Question

The radius k of the equivalent hoop is called the radius of gyration of the given body. Using this formula, find the radius of gyration of (a) a cylinder of radius 1.20 m, (b) a thin spherical shell of radius 1.20 m, and (c) a solid sphere of radius 1.20 m, all rotating about their central axes.

Answer 1

Here is the full question:

The rotational inertia I of any given body of mass M about any given axis is equal to the rotational inertia of an equivalent hoop about that axis, if the hoop has the same mass M and a radius k given by:  

$k=\sqrt{\frac{I}{M} }$

The radius k of the equivalent hoop is called the radius of gyration of the given body. Using this formula, find the radius of gyration of (a) a cylinder of radius 1.20 m, (b) a thin spherical shell of radius 1.20 m, and (c) a solid sphere of radius 1.20 m, all rotating about their central axes.

Answer:

a) 0.85 m

b) 0.98 m

c) 0.76 m

Explanation:

Given that: the radius of gyration  $k=\sqrt{\frac{I}{M} }$

So, moment of rotational inertia (I) of a cylinder about it axis = $\frac{MR^2}{2}$

$k=\sqrt{\frac{\frac{MR^2}{2}}{M} }$

$k=\sqrt{{\frac{MR^2}{2}}* \frac{1}{M} }$

$k=\sqrt{{\frac{R^2}{2}}$

$k={\frac{R}{\sqrt{2}}$

$k={\frac{1.20m}{\sqrt{2}}$

k = 0.8455 m

k ≅ 0.85 m

For the spherical shell of radius

(I) = $\frac{2}{3}MR^2$

$k = \sqrt{\frac{\frac{2}{3}MR^2}{M} }$

$k = \sqrt{\frac{2}{3} R^2}$

$k = \sqrt{\frac{2}{3} }*R$

$k = \sqrt{\frac{2}{3}} *1.20$

k = 0.9797 m

k ≅ 0.98 m

For the solid sphere of  radius

(I) = $\frac{2}{5}MR^2$

$k = \sqrt{\frac{\frac{2}{5}MR^2}{M} }$

$k = \sqrt{\frac{2}{5} R^2}$

$k = \sqrt{\frac{2}{5} }*R$

$k = \sqrt{\frac{2}{5}} *1.20$

k = 0.7560

k ≅ 0.76 m