# A bullet is shot into a block of wood

1. A 2.3kg wood block hangs from the bottom of a 1.0kg, 1.2m long rod. The block and rod form a pendulum that swings on a frictionless pivot at the top end of the rod. A 10g bullet is fired into the block, where it sticks, causing the pendulum to swing out to a 33degree angle.

2. Homework Equations
Angular momentum of a particle = mvr^2
Angular momentum of a rigid body = Iw
Conservation of angular momentum: Li = Lf
Conservation of energy: Ei = Ef

3. The Attempt at a Solution
I have 3 stages set up. 0, which is right before the bullet hits the wood. 1, when the bullet is embedded into the wood and hasn't travelled anywhere yet. 2, when the pendulum has fully swung out.

L0 = L1
(Mbullet)(Vbullet)(R^2) = (Ibullet)(Wbullet) + (Iwood)(Wwood)
(.0144)(Vbullet) = 3.3242w

E1 = E2
(.5)(Ibullet)(w^2) + (.5)(Iwood)(w^2) = (Mbullet)(g)(h) + (Mwood)(g)(h) + PErod
I think the main issue is that I don't know what the potential energy of the rod is, but there certainly may be other issues. I believe h = .194m.

Any help is greatly appreciated!

Last edited:

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If anyone sees any minor errors in my calculations (or major!) please please help, I'm trying to figure this out before a test. Thank you.

gneill
Mentor
Is there some reason why you don't specify the units of the quantities? What's a "1.0 , 1.2--long rod"? Is a "10 bullet" a 10 kilogram bullet? 10 pound? 10 ton? Is a 33 angle 33 degrees, or perhaps 33 grads, or 33 radians (lots of revolutions!)? Marks are lost when units are ignored.

Also, you didn't state what the goal of the question is. What's to be calculated?

Sorry for the lack of units, the question was copy pasted.

A 2.3kg wood block hangs from the bottom of a 1.0kg, 1.2m long rod. The block and rod form a pendulum that swings on a frictionless pivot at the top end of the rod. A 10g bullet is fired into the block, where it sticks, causing the pendulum to swing out to a 33degree angle.

gneill
Mentor
So, what is it that you're trying to calculate?

The initial speed of the bullet.

gneill
Mentor
The angular momentum of a massive particle about a center is mvr, not mvr2.

You can calculate the PE of the rod+block+bullet by considering the elevation of the centers of mass of the rod and the block + bullet combination. (Figure out the Δh's when they're at the 33° position).

If you equate the PE with the mechanical energy (rotational in this case) when the rod+block+bullet is at its lowest point, you can determine the angular velocity there, and hence the angular momentum of the ensemble. The bullet must have supplied this angular momentum.