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Rotational Motion question

2023 · 8 Apr · Shift 1 · Q75
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Rotational Motion question

2023 · 8 Apr · Shift 1 · Q75

JEE MainPhysicsRotational MotionNumerical+4 / −1
The moment of inertia of a semicircular ring about an axis, passing through the center and perpendicular to the plane of ring, is 1xMR2\frac{1}{x} \mathrm{MR}^{2}x1​MR2, where R\mathrm{R}R is the radius and MMM is the mass of the semicircular ring. The value of xxx will be ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 1

  1. Given: A semicircular ring of radius RRR and mass MMM.

    We need the moment of inertia about an axis:

    • passing through the center of the circle,
    • and perpendicular to the plane of the ring.

  2. Key idea: Every small mass element dmdmdm of the semicircular ring lies at the same distance RRR from the center of the circle.

  3. So for the required axis, dI=R2 dmdI = R^2\,dmdI=R2dm

  4. Integrate over the entire semicircular ring: I=∫dI=∫R2 dm=R2∫dmI = \int dI = \int R^2\,dm = R^2\int dmI=∫dI=∫R2dm=R2∫dm

  5. Since the total mass of the semicircular ring is MMM, ∫dm=M\int dm = M∫dm=M hence, I=MR2I = MR^2I=MR2

  6. Compare with the given form: I=1xMR2I = \frac{1}{x}MR^2I=x1​MR2

    Therefore, 1x=1⇒x=1\frac{1}{x} = 1 \quad \Rightarrow \quad x = 1x1​=1⇒x=1

Final Answer

1\boxed{1}1​

Comparison with stored correct answer

Stored correct answer = 111.

My derived answer also is 111, so they agree.

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