A circular disc of radius b has a hole of radius a at its centre(see figure). If the mass per unit area of the disc varies as σ 0 r then, the radius of gyration of the disc about its axis passing through the center is

Solution: (B) d m = σ 2 π r d r = σ 0 r 2 π r d r m =2 π σ 0 ∫ a b d r = 2 π σ 0 ( b - a ) I = 2 π σ 0 ∫ a b r 2 d r = 2 3 π σ 0 b 3 - a 3 Radius of gyration = I m = 2 3 π σ 0 b 3 - a 3 2 π σ 0 b - a = b 3 - a 3 3 ( b - a ) = ( b - a ) ( b 2 + a 2 + a b ) 3 ( b - a ) = b 2 + a 2 + a b 3 .

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PhysicsRotationCalculation of M.I.Hard2 minPYQ_2019

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A circular disc of radius $b$ has a hole of radius $a$ at its centre(see figure). If the mass per unit area of the disc varies as $(\frac{σ_{0}}{r})$ then, the radius of gyration of the disc about its axis passing through the center is

Options:

Answer:

Solution:

Solution: (B)
$d m = σ 2 π r d r$
$= \frac{σ_{0}}{r} 2 π r d r$
$m =2 π σ_{0} \int_{a}^{b} d r = 2 π σ_{0} (b - a)$

$I = 2 π σ_{0} \int_{a}^{b} r^{2} d r = \frac{2}{3} π σ_{0} (b^{3} - a^{3})$
Radius of gyration $= \sqrt{\frac{I}{m}} = \sqrt{\frac{\frac{2}{3} π σ_{0} (b^{3} - a^{3})}{2 π σ_{0} (b - a)}}$
$= \sqrt{\frac{b^{3} - a^{3}}{3 (b - a)}} = \sqrt{\frac{(b - a) (b^{2} + a^{2} + a b)}{3 (b - a)}}$
$= \sqrt{\frac{b^{2} + a^{2} + a b}{3}}$ .

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

⏱ 2mℹ️ Source: PYQ_2019

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