Question

if the radius of gyrating of a solid mass of 10 kg about an axis is 0.40 m then the moment of inertia of disc about that axis is

Answer 1

Explanation:

$i = {mk}^{2}$

$where \: k \: is \: radius \: ofgyration$

$i = 10 \times ( {0.4)}^{2}$

therefore I= 1.6 kg m^2

Answer 2

Given :

The radius of gyration about a axis , k = 0.40 m .

Mass of disc , M = 10 kg .

To Find :

The moment of inertia of disc about that axis .

Solution :

We know , moment of inertia is given by :

$I=Mk^2$

Putting all values in above equation , we get :

$I=10\times 0.4^2\\\\I=1.6\ kg \ m^2$

Therefore , the moment of inertia of disc about that axis is $1.6\ kg \ m^2$ .

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Rotational Motion

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