Answer:
1.605 s
Explanation:
radius of hoop = 0.32 m
moment of inertia [tex]I = 2mr^{2}[/tex]
period P of a hoop is given as
[tex]P = 2\pi \sqrt{\frac{I}{mgr} }[/tex]
since [tex]I = 2mr^{2}[/tex]
[tex]\frac{I}{mgr} = \frac{2r}{g}[/tex]
therefore the period of the hoop reduces to
[tex]P = 2\pi \sqrt{\frac{2r}{g} }[/tex]
where = acceleration due to gravity = 9.81 m/s^2
[tex]P = 2*3.142 \sqrt{\frac{2*0.32}{9.81} }[/tex] = 1.605 s
