# Angular Momentum of a ball of clay

1. Dec 8, 2013

### NATURE.M

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

Consider a thin rod of mass m$_{r}$ and length L hanging from a pivot at its upper end. A ball of clay of mass m$_{c}$ and of horizontal velocity v$_{0}$ strikes the lower end of the rod at a right angle and sticks, causing the rod + ball to rotate. What is the angular velocity of the rod + ball immediately after the collision? (The moment of inertia of the rod about its CENTER OF MASS is ML^2/12. You can treat the clay ball as a particle).

3. The attempt at a solution

Now the solution handout indicates L$_{i}$ = L$_{f}$, which I understand.
But then it indicates L$_{i}$ = I$_{c}$ω$_{c}$$^{2}$, where I$_{c}$ is the moment of inertia of the ball. Now I don't understand why the ball traveling horizontally has angular momentum, unless it represents the angular momentum of the system the instant the ball makes contact with the rod. Otherwise, the angular momentum of the system should be 0 initially. Any clarification is appreciated.

2. Dec 8, 2013

### Simon Bridge

Notice, however, that the ball moving horizontally still has a radius vector to the pivot point. The radius vector has an angular velocity that changes with time. There must be a component of the linear velocity that is perpendicular to the radius vector. ergo: it is a rotating system wrt the pivot - it's just not a rigid rotating body, which is what you've been learning about.

The mass will also have a moment of inertia at all times, whether it is turning or not.
Moment of inertia is usually $L=I\omega$ though isn't it?

3. Dec 8, 2013

### NATURE.M

So then the ball has angular momentum relative to the pivot prior to actual contact? Is this because the ball has angular velocity prior to contact since ω=Rv , where v is the linear speed of the ball and r the distance from pivot, and as you said the ball always has moment of inertia (which makes sense)?

4. Dec 9, 2013

### Simon Bridge

Well it can make sense to say so.. $$L=I\frac{d\theta}{dt}$$ and the object has a position in polar coordinates whatever the shape of it's trajectory. For a line, it only requires that the pivot/origin not be on the trajectory. Strictly you intuition is correct that you really need to have a central force for the angular momentum to be worth the effort. We may do something like this, for instance, if we are dealing with a large cloud of gas ... each particle having it's own linear velocity, but all gravitating towards each other - central force see? But each velocity is instantaneously linear.

Fortunately for you it does not matter because you only care about the angular momentum infinitesimally before the actual impact. It makes no difference if you imagine this is in the instant of time after the impact but before circular motion starts. In both imaginings, v is an instantaneous tangential velocity.

More general treatments are in the future for you.
But sometimes it is fun to get a glimpse of how big this can get.

5. Dec 9, 2013

### NATURE.M

Thanks. I'll keep that in mind.

6. Dec 9, 2013

### Simon Bridge

Since the rod is given as length L, you want to use a different symbol for angular momentum.