# Conservation of angular momentum

1. Dec 23, 2009

### Jokerhelper

1. The problem statement, all variables and given/known data
Hello everybody! A week ago I wrote my (first) university physics exam, which as most entry-level physics courses was a general introduction to mechanical physics. I had a bit of trouble and doubts with a question, which I will paraphrase to the best of my abilities:

A small ball is attached to a rope. A young kid picks up the rope with his hand and uses it to spin the ball at a constant speed over his head. Both the ball and rope rotate on the same horizontal plane. In addition, both gravity and air friction can be ignored, and we were told to consider the ball's motion as being equivalent to the motion of a particle orbiting a centre point.
With this scenario, the boy then decides to reduce the length of the rope used to spin the object by a half.

a) Is the ball's angular momentum conserved?
b) Does the speed of the ball change?

2. Relevant equations
There are no specific equations for this question, but I tried to look at relationships between angular momentum, radius and speed:

$$\frac{d\vec{L}}{dt} = \vec{\tau}_{net} = \vec{r} \times \vec{F}_{net}$$

$$\vec{L} = I \cdot \vec{\omega} = mr^2\vec{\omega}$$

$$v = {\omega}r$$

$r_1 = 2r_2$, where $r_{1}$ and $r_{2}$ represent the length of the radius before and after, respectively.

3. The attempt at a solution
In order for angular momentum to be conserved, the net torque must be equal to zero. Initially, this is definitely true since the ball's speed is constant, which implies that there is no tangential acceleration. Therefore, the cross product between $$\vec{r}$$ and $$\vec{F}_{net}$$ has to be zero. No matter the length of the radius, angular momentum will be conserved as long as there is no change in tangential acceleration.
Because of this, my answer to part A of the question was "Yes, angular momentum is conserved."

Now, to part B. If my rationale from part A is correct, then:

$$\vec{L} = m{r_1}^2\vec{\omega}_1 = m{r_2}^2\vec{\omega}_2$$

$$m({2r_2})^2\vec{\omega}_1 = m{r_2}^2\vec{\omega}_2$$

$$\vec{\omega}_1 = \frac{\vec{\omega}_2}{4} \rightarrow \omega_1 = \frac{\omega_2}{4}$$

And if this is true,

$$\begin{equation*} \begin{split} v_1 = \omega_1r_1 = \frac{\omega_2}{4} \cdot 2r_2 = \frac{1}{2}(\omega_2r_2) = \frac{1}{2}v_2 \\ v_1 > 0 \end{split} \end{equation*}$$

Therefore, the speed did change.

I was wondering if my answers were right, and more importantly if my reasoning was correct. Corrections or suggestions for different ways to approach this question are also appreciated.
Thanks for your attention, and Happy Holidays to all!

2. Dec 23, 2009

### ideasrule

Yup, both parts are correct. Good job, and happy holidays!

3. Dec 24, 2009

### Jokerhelper

Alright. Thanks and cheers again.

4. Dec 24, 2009

### RoyalCat

On a slightly unrelated note, you posted:

$$\vec \tau_{net}= \vec r \times \vec F_{net}$$

That strikes me as a bit fishy, which radius vector would you be referring to? Is there a way to prove that $$\Sigma \vec r_i \times \vec F_i=\vec r_{resultant} \times \vec F_{net}$$ ?