
#1
Jan513, 08:56 PM

P: 51

1. The problem statement, all variables and given/known data
I have a physical pendulum that is rotating about an fixed axis. The period is: [itex]T=2\pi \sqrt{\frac{I}{mgd}}[/itex] I = moment of inertia d = distance between the center of mass and the axis. The problem is: If you add a mass in the end of the pendulum. The period is going to increase or decrease. My attempt: the inertia of a rod (physical pendulum) [itex]I = md^{2} + \frac{1}{12}mL^{2}[/itex] Without the mass, the moment inertia is : [itex]d = \frac {L}{2}[/itex] the center of mass in in the middle of the physical pendulum. [itex]I = m\frac {L^{2}}{4} + \frac{1}{12}mL^{2}[/itex] [itex]I = \frac{1}{3} mL^{2}[/itex] if i add a mass in the end of the pendulum then [itex]I = mL^{2} + \frac{1}{12}mL^{2}[/itex] [itex]I = \frac {13}{12} mL^{2}[/itex] without mass: [itex]T=2\pi \sqrt{\frac{\frac{1}{3} mL^{2}}{mg\frac{L}{2}}}[/itex] [itex] T=2\pi \sqrt{\frac{2L}{3g}} [/itex] with mass: [itex]T=2\pi \sqrt{\frac{\frac {13}{12} mL^{2}}{mgL}}[/itex] [itex]T=2\pi \sqrt{\frac {13L}{12g} }[/itex] If we add a mass in the end of the physical pendulum, then the period is going to increase. Can somebody check if my answer is correct? 



#2
Jan513, 09:14 PM

Emeritus
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P: 7,403

Do you know where the fixed axis is ? 



#3
Jan613, 07:08 AM

P: 51





#4
Jan613, 07:55 AM

Mentor
P: 40,889

Moment of inertia physical pendulumDon't forget that when you add mass the location of the center of mass changes. 



#5
Jan613, 09:12 AM

P: 51

consider the mass of the (added mass) = to the mass of the rod. because d it's the distance from the center of mass and the pivot, then when we add a mass in the end of pendulum then d = L (the center of mass is in the end of the pendulum) if the mass is in the center then d = L^2/2 



#6
Jan613, 09:43 AM

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P: 40,889





#7
Jan613, 09:59 AM

P: 51

so it is [itex]I = mL^{2} + \frac{1}{12}mL^{2} = \frac{13}{13}mL^{2}[/itex] considering the mass of the added mass = mass of the thin rod. 



#8
Jan613, 10:16 AM

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P: 40,889





#9
Jan613, 10:21 AM

P: 51

so my calculus are wrong, because the mass is not all in the end. But if the mass of (added mass) is much greater than the mass of the rod, then the center of the mass is in the end right? 



#10
Jan613, 10:51 AM

Mentor
P: 40,889





#11
Jan613, 07:16 PM

P: 51




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