Question:
The disc of mass M with uniform surface mass density \(\sigma\) is shown in the figure. The centre of mass of the quarter disc (the shaded area) is at the position \(\frac{xa}{3\pi},\ \frac{xa}{3\pi}\) where x is ____________. (Round off to the Nearest Integer).
[a is an area as shown in the figure]
The disc of mass M with uniform surface mass density \(\sigma\) is shown in the figure. The centre of mass of the quarter disc (the shaded area) is at the position \(\frac{xa}{3\pi},\ \frac{xa}{3\pi}\) where x is ____________. (Round off to the Nearest Integer).
[a is an area as shown in the figure]
[a is an area as shown in the figure]
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The centre of mass of a uniform quarter disc of radius a lies at \(\frac{4a}{3\pi}\) from each straight edge.
Updated On: Nov 4, 2026
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Correct Answer: 4
Solution and Explanation
The shaded region is a quarter disc of radius \(a\) (the length marked "a" in the figure) with its centre at the origin.
For a uniform quarter disc: \(x_{cm}=\frac{\int x\,dm}{\int dm}=\frac{\int_0^a\int_0^{\pi/2} r\cos\theta\cdot\sigma r\,d\theta\,dr}{\sigma\pi a^2/4}=\frac{\sigma\frac{a^3}{3}}{\sigma\pi a^2/4}=\frac{4a}{3\pi}\).
By symmetry \(y_{cm}=\frac{4a}{3\pi}\).
Comparing with \(\frac{xa}{3\pi}\): \(x=4\).
Answer: 4
For a uniform quarter disc: \(x_{cm}=\frac{\int x\,dm}{\int dm}=\frac{\int_0^a\int_0^{\pi/2} r\cos\theta\cdot\sigma r\,d\theta\,dr}{\sigma\pi a^2/4}=\frac{\sigma\frac{a^3}{3}}{\sigma\pi a^2/4}=\frac{4a}{3\pi}\).
By symmetry \(y_{cm}=\frac{4a}{3\pi}\).
Comparing with \(\frac{xa}{3\pi}\): \(x=4\).
Answer: 4
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