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A uniform rod of Mass $M$ and length $L$, which is free to rotate about a fixed vertical axis through $O$, is lying on a friction less horizontal table. A particle of equal mass strikes the rod with a velocity $V_0$ and gets stuck to it. The angular velocity of the combination immediately after the collision is
Two different answers by applying conservation of angular momentum and conservation of Energy
Rotational motion When do we use conservation of momentum vs. conservation of energy?
Don't use conservation of energy , since the collision is inelastic. It is always better to use conservation of angular momentum.
Two different answers by applying conservation of angular momentum and conservation of Energy.
My guess is that you did not apply conservation of energy which will give the same answer but rather you applied conservation of kinetic energy which is not the same.
Initially perhaps a better example to consider is the ballistic pendulum where a projectile is fired into the bob of a simple pendulum with the result being that the pendulum bob and the imbedded projectile rise up a certain distance.
The collision between the pendulum bob is inelastic because the total kinetic energy of the bob and imbedded projectile immediately after the collision is less than the kinetic energy of the projectile immediately before the collision.
The difference in kinetic energies is because some of the kinetic energy is converted into heat and sound and also used to permanently deform the projectile and bob.
So unless you are told that the collision is elastic you should not use conservation of kinetic energy.
If there is any sticking of one body to another it is going to be an inelastic collision.
To use conservation of linear or angular momentum there must be no external forces or torques acting however it is often the case that one wishes to apply conservation of momentum along one direction or about a particular axis.
In that case one needs to consider whether or not forces act along that direction or torques act about a particular axis.
In the case of the ballistic pendulum during the collision no horizontal forces act an so the horizontal linear momentum is conserved or you could say that there are no external torques about the point of suspension of the pendulum bob and angular momentum is conserved about the point of suspension.
After the collision things change.
Assuming no air resistance on can say that the sum of the kinetic energy and the gravitational potential energy of the bob and the imbedded projectile is constant.
However the vertical component of the linear momentum of the bob and imbedded projectile changes because the vertical force on them due the the gravitational attraction of the Earth is an external force on the system.
In the same way angular momentum is not conserved about the point of suspension because the gravitational attractive force exerts an external torque.
Going back to you simpler case where you are just considering horizontal motion.
The way the system is described with the axis of rotation being fixed means that the rod could have a horizontal force acting on it due the the fixed axis and so conservation of linear momentum in the horizontal plane should not be used.
However if one considers torques about a vertical axis through the pivot point there are none and so conservation of angular momentum about the pivot point can be used.
If there had been no fixed pivot then there would have been no external horizontal forces and linear momentum conservation could have been used.
The fact that the particle sticks to the rod immediately tells you that the collision is inelastic and conservation of kinetic energy should not be used.
Both approaches should always give you the same answer, when properly implemented. Your question about when to use which is a good one, with a deceptively simple answer: use whichever one makes the equations simpler.
Conservation of energy is convenient in situations where the directions of motion are constrained enough that you don't have to care about them. In a pendulum, you really don't have to worry about integrating momentum across the whole path for a lot of problems. You know it will eventually reach a certain end-state where you can use conservation of energy.
Conservation of momentum is convenient in situations where there may be difficult to model sources of energy involved. In your example, there's an inelastic collision. Those generate heat, and you aren't given enough information to figure out how much heat is generated directly (if you wanted to calculate that heat, you'd actually solve the momentum equations, then plug the results in to the energy equations!).
