From a uniform disc of radius R and mass M, a square plate is removed as shown in figure. The moment of inertia of remaining body about an axis passing through the centre (O) of the original disc and perpendicular to its plane is

Question

From a uniform disc of radius R and mass M, a square plate is removed as shown in figure. The moment of inertia of remaining body about an axis passing through the centre (O) of the original disc and perpendicular to its plane is (A) (MR2/2)(1+(1/3π)) (B) (MR2/2)(1-(1/3π)) (C) MR2 (1+3π) (D) MR2 (2π-1)

Tardigrade Admin · Accepted Answer

Io=(MR2/2)-(M/2π)[(R2/2×6)+(R2/4)]=(MR2/2)[1-(1/3π)]

Q. From a uniform disc of radius R and mass M, a square plate is removed as shown in figure. The moment of inertia of remaining body about an axis passing through the centre (O) of the original disc and perpendicular to its plane is

$\frac{M R ^{2}}{2} (1 + \frac{1}{3 π})$

$\frac{M R ^{2}}{2} (1 - \frac{1}{3 π})$

$M R^{2} (1 + 3 π)$

$M R^{2} (2 π - 1)$

$I_{o} = \frac{M R ^{2}}{2} - \frac{M}{2 π} [\frac{R ^{2}}{2 \times 6} + \frac{R ^{2}}{4}] = \frac{M R ^{2}}{2} [1 - \frac{1}{3 π}]$

Q. From a uniform disc of radius R and mass M, a square plate is removed as shown in figure. The moment of inertia of remaining body about an axis passing through the centre (O) of the original disc and perpendicular to its plane is

2MR2​(1+3π1​)

2MR2​(1−3π1​)

MR2(1+3π)

MR2(2π−1)

Io​=2MR2​−2πM​[2×6R2​+4R2​]=2MR2​[1−3π1​]

$\frac{M R ^{2}}{2} (1 + \frac{1}{3 π})$

$\frac{M R ^{2}}{2} (1 - \frac{1}{3 π})$

$M R^{2} (1 + 3 π)$

$M R^{2} (2 π - 1)$

$I_{o} = \frac{M R ^{2}}{2} - \frac{M}{2 π} [\frac{R ^{2}}{2 \times 6} + \frac{R ^{2}}{4}] = \frac{M R ^{2}}{2} [1 - \frac{1}{3 π}]$