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- Homework Statement
- A physical pendulum which is a mass of 0.25 kg and length 1.0 m with an attached mass (0.25 kg) represented by the white circle which can oscillate about the black circle that goes through one end of the rod.

The mass can be moved to different positions on the rod. The distance from the rotation axis to the mass is x.

Use the period formula below to derive the formula for the period T of the physical pendulum as a function

of the distance x.

- Relevant Equations
- For small-amplitude oscillation, the period of the physical pendulum is theoretically given by

##T=2\pi \sqrt\frac {I}{0.5gd}##

where

##0.5## is the total mass of the physical pendulum (which is composed of the rod and particle), in ##kg##

##g## is the acceleration due to gravity, ##9.8 m/s^2##

Rotation axis

##I## is the rotational inertia of the physical pendulum relative to the rotation axis, in ##kg.m^2##

##d## is the distance from the rotation axis to the centre of mass of the physical pendulum, in ##m##

This is the figure given.

__My Attempt__##T=2\pi \sqrt\frac {I}{0.5gd}##

##\frac {m_r} {l} ##

##dm= \frac {m} {l}dx##

##dI = dm_r x^2##

##dI=(\frac {m_r} {l}dx)x^2##

##I= \int_l^0 (\frac {m_r} {l}dx)x^2 \, dx ##

##I_c.m=\frac {m_r l^2}{3}##

##I_,, = \frac {m_r l^2}{3}+m_p x^2##

Given

##m_r=0.25##

##m+p=0.25##

##l=1##

##I_,, = \frac {(0.25 (1)^2}{3}+(0.25) x^2##

##I_,, = \frac {0.25+0.75x^2}{12}##

From here on I got stuck because I have no idea on how to find the d along with the centre of the mass.

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