Question 110: Centre of Mass and Rotational Motion

Question

The moment of inertia of a circular ring of mass M and diameter r about a tangential axis lying in the plane of the ring is :

Accepted Answer

The moment of inertia of a circular ring with mass $ M $ and diameter $ r $ about a tangential axis lying in the plane of the ring is calculated as follows:
Given the diameter $ r $ of the ring, the radius $ R $ is $ \frac{r}{2} $.
The formula for the moment of inertia about a tangential axis in the plane of the ring is:
$ I_{	ext{tangential}} = \frac{3}{2} M \left(\frac{R}{2}ight)^2 $
Substitute the value of $ R = \frac{r}{2} $ into the equation:
$ I_{	ext{tangential}} = \frac{3}{2} M \left(\frac{r}{2}ight)^2 = \frac{3}{8} Mr^2 $
Therefore, the moment of inertia of the ring about the specified axis is $\frac{3}{8} Mr^2$.

Answer

$\frac{3}{8} \mathrm{Mr}^2$

Answer

$2 \mathrm{Mr}^2$

Answer

$\frac{1}{2} \mathrm{Mr}^2$

Answer

$\frac{3}{2} \mathrm{Mr}^2$

Centre of Mass and Rotational Motion

The moment of inertia of a circular ring of mass M and diameter r about a tangential axis lying in the plane of the ring is :

Choose the correct answer: