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

Answer:

$I = 94.33 kg m^2$

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

$\alpha = \frac{mgd}{I}\theta$

so angular frequency of SHM is given as

$\omega = \sqrt{\frac{mgd}{I}}$

now we will have

$\frac{2\pi}{9} = \sqrt{\frac{4.8(9.8)1.4}{I}$

$I = 94.33 kg m^2$