Question 389: Centre of Mass and Rotational Motion

Question

A circular disc has radius $R_1$ and thickness $T_1$. Another circular disc made of the same material has radius $R_2$ and thickness $T_2$. If the moment of inertia of both discs are same and $\frac{R_1}{R_2}=2$ then $\frac{T_1}{T_2}=\frac{1}{\alpha}$. The value of $\alpha$ is $\_\_\_\_$ .

Accepted Answer

Consider a uniform circular disc of radius $R$, thickness $T$, and made of a material with volume mass density $ho$. The total volume of this disc is $\mathrm{V}=\pi \mathrm{R}^2 \mathrm{~T}$.
The total mass of the disc is density times volume :
$$ \mathrm{M}=ho \cdot \mathrm{V}=ho \pi \mathrm{R}^2 \mathrm{~T} $$
Its moment of inertia (I) about an axis passing through its center and perpendicular to its plane, $\mathrm{I}=\frac{1}{2} \mathrm{MR}^2$ Putting the expression of mass,
$$ \mathrm{I}=\frac{1}{2} 	imes\left(ho \pi \mathrm{R}^2 \mathrm{~T}ight) 	imes \mathrm{R}^2 $$
$$ \mathrm{I}=\frac{1}{2} \pi ho \mathrm{~T} \mathrm{R}^4 $$
We have two discs :
Disc 1 : Radius $R_1$, Thickness $T_1$, Density $ho_1$
Disc 2 : Radius $\mathrm{R}_2$, Thickness $\mathrm{T}_2$, Density $ho_2$
As they are made of the same material. Therefore, their volume mass densities must be equal :
$$ ho_1=ho_2=ho $$
Their moments of inertia are the same :
$$ \mathrm{I}_1=\mathrm{I}_2 $$
$$\Rightarrow $$ $\frac{1}{2} \pi ho \mathrm{~T}_1 \mathrm{R}_1^4=\frac{1}{2} \pi ho \mathrm{~T}_2 \mathrm{R}_2^4$
$$\Rightarrow $$ $\mathrm{T}_1 \mathrm{R}_1^4=\mathrm{T}_2 \mathrm{R}_2^4$
$$\Rightarrow $$ $\frac{\mathrm{T}_1}{\mathrm{~T}_2}=\frac{\mathrm{R}_2^4}{\mathrm{R}_1^4}$
$$\Rightarrow $$ $$ \frac{\mathrm{T}_1}{\mathrm{~T}_2}=\left(\frac{\mathrm{R}_2}{\mathrm{R}_1}ight)^4 $$
It is given that their radii are in ratio, $\frac{\mathrm{R}_1}{\mathrm{R}_2}=2 \Rightarrow \frac{\mathrm{R}_2}{\mathrm{R}_1}=\frac{1}{2}$.
$$ \frac{\mathrm{T}_1}{\mathrm{~T}_2}=\left(\frac{1}{2}ight)^4=\frac{1}{16} $$
$$\Rightarrow $$ $$ \frac{\mathrm{T}_1}{\mathrm{~T}_2}=\frac{1}{16}=\frac{1}{\alpha} \Rightarrow \alpha=16 $$
Therefore, the value of $\alpha$ is 16.

Centre of Mass and Rotational Motion

A circular disc has radius R_1 and thickness T_1. Another circular disc made of the same material has radius R_2 and thickness T_2. If the moment of inertia of both discs are same and \frac{R_1}{R_2}=2 then \frac{T_1}{T_2}=\frac{1}{\alpha}. The value of \alpha is \_\_\_\_ .

Integer / Numeric Answer Type