Question

A metallic ring of mass 1 kg has moment of inertia 1 kg m² when rotating about one of its diameters. It is molten and

remoulded into a thin uniform disc of the same radius. How much will its moment of inertia be, when rotated about its own
axis.​

Answer 1

The moment of inertia of disc about own axis is 1 kg-m².

Step-by-step explanation:

Given that,

Mass of ring m= 1 kg

Moment of inertia of ring at diameter
`(I_(r))_(d)=1\ kg\ m^(2)`

The radius of metallic ring and uniform disc both are equal.

So,
`R_(r)=R_(d)`

We need to calculate the value of radius of ring and disc

Using theorem of perpendicular axes

`(I_(r))_(c)=2* (I_(r))_(d)`

Put the value into the formula

`(I_(r))_(c)=2*1`

`(I_(r))_(c)=2\ kg\ m^2`

Put the value of moment of inertia

`MR_(r)^2=2`

`R_(r)^2=(2)/(M)`

Put the value of M

`R_(r)^2=(2)/(1)`

So,
`R_(r)^2=R_(d)^2=2\ m`

We need to calculate the moment of inertia of disc about own axis

Using formula of moment of inertia

`I_(d)=(1)/(2)MR_(d)^2`

Put the value into the formula

`I_(d)=(1)/(2)*1*2`

`I_(d)=1\ kg\ m^2`

Hence, The moment of inertia of disc about own axis is 1 kg-m².