Bullet hits door, find angular velocity

  • Thread starter aliaze1
  • Start date
  • #1
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Unfortunately I was absent when a similar example was done in class.....

Homework Statement



A 10 g bullet traveling at 400 m/s strikes a 10 kg, 1.0-m-wide door at the edge opposite the hinge. The bullet embeds itself in the door, causing the door to swing open.

What is the angular velocity of the door just after impact?

Homework Equations



0.5mV^2
0.5Iω^2
v = ωr

The Attempt at a Solution



Attempt #1:

I considered the bullet as a particle in linear motion. After collision I considered the mass of the door to be (M = Mdoor+Mbullet). I set up the problem like so:

0.5mv^2 = 0.5Iω^2

the 0.5 is a constant so it cancels..

mv^2 = Iω^2
*note* small 'm' is mass of the bullet, big 'M' is for the door, same goes for 'v' and 'V'

Then I plugged in the moment of inertia I, which is given in a table as 1/3ML^2

the width of the door is 1m, so the L^2 is simply 1

so putting this into the equation, I get:

mv^2 = (1/3M)ω^2

and rearranging and solving for ω, I get:

ω = √[(mv^2)/(1/3M)]|
(square root)

this did not work....


Attempt #2:

I tried to find the linear velocity by using just conservation of linear energy

0.5mV^2 = 0.5MV^2

0.5 cancel, so it is mv^2 = MV^2 and find the linear velocity as so:

V = √((mv^2)/M)|

and then using V = ωr, I solved for ω

but still incorrect.....
 
Last edited:

Answers and Replies

  • #2
174
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pic


Here is an image I made on paint, this is the way I see it, bird's eye view...

http://photo.ringo.com/210/210049338O350481129.jpg [Broken]

here is the link if it doesn't come up for some reason

http://photo.ringo.com/210/210049338O350481129.jpg" [Broken]

http://photo.ringo.com/210/210049338O350481129.jpg [Broken]
 
Last edited by a moderator:
  • #3
960
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when you right 1/3M is that M/3? In any event, I think momenta is what should be conserved.
 
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  • #4
174
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yes, this is the moment of inertia: 1/3 ML^2

but L is 1, so L^2 = 1 and therefore it is 1/3 M, or as you stated, M/3
 
  • #5
Office_Shredder
Staff Emeritus
Science Advisor
Gold Member
3,911
184
conservation of momentum would be the way to go here.

The alternate answer is zero, because the door is locked and the latch holds
 
  • #6
174
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lol

door locked, nice one haha

ok ill try conservation of momentum

so do i do the bullet as linear momentum, and the door as rotational

so it would be:

mv = rMω

?
 
  • #7
174
0
WOOT WOOT!!!!!

Got it!!!

i used conservation of momentum:

mv = Iω (found this part in the txtbook)

and I is given as 1/3ML^2 for a 'thin rod about side'

thanks!
 

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