Clay colliding with rotating stick: conservation of angular momentum

[Kosta1234]

Homework Statement

A thin stick of mass M = 3.9 kg and length L = 1.8 m is hinged at the top. A piece of clay, mass m = 0.3 kg and velocity V = 2.9 m/s hits the stick a distance x = 1.60 m from the hinge and sticks to it. What is the angular velocity of the stick immediately after the collision?
​

Homework Equations

I = 1/3mx^2
I = mx^2

The Attempt at a Solution

in my first attemps I tried to use conservation of energy, but I realized that it's an inelastic collision so the energy is no conserved (right?)
than I tried to use the conservation of angular momentum.
​

there is no angular momentum while the ball flying towards the door.

so can I use:
m*v (of the ball) = Iw (of the door, with the ball)
I = ((m+M)*x^2)/3
and to plug in the moment of inertia to the equation above to find the angular velocity.​

 

[gneill]

Can you show us what you tried in detail?

 

[Kosta1234]

Yes, ofcourse thanks.
Well this is the full question:

A thin stick of mass M = 3.9 kg and length L = 1.8 m is hinged at the top. A piece of clay, mass m = 0.3 kg and velocity V = 2.9 m/s hits the stick a distance x = 1.60 m from the hinge and sticks to it. What is the angular velocity of the stick immediately after the collision?

in my first attemps I tried to use conservation of energy, but I realized that it's an inelastic collision so the energy is no conserved (right?)
than I tried to use the conservation of angular momentum.
there is no angular momentum while the ball flying towards the door.
so can I use?:
m*v (of the ball) = Iw (of the door, with the ball)
I = ((m+M)*x^2)/3
and to plug in the moment of inertia to the equation above to find the angular velocity.

 

[gneill]

Yes, ofcourse thanks.
Well this is the full question:

A thin stick of mass M = 3.9 kg and length L = 1.8 m is hinged at the top. A piece of clay, mass m = 0.3 kg and velocity V = 2.9 m/s hits the stick a distance x = 1.60 m from the hinge and sticks to it. What is the angular velocity of the stick immediately after the collision?

in my first attemps I tried to use conservation of energy, but I realized that it's an inelastic collision so the energy is no conserved (right?)

Right.

than I tried to use the conservation of angular momentum.

Good!

but in the first, there is no angular momentum only the momentum of the ball that flies in a constant speed towards the door.

Think again! A moving object has angular momentum about any point that is not on the line containing the velocity vector of that object.

can I use:
m*v (of the ball) = Iw (of the door, with the ball)

Almost...
Look up the moment of inertia of a point mass

edit: Actually, your second Relevant Equation is what you're looking for!

 

[Kosta1234]

ok I tried this time this:
mv = (1/3Mx^2)*w + (mx^2)*w ?
while x - distance from rotation axis.
w - angular velocity
m - mass of the ball
M - mass of the door

it still doesn't work. but can I say that the linear momentum goes completely to angular momentum?

 

[gneill]

ok I tried this time this:
mv = (1/3Mx^2)*w + (mx^2)*w ?
while x - distance from rotation axis.
w - angular velocity
m - mass of the ball
M - mass of the door

it still doesn't work. but can I say that the linear momentum goes completely to angular momentum?

No, you can't say that. Linear and angular momenta are two different things and can't be equated.

Start by determining the angular momentum of the system (about the pivot point) before the clay lump impacts the rod. What do you know about the angular momentum of an isolated system?

 

[Kosta1234]

ok I got it!
thank you very much!