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Physics - Rotation Question with Solution | TestHub

PhysicsRotationCalculation of M.I.Easy2 minPYQ_2020

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A thin circular plate of mass $M$ and radius $R$ has its density varying as $ρ (r) {=ρ}_{0} r$ with $ρ_{0}$ as constant and $r$ is the distance from its centre. The moment of Inertia of the circular plate about an axis perpendicular to the plate and passing through its edge is $I = a M R^{2}$ . The value of the coefficient $a$ is:

Options:

Answer:

Solution:

$d m = (2 π r d r) \times ρ = 2 π ρ_{0} r^{2} d r$
$M = 2 π ρ_{0} \int_{0}^{R} r^{2} d r = \frac{2}{3} π ρ_{0} R^{3}$
$I = ρ (d m) r^{2} = 2 π ρ_{0} \int r^{4} d r = \frac{2}{5} π ρ_{0} R^{5}$
$I_{c m} = \frac{3}{5} M R^{2}$
$I = I_{c m} + M R^{2} = \frac{8}{5} M R^{2}$

Subject:PhysicsTopic:RotationSubtopic:Calculation of M.I.

⏱ 2mℹ️ Source: PYQ_2020

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