Question

A thin, circular hoop with a radius of 0.32 m is hanging on a nail. Adam notices that the hoop is oscillating back and forth through small angles like a physical pendulum. The moment of inertia of the hoop for the rotational axis passing through the nail is I = 2mr^2. What is the period of the hoop?

Answer 1

1.605 s

Step-by-step explanation:

radius of hoop = 0.32 m

moment of inertia
`I = 2mr^(2)`

period P of a hoop is given as

`P = 2\pi \sqrt{(I)/(mgr) }`

since
`I = 2mr^(2)`

`(I)/(mgr) = (2r)/(g)`

therefore the period of the hoop reduces to

`P = 2\pi \sqrt{(2r)/(g) }`

where = acceleration due to gravity = 9.81 m/s^2

`P = 2*3.142 \sqrt{(2*0.32)/(9.81) }` = 1.605 s