1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

What angular speed is the stick spinning after the collision?

  1. May 4, 2008 #1
    1. The problem statement, all variables and given/known data

    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?

    2. Relevant equations

    I*wi=I*wf

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


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

    3. 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
    please help me ^_^
     
  2. jcsd
  3. May 4, 2008 #2

    alphysicist

    User Avatar
    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?
     
  4. May 4, 2008 #3
    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: May 4, 2008
  5. May 4, 2008 #4

    alphysicist

    User Avatar
    Homework Helper

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

    [tex]I_{1} \omega_{1,i} +I_{2}\omega_{2,i} =I_{1} \omega_{1,f} +I_{2}\omega_{2,f}[/tex]

    should lead to the answer. What do you get?
     
  6. May 4, 2008 #5
    how to relate I1*w1i to bullet??

    i know I2w2i= 0 since stick is at rest
     
  7. May 4, 2008 #6

    alphysicist

    User Avatar
    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?
     
  8. May 4, 2008 #7
    humm

    soo L= R cross P

    p= mv
    so L=m1*v1*r

    is right?
     
  9. May 4, 2008 #8

    alphysicist

    User Avatar
    Homework Helper

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

    [tex]
    I_1 \omega_{1,i} = m_1 v_1 r_1
    [/tex]

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

    (If you had done them separately you would find [itex]I=m r^2[/itex] and [itex] \omega=v/r[/itex], which would have given the same answer of mvr.)
     
  10. May 4, 2008 #9
    ok thank alot alphysicist that is a big help XD

    yep i got it

    w= 4.8 rad/sec
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: What angular speed is the stick spinning after the collision?
Loading...