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Angle of Repose

Moment of ine...

Question

Moment of inertia of uniform triangular plate about axis passing through sides $A B, A C$ and $B C$ 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

A

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

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B

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

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C

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

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D

none of these

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Solution

The correct option is B $I_{C} > I_{P} > I_{B} > I_{H}$

Moment of inertia is more when mass is
farther from the axis. In case of axis $B C$ , mass
distribution is closest to it and in case of axis
$A B$ mass distribution is farthest. Hence
$I_{B C} < I_{A C} < I_{A B}$
$\Rightarrow I_{P} > I_{B} > I_{H}$
Here $I_{A}^{'}$ is moment of inertia of the plate about an axis perpendicular to it and passing through A.
$\Rightarrow I_{C} = I_{A}^{'} + m (y^{2} - x^{2})$
$I_{C} = I_{P} + I_{B} + m (y^{2} - x^{2})$
It means $I_{C} > I_{P} + I_{B}$
also $I_{C} > I_{P}$
$∴ I_{C} > I_{P} > I_{B} > I_{H}$
Final Answer: $(a)$

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Angle of Repose

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