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:
Show Hint
- \( \frac{1}{2} M r^2 \)
- \( \frac{3}{8} M r^2 \)
- \( \frac{3}{2} M r^2 \)
- \( 2 M r^2 \)
The Correct Option is B
Approach Solution - 1
Approach Solution -2
We are asked to find 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.
Concept Used:
We use the Parallel Axis Theorem and the known moment of inertia of a ring about its central axes. For a circular ring of radius \( R \) and mass \( M \):
\[ I_{\text{center, diametral}} = \frac{1}{2}MR^2, \quad I_{\text{center, perpendicular}} = MR^2 \]
The Parallel Axis Theorem states:
\[ I = I_{\text{center}} + Md^2 \]
where \( d \) is the distance between the two parallel axes.
Step-by-Step Solution:
Step 1: Given diameter \( r \), so radius \( R = \dfrac{r}{2} \).
Step 2: The axis about which we need the moment of inertia is tangential and lies in the plane of the ring.
The moment of inertia about the diametral axis (through the center, in the plane of the ring) is:
\[ I_{\text{center, diametral}} = \frac{1}{2}MR^2 \]
Step 3: Using the parallel axis theorem to shift from the center to the tangent point (distance \( R \)):
\[ I_{\text{tangent, in-plane}} = I_{\text{center, diametral}} + MR^2 = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2 \]
Step 4: Substitute \( R = \frac{r}{2} \):
\[ I_{\text{tangent, in-plane}} = \frac{3}{2}M\left(\frac{r}{2}\right)^2 = \frac{3}{8}Mr^2 \]
Final Computation & Result:
The moment of inertia of the circular ring about a tangential axis lying in its plane is:
\[ \boxed{I = \frac{3}{8}Mr^2} \]
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