T = 2π√I/g but I am getting the wrong answer.Pendulum Periods: A & B (L, m)

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SUMMARY

The discussion focuses on calculating the periods of two pendulums, A and B, both with the same length (L) and total mass (m). For Pendulum A, the period is correctly calculated as T = 2π√(L/g). The challenge arises with Pendulum B, where the mass distribution is different, with half the mass in the ball and half in the uniform bar. The moment of inertia (I) for Pendulum B is derived as I = 2/3mL², which is essential for determining its period.

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Mdhiggenz
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Homework Statement


Two pendulums have the same dimensions (length ) L and total mass (m). Pendulum A is a very small ball swinging at the end of a uniform massless bar. In pendulum B , half the mass is in the ball and half is in the uniform bar.

A. Find the period of pendulum A for small oscillations.

B. Find the period of pendulum B for small oscillation

Homework Equations





The Attempt at a Solution



For A. I got 2∏√L/g which is correct however i am struggling for B

What I did was

Pendulum B I= Ibar + Ibar= 1/3 (m/2)L^2+m/2L^2= 2/3mL^2
 
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Hi Mdhiggenz! :smile:

(try using the X2 button just above the Reply box :wink:)
Mdhiggenz said:
In pendulum B , half the mass is in the ball and half is in the uniform bar.

Pendulum B I= Ibar + Ibar= 1/3 (m/2)L^2+m/2L^2= 2/3mL^2

Looks ok! :smile:
 

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