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

AI Thread Summary
The discussion focuses on calculating the periods of two pendulums with the same length and mass but different mass distributions. For Pendulum A, the correct period for small oscillations is determined to be 2π√(L/g). The challenge arises with Pendulum B, where half the mass is in the ball and half in the uniform bar, leading to confusion in calculating the moment of inertia. The proposed moment of inertia for Pendulum B is I = 2/3mL², which is acknowledged as a reasonable approach. The conversation emphasizes the importance of correctly applying the principles of physics to solve for the period of the second pendulum.
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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