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6 May 2012 20:46:26 IST

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rotational motion

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A disk of radius 'r' has a cut or hole of radius r/2.Find the rotational inertia of the remaining part of mass 'm' about the centre of gravity of the remaining part perpendicular to the plane of the disk

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Rohan Vora

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Joined: 7 Feb 2011

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7 May 2012 00:18:37 IST

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First find areal mass density. Mass m is distributed over area of (πR²- πR²/4)=3πR²/4 .Thus density is ρ=4m/3πR².

Now Let mass of complete disc before the hole was made be m1. Mass of dics be m2.

Centre of mass when whole is made is given by

So it comes as

So the we need the moment of inertia about a point P which is at a distance of R/6 away from the hole..

The moment of inertia about P of the entire disc before hole was made :

The moment of inertia of the disc of disc of radius r/2 (which was a part of the disc before hole was made) is

Let Moment of inertia of the disc with hole be

The moment of inertia of the remaing part + moment of inertia of the disc with radius r/2 = moment of inertia of complete disc So now,

m1=ρ.(πR²) and m2=ρ.(πR²)/4

Substituting,we have

There may be some error in calculations but this is how it is done..Like if you find it useful..

Rohan Vora

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Joined: 7 Feb 2011

Posts: 140

7 May 2012 00:21:39 IST

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Correction:We need the moment of inertia about a point P which is at a distance of R/6 from the centre of the disc away from the hole..

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