Question

provides some pertinent background for this problem. A pendulum is constructed from a thin, rigid, and uniform rod with a small sphere attached to the end opposite the pivot. This arrangement is a good approximation to a simple pendulum (period = 0.61 s), because the mass of the sphere (lead) is much greater than the mass of the rod (aluminum). When the sphere is removed, the pendulum no longer is a simple pendulum, but is then a physical pendulum. What is the period of the physical pendulum?

Answer 1

the period of the physical pendulum is 0.498 s

Step-by-step explanation:

Given the data in the question;

`T_{simple` = 0.61 s

we know that, the relationship between T and angular frequency is;

T = 2π/ω ---------- let this be equation 1

Also, the angular frequency of physical pendulum is;

ω = √(mgL /
`I` ) ------ let this equation 2

where m is mass of pendulum, L is distance between axis of rotation and the center of gravity of rod and
`I` is moment of inertia of rod.

Now, moment of inertia of thin uniform rod D is;

`I` =
`(1)/(3)`mD²

since we were not given the length of the rod but rather the period of the simple pendulum, lets combine this three equations.

we substitute equation 2 into equation 1

we have;

T = 2π/ω OR T = 2π/√(mgL/
`I`) OR T = 2π√(
`I`/mgL)

so we can use
`I` =
`(1)/(3)`mD² for moment of inertia of the rod

Since center of gravity of the uniform rod lies at the center of rod

so that L =
`(1)/(2)`D.

now, substituting these equations, the period becomes;

T = 2π/√(
`I`/mgL) OR T =
`2\pi \sqrt{((1)/(3)mD^2 )/(mg((1)/(2))D ) }` OR T = 2π√(2D/3g ) ----- equation 3

length of rod D is still unknown, so from equation 1 and 2 ( period of pendulum ),

we have;

ω
`_{simple` = 2π/
`T_{simple` OR ω
`_{simple` = √(g/D) OR ω
`_{simple` = 2π√( D/g )

so we simple solve for D/g and insert into equation 3

so we have;

T = √(2/3) ×
`T_{simple`

we substitute in value of
`T_{simple`

T = √(2/3) × 0.61 s

T = 0.498 s

Therefore, the period of the physical pendulum is 0.498 s