# Finding angular speed of a system about its center of mass after the impact

1. Nov 30, 2011

### Smartguy94

1. The problem statement, all variables and given/known data

1. The problem statement, all variables and given/known data
On a frictionless table, a glob of clay of mass 0.72 kg strikes a bar of mass 1.34 kg perpendicularly at a point 0.23 m from the center of the bar and sticks to it.

If the bar is 1.22 m long and the clay is moving at 8.3 m/s before striking the bar, what is the final speed of the center of mass?

At what angular speed does the bar/clay system rotate about its center of mass after the impact?

2. The attempt at a solution

Here is what I did

Vcm = sum(mi*vi) / total mass

in this case just one body has velocity:
Mclay = .72 kg
Mbr = 1.34 Kg
Mtotal = 2.06 kg
Vclay = 8.3 m/s

Vcm = 2.901 m/s

which is correct, I got the first part, but here is the second part that I got wrong

now calculate the angular speed:

angular speed - omega

omega = v /r

v - is the linear velocity on the trajectory of the body (tangential velocity)
r - is the distance between the body which rotate and the center of rotation

in your case the centre of rotation is the center of mass and r is the distance of the clay to the centre of mass

D = 1.22 m
d = 0.23 m

the centre of the bar related to one end is D/2
the position of the clay related to the same end is d+D/2

Xcm = [Mclay *(d+D/2)+Mbr*(D/2) ]/Mtotal

Xcm = .6904 m position of the center of mass

the angular velocit of the clay:

omega clay = Vclay / D1

D1=(d+D/2)-Xcm = .1496 m

the center of the bar is situated related to the center of mass at:

Dbar = Xcm - D/2 = .0804 m

If you assume that the bar when it rotate has the same tangential velocity v = 8.3 m/s

omega bar = 8.3 / 0.0804 = 103.249 rad/s

the question is asking about At what angular speed does the bar/clay system rotate about its center of mass after the impact?

and so I add up both the omega of the bar and the clay and got 158.726 rad/s
but it's wrong

Can anyone tell me where my mistake is?

2. Dec 1, 2011

### Simon Bridge

You cannot have the bar and the clay with different omega.
You got that because you assumed you had the same tangential speed for the com bar and the clay - but surely the clay will be moving slower after impact?

This is a conservation of momentum problem.

The com speed formula you used is derived from the conservation of linear momentum - you then need to conserve angular momentum.

What is the total angular momentum just before the clay sticks?
What is the angular momentum of the bar+clay afterwards?