Question

A large stick is pivoted about one end and allowed to swing back and forth with no friction as a physical pendulum. The mass of the stick is 4.8 kg and its center of gravity (found by finding its balance point) is 1.4 m from the pivot. If the period of the swinging stick is 9 seconds, what is the moment of inertia of the stick about an axis through the pivot

Answer 1

`I = 94.33 kg m^2`

Step-by-step explanation:

Let say the rod is slightly pulled away from its equilibrium position

So here net torque on the rod due to its weight is given as

`\tau = mg dsin\theta`

since rod is pivoted at distance of 1.4 m from centre of gravity

so its moment of inertia about pivot point is given as

`Inertia = I`

now we have

`I \alpha = mg d sin\theta`

now for small angular displacement we will have

`\alpha = (mgd)/(I)\theta`

so angular frequency of SHM is given as

`\omega = \sqrt{(mgd)/(I)}`

now we will have

`(2\pi)/(9) = \sqrt{(4.8(9.8)1.4)/(I)`

`I = 94.33 kg m^2`