Question

The radius of gyration of a disc of mass 100g and radius 5cm about an axis passing through

Answer 1

the moment of inertia of a disc passing thrugh the centre of the axis is given by

i=mr^2/2

I=100*25/2

i=50*25

i=1250

radius of gyration is given by k=under root i/m

therefore the radius of a gyration is

k=under root 1250/100

k=under root 12.5

Answer 2

Answer: The correct answer is 3.5 cm.

Explanation:

The expression for the moment of inertia in terms of radius of gyration is as follows;

$I= Mk^{2}$                                                                      ......... (1)

Here, M is the mass of the disc and k is the radius of gyration.

It is given in the problem that the object is disc having mass 100 g and radius 5 cm.

Use the expression for the moment of inertia of disc.

$I= \frac{1}{2} MR^{2}$

Here, R is the radius.

Put $I= \frac{1}{2} MR^{2}$ in the above expression.

$\frac{1}{2} MR^{2}= Mk^{2}$

$k^{2}=\frac{1}{2}(5)^{2}$

$k= 3.5 cm$

Therefore, the value of the radius of gyration of a disc of mass about an axis is 3.5 cm.