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 vanes as (σ0r) , then the radius of gyration of the disc about its axis passing through the centre is
A
a+b2
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B
√a2+b2+ab3
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C
√a2+b2+ab2
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D
a+b3
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Solution
The correct option is B√a2+b2+ab3 Step-1:Draw a diagram of the small mass element.
Considering a circular ring of radius r and thickness dr
Step-2: Find the moment of inertia and mass of the disc.
Moment of inertia of the elemental ring, dI=(dm)r2=(σdA)r2
=(σ0r2πrdr)r2=(σ02π)r2dr
I=∫dI
=∫baσ02πr2dr
=σ02π(b3−a33)
m=∫dm=∫σdA=σ02π∫badr
m=σ02π(b−a)
Find the radius of gyration of the disc.
Radius of gyration,
k=√Im=√(b3−a3)3(b−a)=√(b2+a2+ab3)
Final answer is (d)