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aliaze1
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Unfortunately I was absent when a similar example was done in class...
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?
0.5mV^2
0.5Iω^2
v = ωr
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...
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: