# Off-center impulse equations [closed]

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?

• What work have you done? What physics concept is giving you difficulty? – Bill N Jan 25 '16 at 22:44
• – Floris Jan 26 '16 at 0:35
• @Floris I don't think this question should be closed. The question of "How do we deal with off-center impulses?" although trivial to some does not have an nice accepted answer we can point at. – ja72 Jan 26 '16 at 3:44
• @ja72 there used to be a good set of answers to that specific question but it seems the question in point was deleted (it turned into an ugly flame war and led to several people leaving the site). I like the current answer of yours better than the one you gave to the duplicate I pointed to. – Floris Jan 26 '16 at 6:42
• I am voting to re-open in case this question attracts other even nicer answers. – ja72 Jan 26 '16 at 14:48

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 $v_B$.

• The collision is stated to be elastic so $\epsilon = 1$? – Farcher Jan 26 '16 at 8:20
• @Farcher - Yes. I think general cases are more instructive than specific values. – ja72 Jan 26 '16 at 14:46

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.

• I can calculate the both the final velocity of the center of the bar as well as the balls from conservation of linear momentum. From conservation of energy I can calculate the angular velocity of the bar. Hence it seems like I don't need to consider conservation of the systems angular momentum!? – Jens Jan 25 '16 at 15:35
• @Jens no, you cannot. Linear momentum has two unknowns (final velocity of ball and bar), but is only one equation. – LLlAMnYP Jan 25 '16 at 17:42

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.

• I can calculate the both the final velocity of the center of the bar as well as the balls from conservation of linear momentum. From conservation of energy I can calculate the angular velocity of the bar. Hence it seems like I don't need to consider conservation of the systems angular momentum!? – Jens Jan 25 '16 at 15:37
• Jens no the velocity of the ball after impact requires you to know the reduced mass of the rod, which depends on the location of impact. See @ja72's answer for details. – Floris Jan 26 '16 at 6:45

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?

• Sounds like an answer, but there's this question mark at the end? If you want to improve your question, you should edit the question itself. – Daniel Griscom Jan 26 '16 at 0:16