On a frictionless table, a glob of clay of mass .48kg strikes a bar of mass .94kg perpendicularly at a point .34m from the center of the bar and sticks to it.
1. If the bar is .90m long and the clay is moving at 8.4 m/s befor estrikning the bar, what is the final speed of the center of mass?
2. At what angular speed does the bar/cly system rotate about its cneter of mass after the impact?
Initial Inertia*omega initial =Final Inertia*omega final
The Attempt at a Solution
Wow, this problem is frustrating! At first I was using an example in my textbook to solve this problem:
mvl= Ifinal*omega final = (1/12 ML^2 + ml^2) *omega final
mvl/((1/12 ML^2 + ml^2) = omega final
However that was the wrong approach. Then I realized that the center of mass shifts. I found the new center of mass:
Xcm= (.48 kg)(.34m)/(.48kg+.94kg) = .1149
So the distance the clay has from the center of mass is now .2250
Then I tried mvl/ [(1/12 ML^2 + M(Xcm)^2) + m(new l)^2
(.48kg)(8.4m/s)(.34m)/[(1/12 (.94kg)(.90m)^2 + (.94kg)(.1149)^2) +(.48kg) (.225)^2 = omega final
My answer was 13.7 rad/s which was incorrect. I am lost. I just don't know how to approach this problem. Please help