Off-center impulse equations

Question

A rigid steel bar with mass $M$ is hit sideways (very close to its end) by a steel ball with mass $m$ and velocity $v$. What are the equations of motion after elastic impact and how about conservation of momentum and energy?

Answer 1

A rod of mass $M$ and length $\ell$ has mass moment of inertia $I = \frac{M}{12} \ell^2 $. The impact at a distance of $c = \frac{\ell}{2}$ from the center of mass imparts an impulse $J$, while an equal and opposite impulse $-J$ is applied to the projectile mass $m$.

The projectile is going to bounce with velocity $v_B = v - \frac{J}{m}$. The center of mass of the rod is going to start moving with velocity $v_C = \frac{J}{M}$ while the rod rotation is going to be $\omega = \frac{c J}{I}$.

The linear velocity of the point of impact is thus $v_A = v_C + \omega c = \frac{J}{M} + \frac{c^2 J}{I} $.

The law of impact states that the final separating velocity is a fraction of the initial impacting velocity. $ v_A - v_B = \epsilon v $, where $\epsilon$ is the coefficient of restitution. Putting it all together yields: $$J = (1+\epsilon) \mu v $$$$\mu = \left( \frac{1}{m} + \frac{1}{M} + \frac{c^2}{I} \right)^{-1} $$

The term $\mu$ is called the reduced mass of the system, and it can be viewed as the effective mass of the impact. It converts the impact speed $v$ into momentum $\mu v$. Depending on the bounciness the exchanged momentum (impulse) is between $ J = \mu v \ldots 2 \mu v $. Back substitute $J$ into the equations above to find $v_C$, $\omega$ and $v_B$.

Answer 2

This is a standard rotational motion problem.

Use conservation of linear momentum for the translational motion.
Use conservation of angular momentum about any axis noting that some axes make the algebra easier than others.
Use conservation of kinetic energy as the collision is stated to be elastic.

Answer 3

Let us take the system - Rod + Ball

Let the final velocity of center of mass of rod and ball be $ v_1 $ and $ v_2 $ respectively.

As there is no net external force on the system, the net linear momentum of the system will be conserved. $ mv = Mv_1 + mv_2 $

And, as there is no net external torque about the COM of the system, the angular momentum of the system about it's COM will be conserved.

This will give you one more equation with $ v_1 $, $ v_2 $ and $ v $.

Solve both equations simultaneously to get the results.

Answer 4

The linear motion of the bars center in the direction of the ball will be as if the ball had hit the bar at its center. The ball will bounce off the bar in the exact opposite direction but with reduced speed. The ball will also make the bar rotate around its center with some angular velocity. After impact, the sum of the translational energy of the ball and the bar and the rotational energy of the bar must equal the translational energy of the ball before impact (assuming no friction or deformation energy during impact). In this example, the conservation of momentum only seems to apply to the translational motion and does not include conservation of angular momentum?