0
$\begingroup$

If a ball hits a rod at rest at any position along the rod, the rod will be moving with the same linear velocity in each case. However, if the ball hits the rod away from its center of mass, the rod will also rotate.

How does this not violate the conservation of energy? The incoming ball has the same mass and velocity in each case and thus the same energy is inputted to the system, and the rod will have the same mass and linear velocity in each case, plus excess rotational energy. How can simply choosing where the ball hits the rod add energy to the system? My best guess is that hitting the rod off-center somehow "drains" more energy from the ball and slows it down more after collision, but I dont quite see why this would occur mechanistically.

$\endgroup$
  • $\begingroup$ Why would it have same velocity in all cases? $\endgroup$ – Sidharth Giri Apr 18 at 16:35
  • $\begingroup$ I'm assuming thats the case because linear momentum has to be conserved, and I've seen someone corroborate my thought elsewhere on the internet. $\endgroup$ – jizzle nizzle Apr 18 at 16:39
  • $\begingroup$ In reasoning that the linear velocity of the rod is the same in all cases, you are making the false implicit assumption that the rebound velocity of the ball is the same in all cases. $\endgroup$ – Ben51 Apr 18 at 17:03
  • $\begingroup$ @Ben51 Ah, the ball would feel less resistance the further out from the center of mass it hits, so it would rebound less. And if it rebounds less, that means its kept less of its kinetic energy and poured more into the rod, explaining how the rod gets excess energy to rotate with. $\endgroup$ – jizzle nizzle Apr 18 at 17:15
  • $\begingroup$ Also, the linear motion of the rod is slower following a collision near the end than one near the middle. So not only in the KE of the rebounding ball less after an off center collision, the translational KE of the rod is also less. For elastic collision, the sum of these two and the rotational KE of the rod is always equal to the initial KE of the ball. $\endgroup$ – Ben51 Apr 18 at 17:20
1
$\begingroup$

Energy is always conserved but in problems such as this kinetic energy may not be conserved.

If there are no external forces/torques acting on the ball & rod system then linear/angular momentum will always be conserved.
These two conservation laws give two equations which link the final linear velocity of the centre of mass of the rod (and ball if the ball sticks to the rod), the final linear velocity of the ball and the final angular velocity of the rod (and ball if the ball sticks to the rod) to the initial velocity of the ball.
So in general you have three unknowns and 2 equations linking them and thus cannot solve the problem.

If the ball sticks to the rod then the three unknown become two unknowns and the problem can be solved however the collision will be inelastic and kinetic energy will not be conserved.

In general to get a third equation linking the three unknowns you must add a constraint on the system which, for example, might be that the collision between the ball and the rod is elastic ie kinetic energy is conserved.

$\endgroup$
0
$\begingroup$

In such collisions, net linear and angular momenta are conserved, since the force of interaction is purely internal and cannot cause an impulse. However, this force of interaction can cause some change in elastic potential energy, which results in the supposed discrepancy with the law of conservation of energy. Whether net kinetic energy will be conserved or not depends on the nature of the collision, in which case the coefficient of restitution comes in handy for solving problems.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.