- 17

- 0

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.

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

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!

**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: