Question

Calculate the moment of inertia of thin circular ring about the following axis of rotation :
1. About an axis passing through its centre and perpendicular to its plane.
2. About a tangent perpendicular to its plane.
3. About a tangent in the plane of ring.
4. About a Diameter.

Answer 1

Dear student


Let m be the mass and r be the radius of the ring


let a point mass on the ring be of the mass dm

1) MI of this point mass about the axis passing through centre of ring

=

$$\begin{gathered} dm{r}^2 \\ so,MIofallparticlesconsistingthering \\ =\int dm{r}^2=r^2\int dm=m{r}^2 \end{gathered}$$ about an axis passing through its centre

2)..Now use parallel axis theorem

$$\begin{gathered} M{I}_{new}=M{I}_{old}+mass\times dis\tan cebetweenbothaxes \\ =m{r}^2+m\times r^2=2m{r}^2 \end{gathered}$$ about a tangent perpendicular to plane of ring

4)... Use perpendicular axis theorem.

$$\begin{gathered} M{I}_x+M{I}_y=M{I}_z \\ 2M{I}_x=2M{I}_{diameter}=m{r}^2 \\ M{I}_{diameter}=\frac{m{r}^2}{2} \end{gathered}$$ Ix = Iy since the body is symmetric

3)... Use parallel axis theorem

$$\begin{gathered} M{I}_{\tan gentinplane}=M{I}_{diameter}+mass\times (distbetweendiameterand\tan gent{)}^2 \\ =\frac{m{r}^2}{2}+m{r}^2 \\ =\frac{3m{r}^2}{2} \end{gathered}$$

Note : question 4 has been answered before 3.

Regards