Rotational Motion PYQ — Find x in Moment of Inertia of a Cut Disc 🔥

❓ Question

FOR:

Let $M$ and $R$ be the mass and radius of a (uniform) solid disc. A small disc of radius $R/3$ is removed from the bigger disc (hole cut out) as shown in the figure. The moment of inertia of the remaining part about an axis $AB$ passing through the centre $O$ and perpendicular to the plane of the disc is given as

$$I=\frac{x}{4}M{R}^2\mathrm{.}$$

Find the value of $x$.

(We assume the small disc is concentric with the big disc — i.e. the hole is at the center — this is the standard interpretation unless the figure indicates an off-centre cut.)

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🖼️ Question Image

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✍️ Short Solution

Step 1 — Moment of inertia of the original (full) disc

For a uniform solid disc of mass $M$ and radius $R$,

$$I_{\text{full}}=\frac{1}{2}M{R}^2\mathrm{.}$$

Step 2 — Mass and moment of inertia of the removed small disc

Area (hence mass) scales with $R^2$. The small disc has radius $r=\dfrac{R}{3}$, so its area is $(r/R)^2=(1/3)^2=1/9$ of the big disc. Therefore its mass

$$m=\frac{M}{9}.$$

Moment of inertia of the small disc about its own central axis:

$$I_{\text{small,cm}}=\frac{1}{2}m{r}^2=\frac{1}{2}\cdot \frac{M}{9}\cdot \frac{R^2}{9}=\frac{M{R}^2}{162}\mathrm{.}$$

(If the small disc is concentric, no parallel-axis shift is needed.)

Step 3 — Moment of inertia of the remaining part

Remove the small disc from the full one:

$$I_{\text{remaining}}=I_{\text{full}}-I_{\text{small,cm}} =\frac{1}{2}MR^{2}-\frac{MR^{2}}{162}.$$

Combine terms:

$$\frac{1}{2}-\frac{1}{162}=\frac{81-1}{162}=\frac{80}{162}=\frac{40}{81}.$$

So

$$I_{\text{remaining}}=\frac{40}{81}\;M R^{2}.$$

Step 4 — Put in the given form $\dfrac{x}{4}MR^{2}$

We have

$$\frac{x}{4}MR^{2}=\frac{40}{81}MR^{2} \quad\Rightarrow\quad \frac{x}{4}=\frac{40}{81} \quad\Rightarrow\quad x=\frac{160}{81}.$$

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🧮 Image Solution

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✅ Conclusion & Video Solution

✅ Final Answer:

$$\boxed{x=\dfrac{160}{81}}$$