Answer:
The moment of inertia of disc about own axis is 1 kg-m².
Explanation:
Given that,
Mass of ring m= 1 kg
Moment of inertia of ring at diameter [tex](I_{r})_{d}=1\ kg\ m^{2}[/tex]
The radius of metallic ring and uniform disc both are equal.
So, [tex]R_{r}=R_{d}[/tex]
We need to calculate the value of radius of ring and disc
Using theorem of perpendicular axes
[tex](I_{r})_{c}=2\times (I_{r})_{d}[/tex]
Put the value into the formula
[tex](I_{r})_{c}=2\times1[/tex]
[tex](I_{r})_{c}=2\ kg\ m^2[/tex]
Put the value of moment of inertia
[tex]MR_{r}^2=2[/tex]
[tex]R_{r}^2=\dfrac{2}{M}[/tex]
Put the value of M
[tex]R_{r}^2=\dfrac{2}{1}[/tex]
So, [tex]R_{r}^2=R_{d}^2=2\ m[/tex]
We need to calculate the moment of inertia of disc about own axis
Using formula of moment of inertia
[tex]I_{d}=\dfrac{1}{2}MR_{d}^2[/tex]
Put the value into the formula
[tex]I_{d}=\dfrac{1}{2}\times1\times2[/tex]
[tex]I_{d}=1\ kg\ m^2[/tex]
Hence, The moment of inertia of disc about own axis is 1 kg-m².
