Mass per unit area of a circular disc of radius a depends on the distance r from its centre as σ$$r=A+Br$$ . The moment of inertia of the disc about the axis, perpendicular to the plane and passing through its center is:

Question

Mass per unit area of a circular disc of radius a depends on the distance r from its centre as σ$$r=A+Br$$ . The moment of inertia of the disc about the axis, perpendicular to the plane and passing through its center is:

Infinity Learn · Accepted Answer

Go at distance of r and take an element of thickness dr.​Area of element $$=$$ $$dA=2$$$$π$$rdr​Given : Surface mass density $$=$$ α$$=A+Br$$ Mass of element $$=$$ $$dm=$$αdA​∴Total $$mass=$$∫$$dm=$$∫$$A+Br2$$$$π$$rdrMoment of Inertia about center $$=$$ $$I=$$∫dm $$r2$$ $$=$$∫$$0aA+Br2$$$$π$$$$r3dr$$ $$=2$$$$π$$$$Aa44+Ba55=2$$$$π$$$$a4A4+aB5$$

Mass per unit area of a circular disc of radius a depends on the distance r from its centre as $σ (r) = A + B r$ . The moment of inertia of the disc about the axis, perpendicular to the plane and passing through its center is:

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Mass per unit area of a circular disc of radius a depends on the distance r from its centre as σr=A+Br . The moment of inertia of the disc about the axis, perpendicular to the plane and passing through its center is:

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Mass per unit area of a circular disc of radius a depends on the distance r from its centre as $σ (r) = A + B r$ . The moment of inertia of the disc about the axis, perpendicular to the plane and passing through its center is: