# Angular Momentum problem

1. Jan 23, 2007

### konichiwa2x

Hi,

A rod of mass 'm' and length 'l' is lying on a smooth horizontal surface. A particle of mass 'm' is moving with a velocity $$v_{0}$$ strikes the rod at the centre and sticks to it. What is the velocity of end A after impact? [Solution: $$\displaystyle \frac{v_{0}}{2}$$]

Here is what I did:

Considering both the particle and the rod as the system, no external torque acts. Therefore, angular momentum can be conserved about any point.

$$\displaystyle mv_{0}\frac{l}{2} = I_{A}\omega$$

$$\displaystyle mv_{0}\frac{l}{2} = (\frac{ml^2}{3} + {\frac{ml}{4}}^2)\omega$$

$$\displaystyle \omega = \frac{6v_{0}}{7l}$$

Velocity of end A = $$\displaystyle \frac{\omega}{(\frac{l}{2})} = \frac{12v_{0}}{7l^2}$$

What am I doing wrong?

2. Jan 23, 2007

### Staff: Mentor

Hints: Conservation of linear momentum gives you one part. Conservation of angular momentum gives you the other part. What was the total angular momentum of the system before impact? What was the total linear momentum before impact?

3. Jan 24, 2007

### konichiwa2x

I have done exactly that. The linear momentum before impact is $$mv_{0}$$ and after impact is $$2mv$$ where $$v = v_{0}/2$$. I did conserve angular momentum (refer my original post). Is there anything wrong in what I did? please help.

4. Jan 24, 2007

### Staff: Mentor

You incorrectly assumed that the rod pivots about one end. It doesn't. The angular momentum about the center of mass of the system remains zero. (And the angular momentum about point A remains $mv_{0}l/2$.)

5. Jan 29, 2007

### PhanthomJay

This one is confusing me. Using conservation of linear momentum, and noting that the strike is at the cm of the rod, seems to clearly imply no rotation, and thus, the system, ..all parts of it.. moves forward at V_o/2.
But when using conservation of angular momentum, I have the same question as the poster, I think: Why is it acceptable to calculate the initial angular momentum of the particle ((mV_o(r)) about some ficticious point A, about which there is no fixed pivot point, but when calculating the final angular momentum, it is not OK to calculate the angular momentum of the rod about that point, which leads to an erroneus result for omega? (and which, BTW, was incorrectly calculated using an incorrect formula...using v = wr, not v=w/r, yields v = 3V_o/8 at the c.m, which is still the wrong(?) answer ).

6. Jan 29, 2007

### Staff: Mentor

There is no angular momentum before the impact, and none after the impact with the center of mass of the rod. It's a more interesting question when the mass hits one end of the rod and sticks -- I'm having a harder time visualizing the zero angular momentum aspect of that at the moment....

7. Jan 29, 2007

### Staff: Mentor

Realize that in noting the implication of "the strike is at the cm of the rod" you are implicitly applying conservation of angular momentum.

The erroneous result is not due to using point A as a reference point; it's due to assuming that the rod pivots about one end. In general, the angular momentum of an object about some point A is the sum of:
(a) the object's angular momentum about its center of mass
(b) the angular momentum of the object's center of mass about point A​
Again, I think you are assuming the rod pivots about one end.

In that case, the angular momentum of the system about its center of mass is not zero. The angular momentum about point A is zero, but that just means that the two terms indentified above are equal and opposite.