Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6
Banner 7
Banner 8
Banner 9

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.
Question Image

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya

a

1732MR2

b

932MR2

c

1332MR2

d

732MR2

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

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 

Watch 3-min video & get full concept clarity
score_test_img

courses

No courses found

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!

best study material, now at your finger tips!

  • promsvg

    live classes

  • promsvg

    progress tracking

  • promsvg

    24x7 mentored guidance

  • promsvg

    study plan analysis

download the app

gplay
mentor

Download the App

gplay
whats app icon
personalised 1:1 online tutoring