# Glob of clay

1. Mar 20, 2007

### fruitl00p

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

2. Mar 20, 2007

### Staff: Mentor

What are using as your reference for calculating the angular momentum? If you are using the center of mass of the system--a fine choice!--realize that it is moving. (You found its speed in part 1.) You'll need to calculate the initial angular momenta of the glob of clay and the stick about that point.

3. Mar 20, 2007

### fruitl00p

Doc Al, so the velocity I found in part A should be applied to part B...Oh, well I kept using the initial velocity rather than the final.

So the initial angular momentum = mass of clay* r of clay from center of mass^2* omega. Since omega is velocity/r,
initial momentum = mass of clay*r of clay from center of mass* velocity found in part A.

Now this is the part that confuses me. Should the r for initial angular momentum be the new center of mass distance or the original? Because if it is the original, then I think I would be setting two different "systems" equal to each other.

4. Mar 20, 2007

### Staff: Mentor

In calculating the initial angular momentum of both objects, both r and speed should be with respect to the center of mass of the system.

For example: You are given the initial speed of the clay, but what is its speed with respect to the center of mass of the system? Same point for the stick: what is its speed with respect to the center of mass of the system?

5. Mar 21, 2007

### fruitl00p

Ok, the r and speed should be in respect to the center of mass...When the glob of clay collides with the bar, it shifts the center of mass. The initial angular momentum would be the glob of clay *the final speed of the center of mass found in part 1 *new r. Yes?

Hm...I understand that the final angular momentum deals with the center of mass shift along with the new moment of inertia. Ok , I am more confused now.

6. Mar 21, 2007

### fruitl00p

I tried to do

Lo=Lf
(.48kg)(2.839m/s)(.34)=[(1/12(.94kg)(.90m)^2 + (.94kg)(.1149m)^2) + (.48kg)(.2251kg)^2]*final angular speed

the answer I got was 4.63 rad/s which was incorrect. The 2.839 m/s comes from part one's answer.

I guess I still do not understand with "respect to center of mass"

7. Mar 21, 2007

### Staff: Mentor

You found the speed of the center of mass in part one: 2.839 m/s with respect to the table. So, if the initial speed of the clay is 8.4 m/s with respect to the table, what is its speed with respect to the center of mass? Use that to find its initial angular momentum about the center of mass.

Do the same for the bar.

8. Mar 22, 2007

### fruitl00p

I feel really dumb !
So I need to calculate the speed with respect to the center of mass...I don't have a text book with me at the moment, but I basically do the Mtotal*Vcm =p to find the speed with respect to the center of mass.

Yet what I still don't understand is the fact that initially the only object with any angular momentum is the clay. Before the collision, the center of mass on bar is in the center, so the r in the equation with the clay is the original distance. But if I am to look at it after the collision, then wouldn't the collision immediately shift the center of mass of bar?

9. Mar 22, 2007

### Staff: Mentor

Not true! With respect to the center of mass of the system, the bar has an initial speed and thus some angular momentum about that center of mass.
I think you may be confused about the meaning of "center of mass". By that I mean the center of mass of the entire clay+rod system. The speed of the center of mass, which you calculated for part 1, doesn't change during the collision.

Fill out this chart of initial conditions at the instant before collision:
Location of system center of mass (measured from the center of the rod):
Speed of clay w.r.t. table (given):
Speed of rod w.r.t. table (given):
Speed of system c.o.m. (from part 1):
Speed of clay wrt COM:
Angular momentum of clay about COM:
Speed of rod wrt COM:
Angular momentum of rod about COM:
After the collision, set the total angular momentum equal to $I_{cm}\omega$.