FROM A CIRCULAR DISC OF RADIUS R AND MASS 9M, A SMALL DISC OF RADIUS R/3 IS REMOVED FROM THE DISC. THE M.O.INERTIA OF THE REMAINING DISC ABOUT AN AXIS PERPENDICULAR TO THE PLANE OF THE DISC AND PASSING THRO' IT'S CENTER IS:

Question

Ritvik Gautam · Answer

it will be the same...

Shahid Akbar · Answer

Its will be 4MR*2 BY using parallel axis theorem

Jay Gohil · Answer

Let I1 be the moment of Inertia of bigger circle and I2 of smaller circleI=I1- I2=9mr^2/2-mr^2/18=80mr^2/18=40mr^2/9

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FROM A CIRCULAR DISC OF RADIUS R AND MASS 9M, A SMALL DISC OF RADIUS R/3 IS REMOVED FROM THE DISC. THE M.O.INERTIA OF THE REMAINING DISC ABOUT AN AXIS PERPENDICULAR TO THE PLANE OF THE DISC AND PASSING THRO' IT'S CENTER IS:

FROM A CIRCULAR DISC OF RADIUS R AND MASS 9M, A SMALL DISC OF RADIUS R/3 IS REMOVED FROM THE DISC. THE M.O.INERTIA OF THE REMAINING DISC ABOUT AN AXIS PERPENDICULAR TO THE PLANE OF THE DISC AND PASSING THRO' IT'S CENTER IS:

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FROM A CIRCULAR DISC OF RADIUS R AND MASS 9M, A SMALL DISC OF RADIUS R/3 IS REMOVED FROM THE DISC. THE M.O.INERTIA OF THE REMAINING DISC ABOUT AN AXIS PERPENDICULAR TO THE PLANE OF THE DISC AND PASSING THRO' IT'S CENTER IS:

FROM A CIRCULAR DISC OF RADIUS R AND MASS 9M, A SMALL DISC OF RADIUS R/3 IS REMOVED FROM THE DISC. THE M.O.INERTIA OF THE REMAINING DISC ABOUT AN AXIS PERPENDICULAR TO THE PLANE OF THE DISC AND PASSING THRO' IT'S CENTER IS:

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