The total z-component of the angular momentum of the system immediately before the collision is therefore:

bullet

After the collision, the bullet is attached to the rod and the bullet-rod system rotates as a rigid body around the axis. The z-component of the rotational momentum of a rigid body is L z = Itotω, where I is the moment of inertia. We Þnd the moment of inertia of the bullet-rod system using the superposition principle, summing the moment of inertia of the rod, I , and the bullet, which we consider to be a particle of mass m at a distance L from the rotation axis:

The rotational momentum after the collision is therefore:

Solve: Conservation of angular momentum gives:

Analyze: The subsequent motion of the rod-bullet pendulum can be determined from NewtonÕs second law for rotation around the axis O .

16.4.5 Example: Rotating Rod

Problem: A thin rod of length L and mass M is lying at rest on a frictionless surface. A bullet of the same mass, M , is shot horizontally into the rod, and hits the rod with the horizontal velocity v at a distance y from the center of the rod. The bullet is stuck in the rod. Find the translational and angular velocity of the object immediately after the collision.

Approach: Since the system is not affected by any external forces, the translational and rotational momentum is conserved, and we can use conservation laws to relate the motion before and after the collision.

Identify: The system consists of the rod and the bullet. We describe the motion of the system by the motion of its center of mass and by the rotation of the system around its center of mass. The system is illustrated in Fig. 16.26.

Model: The motion of the system is determined by the external forces acting on the system. Since we can ignore frictional forces in the plane of motion, the only external forces acting are the gravitational force and the normal force from the surface. Since the rod is not moving vertically, the net external forces are zero. From NewtonÕs second law for translational motion, this means that the (translational) momentum is conserved throughout the collision and during the motion afterwards:

The center of mass of the system therefore moves with a constant velocity—before, during, and after the collision! Notice that the translational motion does not depend on where the bullet hits the rodÑthe velocity of the center of mass is the same independently of where the bullet hits, although the position of the center of mass of course will depend on where the bullet hits.

The velocity of the center of mass before the collision depends on the velocity of the objects: The velocity of the rod is, vr = 0, and the velocity of the bullet is v, hence:

Since the system has no velocity component in the y-direction, the y-position of the center of mass remains constant, with the following value:

as illustrated in Fig. 16.26.

Second, we determine the motion relative to the center of mass. The rod may start rotating, depending on where the bullet hits the rod. The rotational motion around the center of mass is related to the torque of the external forces around the center of mass. But, since there are no external forces acting in the plane of motion, the

We apply NewtonÕs second law for rotational motion around the center of mass:

and Þnd that there is no change in the rotational momentum around the center of mass: The rotational momentum around the center of mass of the system is conserved. This is also the case for the z-component of the rotational momentum around the center of mass:

Since we know the rotational momentum around the center of the mass before the collision, we can use this to Þnd the rotational momentum around the center of mass after the collision, and from this we Þnd the angular velocity of the object around the center of mass.

The rotational momentum around the center of mass is the sum of the rotational momentum for each of the componentsÑaround their common center of mass,

Before the collision, the rotational momentum of the rod around the center of mass is 0, since it is not rotating initially. The rotational momentum of the bullet depends on the position r of the bullet relative the center of mass:

Lbullet = rb,cm × M v i = xb,cm i + yb,cm j × M v i = −yb,cm M v k . (16.147)

Notice that only the y-component of rcm contributes. Since the position of the bullet is −y, and the position of the center of mass is −y/2, we Þnd that the y-component of rb,cm is yb,cm = −y − (−y/2)) = −y/2, and