# Center of Mass Velocity of rotating object

1. Dec 2, 2013

### qqchan

1. The problem statement, all variables and given/known data
As part of a carnival game, a 0.523-kg ball is thrown at a stack of 19.5-cm tall, 0.227-kg objects and hits with a perfectly horizontal velocity of 9.9 m/s. Suppose the ball strikes the very top of the topmost object as shown to the right. Immediately after the collision, the ball has a horizontal velocity of 3.15 m/s in the same direction, the topmost object now has an angular velocity of 1.64 rad/s about its center of mass and all the objects below are undisturbed. If the object's center of mass is located 13.0 cm below the point where the ball hits, What is the center of mass velocity of the tall object immediately after it is struck?

2. Relevant equations
L =pr
I=MR2
p=mv
L=Iω

3. The attempt at a solution
p=mv
p = Iω +mballvball
v = (Iω +mballvball)/(mball*r + I)

v = (I*1.64 +0.523*3.15)/(0.523*0.13 + I)
I have calculated this way. However, the teacher said that the method I used to compute the center of mass velocity is wrong. The I value I calculate was correct tho.I would like to know what I did wrong in the process of solving for the center of mass velocity

2. Dec 2, 2013

### haruspex

I assume the first equation represents the linear momentum before impact, so m there stands for mball and v for the initial speed of the ball?
The second one is strange. Iω is an angular momentum. You can't mix that with linear momentum. Is I the moment of inertia of the object about its centre? I don't think it would be merely rotating about its centre after impact.
I don't see how you get this equation fro the other two.
In fact, there seems to be some missing information in the OP. How wide are the objects? Or are they to be taken as very thin? If they have width, the velocity you are after will not be purely horizontal.

3. Dec 2, 2013

### qqchan

yes, that is right. and I is the moment of inertia of the object about its centre. I could calculate I by using the conservation of linear momentum. However, I do not know what they mean by centre of mass velocity.

4. Dec 2, 2013

### haruspex

How? As I explained, this equation is dimensionally wrong:
p and mballvball are linear momenta, whereas Iω is an angular momentum.
It's the linear velocity of the mass centre. Suppose the ball comes from the left. If the top object is essentially a rectangle (in the vertical plane containing the trajectory of the ball) then when it is struck it will rotate about its lower right corner. If the distance from that corner to its mass centre is d then its instantaneous centre of mass velocity will be dω.
The OP mentions a diagram. Can you post it? If not, what does the object look like in the diagram?

5. Dec 4, 2013

### qqchan

Here is the image.

I decided to accomplish it this way, but it still didn't work.

v = (mball*vball + mtall*0.13*ω)/(mball + mtall)

I believe I have made it all to the same dimensions

6. Dec 4, 2013