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Q.

A uniform circular disc of radius ' R ' and mass ' M ' is rotating about an axis perpendicular to its plane and passing through its centre. A small circular part of radius R/2 is removed from the original disc as shown in the figure. Find the moment of inertia of the remaining part of the original disc about the axis as given above.
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An Intiative by Sri Chaitanya

a

1732MR2

b

932MR2

c

1332MR2

d

732MR2

answer is D.

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Detailed Solution

Step 1: Moment of Inertia of the Original Disc

The moment of inertia of a uniform circular disc of mass MM and radius RR about its central axis (perpendicular to its plane) is given by:

Ioriginal=12MR2I_{\text{original}} = \frac{1}{2} M R^2

 Step 2: Moment of Inertia of the Removed Part

The removed part is a smaller disc of radius R/2R/2.
Since, the original disc has uniform mass distribution, the mass of the smaller disc (proportional to its area) is:

Mremoved=M×π(R/2)2πR2=M×14=M4M_{\text{removed}} = M \times \frac{\pi (R/2)^2}{\pi R^2} = M \times \frac{1}{4} = \frac{M}{4}

The moment of inertia of a smaller disc about its own center is:

Iremoved,center=12Mremoved(R2)2I_{\text{removed,center}} = \frac{1}{2} M_{\text{removed}} \left(\frac{R}{2}\right)^2

 Iremoved,center=12×M4×R24=132MR2I_{\text{removed,center}} = \frac{1}{2} \times \frac{M}{4} \times \frac{R^2}{4} = \frac{1}{32} M R^2

However, this small disc was originally positioned at a distance R/2R/2 from the center of the original disc. Using the parallel axis theorem, the moment of inertia of the removed part about the axis of the original disc is:

Iremoved=Iremoved,center+Mremovedd2I_{\text{removed}} = I_{\text{removed,center}} + M_{\text{removed}} d^2

 Iremoved=132MR2+(M4×R24)I_{\text{removed}} = \frac{1}{32} M R^2 + \left(\frac{M}{4} \times \frac{R^2}{4}\right)

 Iremoved=132MR2+116MR2I_{\text{removed}} = \frac{1}{32} M R^2 + \frac{1}{16} M R^2

 Iremoved=132MR2+232MR2=332MR2I_{\text{removed}} = \frac{1}{32} M R^2 + \frac{2}{32} M R^2 = \frac{3}{32} M R^2 

Step 3: Moment of Inertia of the Remaining Part

The moment of inertia of the remaining part is:

Iremaining=IoriginalIremovedI_{\text{remaining}} = I_{\text{original}} - I_{\text{removed}}

 Iremaining=12MR2332MR2I_{\text{remaining}} = \frac{1}{2} M R^2 - \frac{3}{32} M R^2

 Iremaining=1632MR2332MR2I_{\text{remaining}} = \frac{16}{32} M R^2 - \frac{3}{32} M R^2 

Iremaining=1332MR2I_{\text{remaining}} = \frac{13}{32} M R^2 

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A uniform circular disc of radius ' R ' and mass ' M ' is rotating about an axis perpendicular to its plane and passing through its centre. A small circular part of radius R/2 is removed from the original disc as shown in the figure. Find the moment of inertia of the remaining part of the original disc about the axis as given above.