# Rod swinging and hitting a ball

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1. Nov 29, 2017

### erfz

1. The problem statement, all variables and given/known data
There is a uniform rod of mass m = 1 kg and length L = 0.2 m fixed to the wall by an axis passing through its end.
A uniform ball of mass M = 0.1 kg and radius R = 2.85 cm is on the ground, below the axis of the rod, such that the rod's unfixed end is at the height of the ball's center-of-mass if the rod were to be at an angle of 0 degrees with the vertical.
The rod is set at angle 45 degrees to the left of the vertical and let go -- find the angular velocity of the rod before and after the collision, as well as the ball's translational and angular speeds after the collision. Assume the ball rolls without slipping.

I have written my approach below. I don't have the answer at hand, but I'm sure something is wrong here since the translational speed of the ball I get is 0.

2. Relevant equations
$\vec{L} = \vec{r} \times m\vec{v}_{cm} + I_{cm}\vec{\omega}$
$L = I\omega$

3. The attempt at a solution
I viewed this as an instantaneous collision and set the axis of rotation at the axis through the rod's end. So, the first step is finding the initial omega $\omega_i$ right before the collision using energy conservation about the rod's COM: $$m_{rod}g\frac{L}{2}(1-cos ~\theta) = 1/2 I_{rod}\omega_i^2$$ The moment of inertia of the rod about its end is $1/3mL^2$. Plugging in the numbers, I get that $\omega_i = 6.6 ~\text{rad/s}$

The angular momentum expression is: $$I_{rod}\omega_i = I_{rod}\omega_f + MLv_{ball,cm} + I_{ball,cm}\omega_{ball}$$ This second and third terms on the right come from the first equation I listed in "Relevant equations." I am defining counter-clockwise to be positive, so the translational contribution to the angular momentum of the ball (the second term on the right) should also be positive. However, the third term should be negative overall since the ball will be rotating clockwise. Since the ball isn't slipping, I can replace $\omega_{ball} = -v_{ball,cm}/R$. So then: $$I_{rod}\omega_i = I_{rod}\omega_f + MLv_{ball,cm} - I_{ball,cm}v_{ball,cm}/R$$ I still need another equation to relate $\omega_f$ and $v_{cm}$. I believe I can use linear momentum conservation for this: $$m_{rod}v_{rod,cm} = m_{rod}v'_{rod,cm} + Mv_{ball,cm}$$ The velocity of the rod's center-of-mass, I believe, should be such that $v_{rod,cm} = \omega r$, where r is the distance from the axis to the center-of-mass of the rod (i.e. $L/2$). So then this becomes $$m_{rod}\omega_i L/2 = m_{rod}\omega_f L/2 + Mv_{ball,cm}$$ Plugging in the values here, I get that relation that $$\omega_f = \omega_i - v_{ball,cm}$$ Now, if I substitute this into the angular momentum expression and solve, I get that $v_{ball,cm}$ is essentially 0, which intuitively cannot be correct. What is wrong with my approach?

Last edited by a moderator: Nov 29, 2017
2. Nov 29, 2017

### haruspex

That is almost impossible. You would have to allow the collision process to take a significant time and for the coefficient of friction with the ground to be very high. But glossing over that...
No. There is a frictional impulse from the ground (or the ball would not be immediately rolling) and a horizontal impulse from the hinge.
You need to know whether the impact is completely elastic or completely inelastic, or, in between, the coefficient of restitution.

3. Nov 29, 2017

### erfz

What is the reasoning behind that? Such a scenario seems very possible in real life without requiring a great time or large coefficient of friction.
I see. But even if the impact was elastic, how would that allow me to account for the frictional and hinge impulses?

4. Nov 29, 2017

### haruspex

In an impact you get large forces acting for very short times. Often it is very hard to know what the actual force profile looks like over that time. It is much easier to calculate the total momentum and work with that.
In the present problem, the impact is horizontal, so does not affect the normal force between ball and ground, so does not affect the maximum frictional force. The peak force in the impact would almost always overwhelm the frictional force.
This might not be apparent because if the frictional coefficient is high then it might quite quickly develop rolling contact.
In other situations, the impact might have a component normal to the ground, thereby generating a corresponding normal impulse reaction and corresponding impulsive friction.
You just need another equation. If completely elastic, work is conserved; if completely inelastic then the linear motions will have the same velocity immediately after impact (thereby maximising work lost).

5. Nov 30, 2017

### erfz

@haruspex So for example, if we assume that collision is (instantaneously?) completely inelastic, then is it true that $m \omega_i L/2 = (m + M)v = (m + M) \omega_f L/2$? Wouldn't this mean that I don't even need the angular momentum expression?

