Two questions on rotation/angular kinematics

[killazys]

Homework Statement

A uniform stick 1.0 m long with a total mass of 300 g is pivoted at its center. A 3.5 g bullet is shot through the stick midway between the pivot and one end. The bullet approaches at 250 m/s and leaves at 160 m/s. With what angular speed is the stick spinning after the collision?

Homework Equations

K_{i} = K_{r} + K_{f}

K_{r} = \frac{1}{2}*\frac{mL}{12}*(\omega)^{2}

Where K_{i} and K_{f} are of the bullet, and K_{r} is of the stick.

The Attempt at a Solution

Plug in and solve. The problem probably is that the bullet hitting right at the 0.25m mark has an effect, one that I am unaware of :'(
Attempted first answer was 71.875 rad/s.

Homework Statement

Fig. 11-25 shows two masses connected by a cord passing over a pulley of radius R0 and moment of inertia I. Mass M1 slides on a frictionless surface, and M2 hangs freely.
[PLAIN]http://img217.imageshack.us/img217/8194/1125u.gif

Determine a formula for the angular momentum of the system about the pulley axis, as a function of the speed v of mass M_{1} or M_{2}.

Homework Equations

L = I\omega

\omega = \frac{v}{r}

\alpha = \frac{a}{r}

a = \frac{v^2}{r}

The Attempt at a Solution

Don't even know.

 

[collinsmark]

Hello killazys,

Welcome to Physics Forums,

One one homework problem is allowed per thread. So I'll start on your first one.

K_{i} = K_{r} + K_{f}

K_{r} = \frac{1}{2}*\frac{mL}{12}*(\omega)^{2}

Where K_{i} and K_{f} are of the bullet, and K_{r} is of the stick.
[...]
Plug in and solve. The problem probably is that the bullet hitting right at the 0.25m mark has an effect, one that I am unaware of :'(
Attempted first answer was 71.875 rad/s.

Conservation of kinetic energy does not apply to this problem. Much of the energy is lost as heat during the collision between the bullet and meter long stick.

However, you can use conservation of angular momentum.

Hints: What is the angular momentum of the bullet (about the pivot point) just before it hits the stick? What is the angular momentum of the bullet (also about the pivot point) just after the collision?

 

[killazys]

Second attempt at solution:

L_{i} = L_{f} + L_{s}

mr^{2}*\frac{v_{i}}{r} = mr^{2}*\frac{v_{f}}{r}

Where the radii are both 0.25m, and the bullet is treated as a point mass. Then,

L_{s} = 2.1875 - 1.4

\frac{mL^{2}}{12}*\omega = .7875

\omega = 31.5

It's wrong, though.

 

[collinsmark]

Second attempt at solution:

L_{i} = L_{f} + L_{s}

mrv_{i} = mrv_{f} \color{red}{\ + \ L_s}

Slightly reformatted/simplified from true quote. Red text is mine.

Where the radii are both 0.25m, and the bullet is treated as a point mass. Then,

L_{s} = 2.1875 - 1.4

You're off by a factor of 10 here. I'm not sure what's going on, but I'm guessing you are missing a 0 in the mass. 3.5 g = 0.0035 kg, not 0.035 kg.

 

[killazys]

Thank you for your help! Should I make a new thread for the second question, then?

 

[collinsmark]

Thank you for your help! Should I make a new thread for the second question, then?

Yes, start a new thread.

But do show some work. Then we can help you at the place where you are getting stuck.

Here is something to think about to get you started: The total angular momentum of the system about some axis is the sum of the individual angular momentums about the same axis of all the components of the system.