Askiitians Tutor Team
Last Activity: 9 Months ago
To find the moment of inertia of a uniform semicircular disc about an axis passing through its center of mass and perpendicular to its plane, let’s analyze the problem step by step.
1. Center of mass of the semicircular disc:
The center of mass of a semicircular disc is located at a distance of 4R3π\dfrac{4R}{3\pi} from the flat edge (diameter) along the axis of symmetry.
2. Moment of inertia of the entire disc about its center:
The moment of inertia of a full circular disc about an axis perpendicular to its plane and passing through its center is:
Ifull=12MR2I_{\text{full}} = \dfrac{1}{2}M R^2
For the semicircular disc, the mass is the same as MM, but the moment of inertia will be proportional to the area. Since the semicircular disc is half of the full disc, the moment of inertia of the semicircular disc about the same axis (perpendicular to the plane and passing through the center of the full disc) is:
Isemi=12⋅12MR2=14MR2I_{\text{semi}} = \dfrac{1}{2} \cdot \dfrac{1}{2} M R^2 = \dfrac{1}{4}M R^2
3. Use the parallel axis theorem to find the moment of inertia about the center of mass:
The parallel axis theorem states:
Icm=Isemi−Md2I_{\text{cm}} = I_{\text{semi}} - M d^2
where:
• Isemi=14MR2I_{\text{semi}} = \dfrac{1}{4}M R^2 (moment of inertia about the center of the full disc),
• d=4R3πd = \dfrac{4R}{3\pi} (distance of the center of mass of the semicircular disc from the center of the full disc).
Substitute the values:
Icm=14MR2−M(4R3π)2I_{\text{cm}} = \dfrac{1}{4}M R^2 - M \left( \dfrac{4R}{3\pi} \right)^2
4. Simplify the expression:
Icm=MR24−M⋅16R29π2I_{\text{cm}} = \dfrac{M R^2}{4} - M \cdot \dfrac{16R^2}{9\pi^2}
Factor out MR2M R^2:
Icm=MR2(14−169π2)I_{\text{cm}} = M R^2 \left( \dfrac{1}{4} - \dfrac{16}{9\pi^2} \right)
5. Identify the correct option:
From the given choices, the correct answer matches the second option:
MR22−M(4R3π)2\boxed{\dfrac{M R^2}{2} - M \left( \dfrac{4R}{3\pi} \right)^2}
The correct answer is (b).
