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Physics - Rotation Question with Solution | TestHub

PhysicsRotationCalculation of M.I.Easy2 minPYQ_2020

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Line $PQ$ is parallel to $y$ -axis and moment of inertia of a rigid body about $P Q$ line is given by $I = 2 x^{2} - 12 x + 27,$ where $x$ is in meter and $I$ is in $k g$ $m^{2}$ . The minimum value of $I$ is:

Options:

Answer:

Solution:

$I = 2 x^{2} - 12 x + 27$
$\frac{d I}{d x} = 4 x - 12$
For minimum or maximum
$\frac{d I}{d x} = 0 \Rightarrow x = 3$

$\frac{d^{2} I}{d x^{2}} = 4.$

Hence $I$ is minimum at $x = 3$
$I = 9 k g m^{2}$

Stream:JEESubject:PhysicsTopic:RotationSubtopic:Calculation of M.I.

⏱ 2mℹ️ Source: PYQ_2020

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