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?

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nuhulawal20 nuhulawal20 · Answer 1 · 2021-04-25T13:53:04+02:00

Answer:

the period of the physical pendulum is 0.498 s

Explanation:

Given the data in the question;

[tex]T_{simple[/tex] = 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 / [tex]I[/tex] ) ------ 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 [tex]I[/tex] is moment of inertia of rod.

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

[tex]I[/tex] = [tex]\frac{1}{3}[/tex]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/[tex]I[/tex]) OR T = 2π√([tex]I[/tex]/mgL)

so we can use [tex]I[/tex] = [tex]\frac{1}{3}[/tex]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 = [tex]\frac{1}{2}[/tex]D.

now, substituting these equations, the period becomes;

T = 2π/√([tex]I[/tex]/mgL) OR T = [tex]2\pi \sqrt{\frac{\frac{1}{3}mD^2 }{mg(\frac{1}{2})D } }[/tex] 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[tex]_{simple[/tex] = 2π/[tex]T_{simple[/tex] OR ω[tex]_{simple[/tex] = √(g/D) OR ω[tex]_{simple[/tex] = 2π√( D/g )

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

so we have;

T = √(2/3) × [tex]T_{simple[/tex]

we substitute in value of [tex]T_{simple[/tex]

T = √(2/3) × 0.61 s

T = 0.498 s

Therefore, the period of the physical pendulum is 0.498 s