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1 Mar 2012 23:10:44 IST

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IIT Level- Moment of Inertia Question

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Find the M.I. of a uniform semicircular disc of mass M and radius R about an axis perpendicular to its plane and passing through point P (on its edge).

Comments (13)

Abhinav Sharma

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2 Mar 2012 09:17:18 IST

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Point P is exactly at a point opposite to its centre halfway on its circumference, I am unable to upload a picture.

kartikAIR1

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2 Mar 2012 19:33:46 IST

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ANSWER Is MR2{(9pi-16)/6pi}

Abhinav Sharma

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2 Mar 2012 19:44:33 IST

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I know that, but I am not able to get there, care to explain the answer please.

Arjun Virmani

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2 Mar 2012 19:57:06 IST

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MI of disc abt the axis from center = MR² / 4

MI of half disc about the same axis (say I )= MR² / 4 (cos its just the overlap of the two halves)

For Applying Parallel axis theorem u must know the location of P

I = MIcm + ms²

where s is the distnce between Centre of Mass and centre of circle

and then apply

MI (thru P) = MIcm + md²

where d is the distance between Centre of mass and P.

kartikAIR1

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2 Mar 2012 20:01:05 IST

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CONSIDER a point O as centre of disc. centre of mass is at a distance 4r/3pi from point O.

now using parallel axis theorem,

let I1 b moment of inertia about P.

let I2 b MI about COM.

let I3 b MI about O.

using parallel axis theorem,

I3=I2 + m(4r/3pi)sq.

MRsq/2 = I2 + m(4r/3pi)sq............(1)

similarly,

I1= I2 + m(r-4r/3pi)sq. ..................(2),

now eliminate I2 from these 2 equations.

you get the answer

kartikAIR1

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2 Mar 2012 20:07:17 IST

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THE STUPID ANSWER. TOTALLY WRONG.

MI OF A DISC ABOUT CENTRE IS (MR^2)/2.

haha...

kartikAIR1

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2 Mar 2012 20:09:24 IST

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above comment for ARJUN VIRMANI post

nageen

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3 Mar 2012 18:02:43 IST

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the MI of the disc given by arjun virmani is correct...its given for the semi circular disc

kartikAIR1

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3 Mar 2012 18:47:23 IST

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So what? mi of semicircular disc about perpendicular axis passing through is mrsq. by a2. this is ridiculous that people r supporting wrong ans..

nageen

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6 Mar 2012 16:41:51 IST

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sorry...MI of semi circular disc is mr^2/2 as told by karthik

sudhindra shenoy

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13 Mar 2012 21:50:52 IST

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use

Saurabh Dubey

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19 Mar 2012 22:39:32 IST

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I=I(CM)+MD^2The center of mass of a disc is 2r/3pie.and its MI is same as the disc(MR^2/2).calculate the distance of the point from the center of mass and use the above formula

kartikAIR1

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19 Mar 2012 23:24:05 IST

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Follow my answer,

Its 100 % virus FREE

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