Posted by **hania** on Sunday, March 25, 2012 at 11:10am.

A 1-kg mass (the blue mass) is connected to a 8-kg mass (the green mass) by a massless rod 67 cm long, as shown in the figure. A hole is then drilled in the rod 40.2 cm away from the 1-kg mass, and the rod and masses are free to rotate about this pivot point, P. Calculate the period of oscillation for the masses if they are displaced slightly from the stable equilibrium position.

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**drwls**, Sunday, March 25, 2012 at 11:37am
The lever arm from the pivot axis to the 1 kg mass is 40.2 cm, and to the 8 kg mass is 26.8 cm. The moment of inertia about the pivot axis is

I = 1*(0.402)^2 + 8*(0.268)^2

= 0.736 kg/m^2

You will also need to know the distance L from the center of mass to the pivot axis. In this case, it is 17.23 cm = 0.1723 m. (The center of mass is 9.57 cm from the 8 kg mass).

The equation you need to use for the period can be found at

http://hyperphysics.phy-astr.gsu.edu/hbase/pendp.html

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**hania**, Sunday, March 25, 2012 at 11:44am
how did you figure out distance from the center of mass?

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**drwls**, Sunday, March 25, 2012 at 12:42pm
Require that the moment about the center of mass be zero, and solve for the location.

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**hania**, Sunday, March 25, 2012 at 12:54pm
if the mass is 9 kg instead of 8, would the L be 21.9

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