First slide

Moment of intertia

Question

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:

Moderate

A

$I_{C} > I_{P} > I_{B} > I_{H}$
B

$I_{H} > I_{B} > I_{C} > I_{P}$
C

$I_{P} > I_{H} > I_{B} > I_{C}$
D

none of these

Solution

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. Hence
$I_{BC} < I_{AC} < I_{AB} \Rightarrow I_{P} > I_{B} > I_{H}$
$I_{C} = I_{CM} + {my}^{2} = I_{B}^{'} - {mx}^{2} + {my}^{2}$
Here $I_{B}^{'}$ is moment of inertia of the plate about an axis perpendicular to it and passing through B.
$\Rightarrow I_{C} = I_{B}^{'} + m (y^{2} - x^{2}) = I_{P} + I_{B} + m (y^{2} - x^{2})$
It means $I_{C} > I_{P} + I_{H} also I_{C} > I_{P}$
$∴ I_{C} > I_{P} > I_{B} > I_{H}$

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