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Find density of a ball on a pendulum 
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#1
Oct1810, 03:20 PM

P: 1

First of all, sorry for my bad English as it isn't my native language. Also, sorry for poor formating.
1. The problem statement, all variables and given/known data A pendulum consists of a rod of length [tex]L=1 m[/tex] and mass [tex]m_1=0.5 kg[/tex] and a ball (radius [tex]r=5 cm[/tex]) attached to one end of the rod. The axis of oscillation is perpendicular to the rod, and goes through the other end of the rod. The period of oscillation is [tex]T=2s[/tex]. Find the density [tex]\rho[/tex]of the ball. 2. Relevant equations [tex]T=2 \pi \sqrt{\frac{I}{mgx}}[/tex] [tex]m= \rho V= \rho \frac{4}{3}r^3 \pi[/tex] 3. The attempt at a solution For this pendulum, the period is: [tex]T=2 \pi \sqrt{\frac{I}{mgx}}[/tex] where [tex]x[/tex] is the distance between the axis of oscillation and the center of mass and [tex]I[/tex] is the moment of inertia. So, according to parallel axis theorem, the moment of inertia for the rod [tex]I_{rod}[/tex] is equal to: [tex]I_{rod}=\frac{1}{3}m_1 L^2 + m_1 (\frac{1}{2}L  x)^2[/tex] Similarly, the moment of inertia of the ball [tex]I_{ball}[/tex] is equal to: [tex]I_{ball}=\frac{2}{5}m_2 r^2 + m_2 (L  x + r)^2[/tex] Considering that [tex]I=I_{rod} + I_{ball}[/tex], we are left with two unknowns  [tex]x[/tex] and [tex]\rho[/tex]. That means we need another equation, and that's the part that isn't clear to me. I suspect it has to do something with the fact that this is pretty much a onedimensional problem and certain properties of center of mass. Unfortunately, I do not know what to do next. Thanks a lot, Bruno 


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