Question

(11) A thin uniform rod of length 1 m and mass 1 kg
is rotating about an axis passing through its
centre and perpendicular to its length. Calculate
the moment of inertia and radius of gyration of
the rod about an axis passing through a point
midway between the centre and its edge,
perpendicular to its length.
(Ans.: 0.1458 kgm2,0.3818 m)​

Answer 1

Answer:

0.3818 m

Explanation:

Given A thin uniform rod of length 1 m and mass 1 kg is rotating about an axis passing through its centre and perpendicular to its length. Calculate the moment of inertia and radius of gyration of the rod about an axis passing through a point midway between the centre and its edge, perpendicular to its length.

Mass of thin rod = 1 kg, length of rod = l = 1 m

Now we need to find I and l ,  

Moment of inertia of a thin rod about a transverse axis passing through its centre is  

             I g = I = Ml^2 / 12

By parallel axes theorem we have

I = Ig + Mh^2

I = Ml^2 /12 + Ml^2 / 16

I = 4 Ml^2 + 3 Ml^2 / 48

I  = 7 Ml^2 / 48

I = 7x1 x 1 / 48 = 7 / 48

I = 0.1458 kg m^2

Now K^2 = l /m

     K^2 = 0.1458 / 1

 K = √0.1458

K = 0.3818 m

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