(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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?

2. Relevant equations

Lo=Lf

Initial Inertia*omega initial =Final Inertia*omega final

Mtotal*Xcm= mi*xi...

3. The attempt at a solution

Wow, this problem is frustrating! At first I was using an example in my text book 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

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Glob of clay

**Physics Forums | Science Articles, Homework Help, Discussion**