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

## 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

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alphysicist
Homework Helper
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 dont make sense >.<

Last edited:
alphysicist
Homework Helper
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

alphysicist
Homework Helper
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?

alphysicist
Homework Helper
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 alot alphysicist that is a big help XD

yep i got it