# Angular Momentum of a paint ball

1. Nov 10, 2013

### Nicolaus

1. The problem statement, all variables and given/known data
A paint ball fires a ball of putty at a pendulum at a speed of 14 m/s, with a mass of 53g, at an angle of 42 degrees below the horizontal. The pendulum is made of a thin bar 51 cm long and mass of 310 g. The sphere fixed to the end of the pendulum is 17 cm in radius and has a mass of 190g. The pendulum is initially at rest in vertical position, and pivots about a free hinge at its top. The putty sticks to the pendulum at the point L - L/5 on the bar; so the putty is closer to the sphere at the bottom. Find the max. angle that pendulum makes with vertical bar after collision. (Consider putty as point mass).

2. Relevant equations
Conservation of angular momentum:
Angular Momentum (L) = Iw
Li = Lfinal

3. The attempt at a solution
Moment of inertia of apparatus after putty sticks:
I = m(putty)[L - L/5]^2 + m(bar)L^2 / 3 + 2m(sphere)r^2 / 5 = 0.038

Initial Angular momentum of putty:
Li = Lmvsin(theta) = (L-L/5)(0.0053)(14)sin(90+42)

Lf = I(calculated above)wf
wf = v/L
Equate both Li = Lf, isolate v, then using energy conservation to find max height, then find angle.

Is this setup correct?

2. Nov 10, 2013

### Simon Bridge

That's a common strategy - yep. Just some pointers:

You may want to consider that the rotating KE is $K_{rot}=L^2/2I$ and sin(90+A)=cos(A).
On impact with the rod - there is a horizontal and vertical component to the initial momentum isn't there? The horizontal component becomes angular momentum - you may want to say what happens to the vertical component.

If you want to use L for the length of the rod, you don't want to use it for angular momentum.
(In general, try to avoid using the same letter for more than one physical thingy.)

It is best practice to avoid substituting actual values until you have finished your algebra.

Last edited: Nov 10, 2013