A circular disc of radius 2m has a hole of radius 1m at its centre. Then, find the radius of gyration of the disc about the axis passing through its centre and perpendicular to its plane. Given that mass per unit area of the disc varies as σ0r.
A
√213m
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B
√313m
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C
√203m
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D
√413m
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Solution
The correct option is A√213m Let us take a small elemental ring of mass dm, radius r and width dr as shown in figure.
Moment of inertia of ring dI=r2dm ⇒dI=r2σ02πdr...(i) [Given, mass per unit area varies as σ0r, For the elemental ring, mass =dm, area =2πrdr ⇒dm2πrdr=σ0r...(ii)] Integrating Eq.(i) with limits of radius r=1m→2m I∫0dI=2σ0π2∫1r2dr ⇒I=2σ0π[r33]21 ⇒I=14σ0π3 For finding radius of gyration about the axis through the centre of the disc, I=mK2 Putting the value of I, we get ⇒mK2=14σ0π3...(iii) Since, m=m∫0dm and substituting the value of dm from Eq.(ii) ∫m0dm=r=2∫r=1σ02πrdrr ⇒m=2πσ0
Putting in Eq.(iii), we get, ⇒2πσ0K2=14σ0π3 ⇒K=√73 ∴K=√213m The radius of gyration of the disc about its axis passing through the centre is √213m.