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:
[tex]k=\sqrt{\frac{I}{M} }[/tex]
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 [tex]k=\sqrt{\frac{I}{M} }[/tex]
So, moment of rotational inertia (I) of a cylinder about it axis = [tex]\frac{MR^2}{2}[/tex]
[tex]k=\sqrt{\frac{\frac{MR^2}{2}}{M} }[/tex]
[tex]k=\sqrt{{\frac{MR^2}{2}}* \frac{1}{M} }[/tex]
[tex]k=\sqrt{{\frac{R^2}{2}}[/tex]
[tex]k={\frac{R}{\sqrt{2}}[/tex]
[tex]k={\frac{1.20m}{\sqrt{2}}[/tex]
k = 0.8455 m
k ≅ 0.85 m
For the spherical shell of radius
(I) = [tex]\frac{2}{3}MR^2[/tex]
[tex]k = \sqrt{\frac{\frac{2}{3}MR^2}{M} }[/tex]
[tex]k = \sqrt{\frac{2}{3} R^2}[/tex]
[tex]k = \sqrt{\frac{2}{3} }*R[/tex]
[tex]k = \sqrt{\frac{2}{3}} *1.20[/tex]
k = 0.9797 m
k ≅ 0.98 m
For the solid sphere of radius
(I) = [tex]\frac{2}{5}MR^2[/tex]
[tex]k = \sqrt{\frac{\frac{2}{5}MR^2}{M} }[/tex]
[tex]k = \sqrt{\frac{2}{5} R^2}[/tex]
[tex]k = \sqrt{\frac{2}{5} }*R[/tex]
[tex]k = \sqrt{\frac{2}{5}} *1.20[/tex]
k = 0.7560
k ≅ 0.76 m
