Moment of inertia of uniform triangular plate about axis passing through sides AB, AC, BC are Ip, IB and IH respectively and about an axis perpendicular to the plane and passing through point C is IC. Then:

Question

Moment of inertia of uniform triangular plate about axis passing through sides AB, AC, BC are Ip, IB and IH respectively and about an axis perpendicular to the plane and passing through point C is IC. Then:

Infinity Learn · Accepted Answer

Moment of inertia is more when mass is farther from the axis. In case of axis BC, mass distribution is closest to it and in case of axis AB mass distribution is farthest. HenceIBC< IAC< IAB ⇒IP> IB > IHIC $$=$$ $$ICM+my2$$ $$=$$ IB'$$-mx2+my2Here$$ IB' is moment of inertia of the plate about an axis perpendicular to it and passing through B.⇒IC $$=$$ IB'$$+m(y2-x2)$$ $$=$$ $$IP+IB+m(y2-x2)It$$ means IC > $$IP+IH$$ also IC > IP∴ IC > IP > IB > IH

Moment of inertia of uniform triangular plate about axis passing through sides AB, AC, BC are $I_{p}, I_{B} and I_{H}$ respectively and about an axis perpendicular to the plane and passing through point C is $I_{C}$ . Then:

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