Conservation of angular momentum of a stick

[234jazzy2]

Homework Statement

A uniform stick 1.00 m long with a total mass of 270 g is pivoted at its center and is initially stationary. A 30 g piece of clay is thrown at the stick midway between the midpoint of the stick (pivot) and one end. The clay piece is going at 50 m/s and sticks to the stick. What is the angular velocity of the stick after the collision?

Homework Equations

Conservation of angular momentum.
Li = Lf

The Attempt at a Solution

I = moment of inertia for stick
ms = mass of stick
mb = mass of bullet
v = velocity of the bullet
r = distance between the pivot and where the bullet hit = 0.5 m
Iω(rod) + (mb)vr = Iωf(rod) + (mb)ωf
Iω(rod) = 0
ωf = ((mb)vr)/(I + (mb))
I = (1/12)(ms)r^2
ωf = 14.286 rad/s

I assumed the angular velocity is the same for the mass and the stick, since they are stuck together.

Thanks.

 

[mattbeatlefreak]

A 30 g piece of clay is thrown at the stick midway between the midpoint of the stick (pivot) and one end.
r = distance between the pivot and where the bullet hit = 0.5 m

You are correct in saying that the r value is the distance between the pivot and the point where the clay hits. However, the r value is not 0.5 meters here. The entire stick is 1.00 m, and the clay sticks at the point halfway from the pivot to the end. The pivot is in the middle.

 

[soviet1100]

The Attempt at a Solution

I = moment of inertia for stick
ms = mass of stick
mb = mass of bullet
v = velocity of the bullet
r = distance between the pivot and where the bullet hit = 0.5 m
Iω(rod) + (mb)vr = Iωf(rod) + (mb)ωf
Iω(rod) = 0
ωf = ((mb)vr)/(I + (mb))
I = (1/12)(ms)r^2
ωf = 14.286 rad/s

Thanks.

Your conservation equation has an error. Corrected, it stands as: $m_b v r = (I + m_b r^2)\omega_{f}$

The stick in the initial configuration has no angular momentum. All the angular momentum in the final configuration is contributed by the bullet.

[EDIT]: Also, how is r = 0.5 m. If the stick is pivoted about the centre and the origin of the coordinate system has been chosen to be this point, then r is midway between the pivot/origin and the end (r=0.25 m)

 

[234jazzy2]

You are correct in saying that the r value is the distance between the pivot and the point where the clay hits. However, the r value is not 0.5 meters here. The entire stick is 1.00 m, and the clay sticks at the point halfway from the pivot to the end. The pivot is in the middle.

So,
ωf = ((mb)vr1)/(I + (mb)) , where r1 = 0.25 m (half of the half...)
I = (1/12)(ms)r^2 (r = 0.5 m (half of the stick)
ωf = 10.526 rad/s

 

[234jazzy2]

Your conservation equation has an error. Corrected, it stands as: $m_b v r = (I + m_b r^2)\omega_{f}$

The stick in the initial configuration has no angular momentum. All the angular momentum in the final configuration is contributed by the bullet.

EDIT: NVM read it wrong

How did you get $m_b r^2$?

 

[mattbeatlefreak]

EDIT: NVM read it wrong

How did you get $m_b r^2$?

The angular momentum of an object is Iω. I is the moment of inertia. The moment of inertia of a point mass is mr².

 

[soviet1100]

Yea, in the line below i stated that Iω(rod) = 0. I just right it so i have all the objects in the system and then set things to zero.

In your initial post, you wrote :

Iω(rod) + (mb)vr = Iωf(rod) + (mb)ωf

The second term on the right side of this equation is incorrect. It should be $m_b \omega_f r^2$. Remember, angular momentum has dimensions M L^2 T^(-1).

 

[234jazzy2]

The angular momentum of an object is Iω. I is the moment of inertia. The moment of inertia of a point mass is mr².

Oh ok, I didn't think you could assume that clay is a point mass. If you do that, the angular velocity if 50 rad/s... Is that correct?

 

[mattbeatlefreak]

Oh ok, I didn't think you could assume that clay is a point mass. If you do that, the angular velocity if 50 rad/s... Is that correct?

This is not the value I am finding for the final angular momentum.
Make sure to use m_bvr=(I+m_br²)ω_f and to use I as the moment of inertia of the rod, which you previously said, was (1/12)m_rL².
Also make sure to use all lengths in meters and masses in kilograms so you get the correct final units.

edit: the assumption that the clay is a point mass is a common occurrence in these types of problems. If a bullet or "piece" of clay is hitting something and the problem doesn't give you a value for its moment of inertia, it is usually safe to assume that they want you to treat it as a point mass.

 

[234jazzy2]

This is not the value I am finding for the final angular momentum.
Make sure to use m_bvr=(I+m_br²)ω_f and to use I as the moment of inertia of the rod, which you previously said, was (1/12)m_rL².
Also make sure to use all lengths in meters and masses in kilograms so you get the correct final units.

edit: the assumption that the clay is a point mass is a common occurrence in these types of problems. If a bullet or "piece" of clay is hitting something and the problem doesn't give you a value for its moment of inertia, it is usually safe to assume that they want you to treat it as a point mass.

Is it 15.38 rad/s?

 

[mattbeatlefreak]

Is it 15.38 rad/s?

That's what I get

 

[234jazzy2]

That's what I get

Thanks.