Alarge 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

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

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

now we have

now for small angular displacement we will have

so angular frequency of SHM is given as

now we will have

the closer the planet is to the sun, the larger the gravity

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer