thickness $\frac{t}{4}$

Explanation:

Moment of inertia of circular disc of radius r :

$I =\frac{1}{2} mr^{2}$

Mass = Volume × Density

Volume of disc =  πr 2 t x σ

Here, t is the thickness of the disc.

As density is same for both the rods, we have:  

Moment of inertia:-  Measurement of a body's resistance to angular acceleration on a given axis equal to the sum of products of each element of mass in the body and the square of the distance of the element from the axis

I ∝ thickness × radius  

$I =\frac{1}{2} mr^{2}$

$m \propto \mathrm{t} \propto \mathrm{R}^{2}$

For disc IA  =  $\frac{1}{2} mr^{2}$

$\text { For disc Is }\left[\frac{1}{2} \pi(4 R)^{2} \cdot \frac{t}{4}\right][4 R]^{2}$

I A < I B  is the relation between the moments of inertia IA and IB of two circular radius A and B with radius r and 4r and thickness t and $\frac{t}{4}$ .