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Question

Physics Question on Moment Of Inertia

The moment of inertia of uniform semicircular disc of mass MM and radius rr about a line perpendicular to the plane of the disc through the centre is:

A

14Mr2\frac{1}{4}Mr^{2}

B

25Mr2\frac{2}{5}Mr^{2}

C

Mr2Mr^{2}

D

12Mr2\frac{1}{2}Mr^{2}

Answer

12Mr2\frac{1}{2}Mr^{2}

Explanation

Solution

The mass of complete (circular) disc is The moment of inertia of disc about the given axis is I=2Mr22=Mr2I=\frac{2\,Mr^{2}}{2} =Mr^{2} Let, the moment of inertia of semicircular disc is I1I_{1} The disc may be assumed as combination of two semicircular parts. Thus, I1=II1I_{1}=I-I_{1} I1=I2=Mr22\therefore I_{1}=\frac{I}{2}=\frac{Mr^{2}}{2}