1. Tardigrade
  2. Question
  3. Physics
  4. A circular disc X of radius R is made from an iron plate of thickness t, and another disc Y of radius 4R is made from an iron plate of thickness t/4. Then the relation between the moment of inertia IX and IY is

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. A circular disc $X$ of radius $R$ is made from an iron plate of thickness $t$, and another disc $Y$ of radius $4R$ is made from an iron plate of thickness $t/4$. Then the relation between the moment of inertia $I_{X}$ and $I_{Y}$ is

System of Particles and Rotational Motion

Solution:

Moment of Inertia of disc $I=\frac{1}{2}MR^{2}$
$=\frac{1}{2} (\pi R^{2} t\,\rho) R^{2} = \frac{1}{2} \pi \,t\,\rho\,R^{4}$
[As $M=V \times \rho=\pi R^{2}\,t\,\rho$ where t = thickness, $\rho$ =density]
$\therefore \frac{I_{y}}{I_{x}}=\frac{t_{y}}{t_{x}} \left(\frac{R_{y}}{R_{x}}\right)^{4}$ [If $\rho$ =constant]
$\Rightarrow \frac{I_{y}}{I_{x}}=\frac{1}{4}\left(4\right)^{4}=64$ [Given $R_{y}=4R_{x}, t_{y}=\frac{t_{x}}{4}]$
$\Rightarrow I_{y}=64 I_{x}$