# What angular speed is the stick spinning after the collision?

• saturn67
In summary, a 1.0 m long stick with a mass of 220 g is pivoted at its center. A 3.2 g bullet traveling at 250 m/s collides with the stick midway between the pivot and one end, and leaves at a speed of 140 m/s. The angular speed of the stick after the collision is 4.8 rad/sec, found by conserving angular momentum.

## Homework Statement

A uniform stick 1.0 m long with a total mass of 220 g is pivoted at its center. A 3.2 g bullet is shot through the stick midway between the pivot and one end The bullet approaches at 250 m/s and leaves at 140 m/s

With what angular speed is the stick spinning after the collision?

## Homework Equations

I*wi=I*wf

I of stick = (1/12)ML^2

K= (1/2)(Mb*Vf^2+(1/2)Iw^2

## The Attempt at a Solution

(1/2)*Mb*Vi^2 = (1/2)(Mb*Vf^2+(1/2)Iw^2

i use the above equaltion but it wrong
anyone got any idea how to solve this XD

Hi saturn67,

Since this is not an elastic collision, kinetic energy is not conserved and so

(1/2)*Mb*Vi^2 = (1/2)(Mb*Vf^2+(1/2)Iw^2

is not true.

What is conserved in this collision?

angular momentum conserved?

I*wi = I*wf

but bullet don't have angular momentum?

so 0=I*wf

so it don't make sense >.<

Last edited:
Yes, that's right. So your first equation (written out for two separate objects)

$$I_{1} \omega_{1,i} +I_{2}\omega_{2,i} =I_{1} \omega_{1,f} +I_{2}\omega_{2,f}$$

how to relate I1*w1i to bullet??

i know I2w2i= 0 since stick is at rest

We can treat the bullet as a point particle. What is the moment of inertia of a point particle at the point where it hits the stick? We know its linear speed; how is that related to its angular speed?

humm

soo L= R cross P

p= mv
so L=m1*v1*r

is right?

saturn67 said:
humm

soo L= R cross P

p= mv
so L=m1*v1*r

is right?

Yes, that's a good way to do it. So you know:

$$I_1 \omega_{1,i} = m_1 v_1 r_1$$

and you can do the same thing for the angular momentum of the bullet after the collision. Once you find $I_2\omega_{2,f}$, you can then solve for $\omega_{2,i}$.

(If you had done them separately you would find $I=m r^2$ and $\omega=v/r$, which would have given the same answer of mvr.)

ok thank a lot alphysicist that is a big help XD

yep i got it