Even if this is true, I suspect that friction does work to make the ball roll and that $v$ is not precisely $v_{cm,ball}.$ In particular, $v_{cm,ball} < v$. Realistically, how different would they be and what determines this difference?

Last edited by a moderator: Nov 30, 2017
6. Nov 30, 2017

### haruspex

No, reread my post. You get an equation about velocities, not momenta.

7. Nov 30, 2017

### erfz

So then $\omega_f L/2 = v_{ball, cm}$?
And if the collision were perfectly elastic, then $1/2I_{rod}\omega_i^2 = 1/2I_{rod}\omega_f^2 + 1/2Mv_{ball,cm}^2 + 1/2I_{ball}(\frac{v_{ball,cm}}{R})^2$?

8. Nov 30, 2017

### haruspex

Not quite. How fast would the tip of the rod be moving in that case?

9. Nov 30, 2017

### erfz

$v_{tip} = \omega_f L = v_{ball,cm}$? Also, see my edit if the collision is elastic.

10. Nov 30, 2017

### haruspex

yes, depending on the axes for the two moments of inertia.

11. Nov 30, 2017

### haruspex

Yes.

12. Nov 30, 2017

### erfz

How do you mean? In my case I would choose the axis passing through the tip of the rod.

Also, as for the coefficient of restitution, I see that it is defined as $e = \frac{v_b - v_a}{u_a - u_b}$ in one-dimension, where $v$ denotes final velocities and $u$ denotes initial velocities of two objects. If I call the velocities of the rod $v_b$ and $u_b$, does this refer to the velocities of its center-of-mass or of its tip? Also, is it possible to measure this coefficient in such a problem without having the described collision occur? How could I even measure it, practically speaking?

13. Nov 30, 2017

### haruspex

I assume you mean through the hinge, but the value is the same.
Since that is the axis of the rod's rotation, not its mass centre, it is not clear from your equation what Iball means. Again, you could choose mass centre of the ball or its instantaneous centre of rotation. I'm pretty sure which you mean, but it should be made clear in the equation.
Clearly not the mass centre since that again would lead to the rod penetrating the ball when e=0. It would have to be the tip.
However, it is not entirely clear that the formula can be directly applied in this case. You could check by plugging in e=1 and seeing if you get the same solution as you got for perfect elasticity. If not, let me know and I will see if I can come up with a working formula.

14. Dec 1, 2017

### erfz

I am not able to sit down and check this right now but I do want to bring up another question I thought of.
Since the hinge exerts a horizontal force, making momentum not conserved, doesn't it also do work? Wouldn't energy not be conserved? So then how can the collision be perfectly elastic... Is this from assuming that the work done by the hinge is negligible?

15. Dec 1, 2017

### haruspex

The hinge does not move. Work = force times distance moved in the direction of the force.

16. Dec 1, 2017

### erfz

Ah of course

17. Dec 3, 2017

### erfz

@haruspex
Using energy conservation (assuming elastic), I get that $\omega_f = 4.49 ~rad/s$ and that $v_{ball, cm} = 1.49 ~m/s$.
Using $$e = \frac{v_{tip, f} - v_{ball, cm}}{0 - v_{tip, i}} = 1$$ and of course $v_{tip} = \omega L$, I get that $\omega_f = 3.69 ~rad/s$ and that $v_{ball, cm} = 2.06 ~m/s$. So it does not seem like $e$ can be used directly here, unless I've made a mistake.

18. Dec 3, 2017

### haruspex

Working purely algebraically (no numbers) I get that the usual coefficient of restitution formula works, using the velocity of the tip of the rod.
Please try it the same way (no numbers) and post your working. You should be able to leave the rod's MoI as just "I", not needing to substitute mL2/3.

19. Dec 4, 2017

### erfz

I have decided to upload the image of my written algebraic work. Hopefully it is neat enough for you to read.
On the bottom right, I have used the angular momentum expression, multiplied both sides by $\omega_i + \omega_f$ to utilize the relationship with the coefficient of restitution, and combined. I get a form strikingly similar to the energy expression, but not quite. The denominator of the ball's rot. KE contains $LR$ rather than $R^2$. Interestingly, the numbers I found before work here.

Where have I gone wrong?

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20. Dec 4, 2017

### haruspex

I had forgotten that the ball is supposed to start rolling instantly. This might be why the standard equation does not work. (My original reason for thinking it might not work was connected with the hinge on the rod.)
I'll take the rolling into account and get back to you, but it might be 12 hours or so before I get the chance.