A bullet hits a rod; find the angular velocity

  • Thread starter tylertwh
  • Start date
  • #1
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Homework Statement



A uniform rod of length L rests on a frictionless horizontal surface. The rod pivots about a fixed frictionless axis at one end. The rod is initially at rest. A bullet traveling parallel to the horizontal surface and perpendicular to the rod with speed v strikes the rod at its center and becomes embedded in it. The mass of the bullet is one-fourth the mass of the rod.
Mass(bullet) = (1/4)Mass(rod)

A)What is the final angular speed of the rod?

B)What is the ratio of the kinetic energy of the system after the collision to the kinetic energy of the bullet before the collision?

Homework Equations



v = ω*L
KE = (1/2)Iω2

The Attempt at a Solution



I know that v = ω*L is correct. I am missing the coefficient in front of it. The only thing that I can think of is the fact that the mass of the bullet being 1/4 that of the rod will somehow be put into this equation as well as the equation for part B.

So for part A...
I am not sure of any equation that will get me the proper coefficient for the answer of ω = v/L

For part B...

Bullet
KE = (1/2)Iω2
KE = (1/6)mL2ω2
KE = (1/6)mv2

System
KE = (1/2)Iω2
KE = (1/6)mL2ω2
.....since mass of the bullet is 1/4 of the system....
KE = (1/24)mL2ω2
KE = (1/24)mv2

When we put {KE = (1/24)mv2}/{KE = (1/6)mv2}
1/4

But that is wrong as well...
 

Answers and Replies

  • #2
44
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Try the conservation of angular momentum for part a. Until you get the final angular speed of the rod/bullet system I don't think you can calculate the energy fraction.
 
  • #3
22
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how would that work when the bullet hits the rod at a 90 degree angle? I dont know what I would put for the ω(bullet). Unless somehow I incorperated ω=v/r, but even then I am not sure what I would put for the r?

Iω(initial) = Iω(final)
mv2ω(initial) = (1/3)mv^2*[v/r]

ω(initial) = ?
 
  • #4
44
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w=v/r, yes. Doesn't the bullet strike the rod at its center (i.e. halfway along its length)?
 
  • #5
22
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Iω(initial) = Iω(final)
mv2ω(initial) = (1/3)mv2*[v/r]
(mv3)/r = ((1/3)mv2ω
ω = 3v/r = 3v/L
 
  • #6
22
0
Iω(initial) = Iω(final)
mv2ω(initial) = (1/3)mv2*[v/r]
(mv3)/r = ((1/3)mv2ω
ω = 3v/r = 3v/L
wait.
r => (1/2)r....
so the answer would actually be 6v/L = ω
 
  • #7
22
0
well... I used all my attempts for that one.. the answer is (6/19)v/L

Could you possibly explain why?
 
  • #8
22
0
figured all of it out! no more help needed
 

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