Question

The radius of gyration of a disc of mass 50g and radius 0.5 cm about an axis passing through its centre of gravity and perpendicular to plane is

Answer 1

Answer:

Explanation:

=> Here, it is given that,

The radius of a disc = 0.5 cm.

Mass of a disc = 50g

=> Required moment of inertia of the disc is :

I = MR²/2

= MK²

Thus, K² = R²/2

K = R /√2

= 2.5 /√2

= 1.5√2 / 2

= 1.767

= 1.77 cm

Answer 2

Radius of gyration of disc will be 1.77 cm.

Explanation:

In the statement it is given that:

Mass of disc = m = 50

Radius of disc = R = 2.5 cm  

The moment of inertia of the disc is given by              

                    I = MR^2/2   ……….. (i)

Also,  

                                        I = MK^2   ……….. (ii)

By comparing equation (i) and (ii)  and solving for K

                                        MR^2/2 = MK^2

                                                                               

K =R/√2

K =2.5/√2

K = 1.77 cm