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Mechanics

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 Joined: 10 Jun 2009 Post: 20
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).

New kid on the Block

Joined: 10 Jun 2009
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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.

Blazing goIITian

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2 Mar 2012 19:33:46 IST
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New kid on the Block

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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.

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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.

Blazing goIITian

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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.

Blazing goIITian

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2 Mar 2012 20:07:17 IST
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MI OF A DISC ABOUT CENTRE IS (MR^2)/2.

haha...

Blazing goIITian

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2 Mar 2012 20:09:24 IST
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above comment for ARJUN VIRMANI post

Hot goIITian

Joined: 27 Dec 2011
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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

Blazing goIITian

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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..

Hot goIITian

Joined: 27 Dec 2011
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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

Blazing goIITian

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13 Mar 2012 21:50:52 IST
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use

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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

Blazing goIITian

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19 Mar 2012 23:24:05 IST
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