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What angular speed is the stick spinning after the collision?

  • Thread starter saturn67
  • Start date
  • #1
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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
please help me ^_^
 

Answers and Replies

  • #2
alphysicist
Homework Helper
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1
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?
 
  • #3
54
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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:
  • #4
alphysicist
Homework Helper
2,238
1
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?
 
  • #5
54
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how to relate I1*w1i to bullet??

i know I2w2i= 0 since stick is at rest
 
  • #6
alphysicist
Homework Helper
2,238
1
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?
 
  • #7
54
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humm

soo L= R cross P

p= mv
so L=m1*v1*r

is right?
 
  • #8
alphysicist
Homework Helper
2,238
1
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:

[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.)
 
  • #9
54
0
ok thank alot alphysicist that is a big help XD

yep i got it

w= 4.8 rad/sec
 

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