Can someone help me with Equilibirum?

jinman
I don't understand how to do equilibrium problems. I don't know how to set them up and where to start. here is an example. Can anyone please help?

Homework Statement

A uniform thin rod of length 0.50m and mass 4.0 kg can rotate in a horizontal plane about a vertical axis through its center. The rod is at rest when a 3.0g bullet traveling in the rotation plane is fired into one end of the bullet's path makes angle = 60.0 degrees with the rod. If the bullet lodges into the and the angular velocity of the rod is 10 rad/s immediately after the collision, what is the bullet's speed just before impact.

1/2mv^2=1/2Iw^2
I=1/12ML^2

The Attempt at a Solution

K-initial=K-final

1/2mv^2=1/2Iw^2

solve for v>>>>

v=sq. root[Iw^2/m]

v=sq. root[(1/12*ML^2)w^2/m]

v=sq.root.[(1/12*4.003*.50^2)*10^2/.003]

v=52.7m/s

Staff Emeritus
Gold Member
Welcome to PF jinman,

The collision isn't elastic and therefore kinetic energy isn't conserved. What is always conserved?

Mentor
This is not an equilibrium problem (not sure what you mean by that, anyway).

I assume the bullet lodges itself in the rod, making this a perfectly inelastic collision. Kinetic energy is not conserved. But something else is. What?

jinman
I don't know why i put Equilibrium. It must be momentum.

P=P

mv=Iw??

Mentor
I don't know why i put Equilibrium. It must be momentum.

P=P

mv=Iw??
Don't mix up linear momentum (p) with angular momentum (Iw). In this case, only one of them is conserved. Which one?

jinman
I would say linear.

m1v1=m2v2

v1=m2(wr)/m1

v1=4.003(10*.25)/ .003

v1*sin 60=2888.91m/s?

Mentor
No, linear momentum is not conserved. I assume the rod is restrained by some fixed axis to only rotate about its center. That axis exerts a force on the rod, thus linear momentum is not conserved.

jinman
Ok.

I1w1=I2w2

w1=I2w2/I1

OK so I have to find the angular velocity of the bullet before the collision? How can i find the Inertia of the bullet?

Mentor
http://hyperphysics.phy-astr.gsu.edu/Hbase/amom.html" [Broken]

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