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Question

A circular disc of radius b has a hole of radius a at its center. if the mass per unit area of the disc varies as σ0/r


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Solution

Step 1: Given data:

The radius of gyration:

  1. It is the distance from the center of mass of a body at which the whole mass is concentrated.
  2. It does not change its moment of rotational inertia about an axis through the center of mass.
  3. It is an act or instance of gyrating.
  4. Something such as a coil or shell that is gyrate.

A diagram to represent the radius of gyration:

solution

The formula of the radius of gyration:

I=MK2

Step 2: To find the radius of gyration:

We have,

dI=(dm)r2

Now,

dI=(σdA)r2

σr2πrdrr2

So, =(σ2π)r2dr

Now, I=dI=abσ2πr2dr=σ2πb3-a33m=dm=σdA=σ2πabdrm=σ2π(b-a)

The radius of gyration:

k=Im=(b3-a3)3(b-a)k=a3+b3+ab3

Hence, then the radius of gyration of the disc about its axis passing through the centre is k=a3+b3+ab3.


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