What angular speed is the stick spinning after the collision?

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Homework Help Overview

The problem involves a uniform stick that is pivoted at its center and is impacted by a bullet, leading to a question about the angular speed of the stick after the collision. The context includes concepts of angular momentum and the dynamics of inelastic collisions.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the conservation of angular momentum and the relationship between linear and angular quantities. There are attempts to apply equations related to kinetic energy and angular momentum, with some questioning the validity of certain assumptions regarding the collision type.

Discussion Status

Participants are actively engaging with the problem, exploring different aspects of angular momentum and how to relate the bullet's motion to the stick's angular speed. Some guidance has been offered regarding the treatment of the bullet as a point particle and the use of angular momentum equations.

Contextual Notes

There is a recognition that kinetic energy is not conserved in this collision, prompting discussions about what physical quantities are conserved instead. Participants are also navigating the implications of the bullet's impact on the stick's motion.

saturn67
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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^2K= (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 ^_^
 
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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)

[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?
 
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:

[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.)
 
ok thank a lot alphysicist that is a big help XD

yep i got it

w= 4.8 rad/sec
 

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