Suppose there is a uniform circular disc of mass M kg and radius r m shown in figure. The shaded regions are cut out from the disc. The moment of inertia of the remainder about the axis A of the disc is given by $\frac{x}{256} M r^{2}$ The value of $x$ is ______. [2026]

Question

Smart Achievers · Accepted Answer

(109)

$M = σ π R^{2}$

$σ π R^{2} = 16 m$

$m = \frac{σ π R^{2}}{16}$

$I_{system} = \frac{M R^{2}}{2} - 2 (\frac{m R^{2}}{2 \times 16} + \frac{9 m R^{2}}{16})$

$= \frac{M R^{2}}{2} - 2 \times \frac{19 m R^{2}}{32}$

$= \frac{M R^{2}}{2} - \frac{19}{16} m R^{2}$

$= \frac{M R^{2}}{2} - \frac{19}{256} M R^{2} because m = \frac{M}{16}$

$= \frac{(128 - 19) M R^{2}}{256}$

$= \frac{109 M R^{2}}{256}$

Solution

Q.
Suppose there is a uniform circular disc of mass M kg and radius r m shown in figure. The shaded regions are cut out from the disc. The moment of inertia of the remainder about the axis A of the disc is given by $\frac{x}{256} M r^{2}$ The value of $x$ is ______. [2026]

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Solution

Q. Suppose there is a uniform circular disc of mass M kg and radius r m shown in figure. The shaded regions are cut out from the disc. The moment of inertia of the remainder about the axis A of the disc is given by x256Mr2 The value of x is ______. [2026]

Ans. (109) M=σπR2 σπR2=16m m=σπR216 Isystem=MR22-2(mR22×16+9mR216) =MR22-2×19mR232 =MR22-1916mR2 =MR22-19256MR2 because m=M16 =(128-19)MR2256 =109MR2256

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Q.
Suppose there is a uniform circular disc of mass M kg and radius r m shown in figure. The shaded regions are cut out from the disc. The moment of inertia of the remainder about the axis A of the disc is given by $\frac{x}{256} M r^{2}$ The value of $x$ is ______. [2026]