physics
posted by hania on .
A 1kg mass (the blue mass) is connected to a 8kg 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 1kg 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.

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.phyastr.gsu.edu/hbase/pendp.html 
how did you figure out distance from the center of mass?

Require that the moment about the center of mass be zero, and solve for the location.

if the mass is 9 kg instead of 8, would the L be 21.9