# Conservation of angular momentum help

1. Nov 17, 2007

### sunumen

I haven't any suggested answer on this problem

The most important thing is that I really don't know how to start to deal with this problem.

< I am lack of knowledge on angular momentum<<I don't know is this needed in solving this problem>> >

I guess that using momentum in the first part finding angular velocity
and use conservation of energy in solving angular displacement??

Can anyone give me some hints?
Thankyou.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Nov 18, 2007

### chaoseverlasting

Yes. You need to know conservation of angular momentum. $$\vec{L}=\vec{L_{cm}}+\I\vec{w}$$.

More simply, as the net torque on the system is negligible, $$m_1v_1r_1=m_2v_2r_2$$. From this you find $$v_2$$ and the kinetic energy of the system, from which you can find the height and the angle obtained etc.

3. Nov 19, 2007

### sunumen

I have tried to solve the problem in this way.
m$$_{1}$$v$$_{1}$$r$$_{1}$$=m$$_{2}$$v$$_{2}$$r$$_{2}$$
0.02x300x0.4 = 1.02x v$$_{2}$$x(0.4+0.5)
v$$_{2}$$=2.614m/s

angular velocity= v/r = 2.614/0.9

===========================
K.E. loss = P.E. gain
1/2 mv$$^{2}$$ = mgh

1/2 x 2.614$$^{2}$$ = 9.8 x h
h=0.349m

cos (angle) = (0.9-h)/0.9
angle = 52.24 degree

===========================
There are some questions I want to clear
I hope that someone can answer me~~Thx

1~the 1.02(kg) which I highlighted is the sum of the bullet and mass.
Am I incorrect to consider both of them?
I should use 1kg only or 1.02 kg?
If I use 1.02 kg , I think I should consider the center of mass after the bullet embedded....but it is quite troublesome.

2~is there any other solution to find out the maximum angular displacement?
I feel that using conservation of energy may not get the exact solution

moreover, Is there a solution that the angular I found in part a can be used again to solve the angular displacement??

<Speak again, I haven't learn angular momentum before >.< >

Thx any help.