# Homework Help: The fall of a solid rod

1. Mar 18, 2013

### sergiokapone

1. The problem statement, all variables and given/known data

Rod AB of length $2l$, inclined to the horizontal at an angle $\phi$, falls, without rotating, with some height $h$ on a horizontal table and hits the table surface elastically first left, then right end. During the strike the right end, the rod again consist angle with the horizone $\phi$. Find the height $h$.

2. Relevant equations

Energy conservation law $mgh=I_{CM}\dfrac{\omega^2}{2}$ (1)
where $I_{CM}$ - moment of inertia about the center of mass of the rod.
Motion of center mass $m\dfrac{dv_{CM}}{dt}=mg-N$ (2)
where $N$ - normal reaction, acting on the left edge of the rod.
Rotational motion $I_{CM}\dfrac{d\omega}{dt}=N l cos \phi$ (3)

3. The attempt at a solution
And now no idea, how to solve.

Of course, the height can be found from the law of conservation of energy. In an elastic collision, the kinetic energy of translational motion partially transferred to the rotational kinetic energy, and the center of mass is not raised at the previous height, but on the other, where the selected zero potential energy. Therefore, in (1) $\omega$- the final angular velocity, I do not know how to find.

Last edited: Mar 18, 2013
2. Mar 19, 2013

### haruspex

That assumes there's no linear motion immediately after the first bounce. That cannot be true. In order for the two bounces to be symmetric the first bounce must produce an upward movement at the mass centre.
Suppose the bounce involves a vertical impulse J at point of contact. You can write down three equations: linear momentum, angular momentum, energy. You have four unknowns: impulse, linear speed before, linear speed after, angular speed after.
Next, consider the time t from first to second contact. You have two equations: angle of rod at time t, height of right hand end of rod at time t (=0). You now have enough equations to find all the unknowns.

3. Mar 19, 2013

### sergiokapone

I meant that after the impact, the linear upward motion disappears at a height where the potential energy is zero, so we obtain (1).

4. Mar 19, 2013

### sergiokapone

I do not understand how to write them. Neither linear nor angular momentum is not conserve in this case.

I've thought that I could be wrong on that account, at what height it will rise after the first strike.Okay, I'll try to consider this time t. Thank you.

5. Mar 19, 2013

### haruspex

But the equation refers to h, the original height. The height it will rise to after the bounce will be less than that.

6. Mar 19, 2013

### haruspex

By introducing the impulse J you can write the two momentum equations. E.g. J l cos(ϕ) is the imparted angular momentum about the mass centre.

7. Mar 20, 2013

### sergiokapone

I agree with you, then I made a mistake.

From eqns (2) and (3) after integrating I can get:

$I_{CM}\omega=mgl\int cos \phi dt - mV_{CM}lcos \phi '$
where $\omega$ - constant angular velocity after reflection, $V_{CM}$ - initial velocity of center of mass after reflection, $\phi '$ - here, the angle after reflection.

And of course, depending on the angle of the time did not know, and the angle of the reflection too.

8. Mar 20, 2013

### haruspex

You don't need to do any integration. It's all a matter of energy, momentum, and motion under gravity.
Let m be the mass of the rod, the linear speed be u before bounce and v after, and the impulse from the ground be J. Write the equation for conservation of momentum in the vertical direction. (I don't care whether you measure things up or down as long as it's consistent.)
Let I be the moment of inertia and ω be the clockwise angular speed after bounce. Using J again, and taking moments about the mass centre, write the equation for angular momentum. Thirdly, write the equation for conservation of energy through the bounce event. When you've done that we'll move onto the period between the two bounces.

9. Mar 20, 2013

### sergiokapone

I thought that the momentum in the vertical direction is not conserved. By itself, the rod is not a closed system. If the momentum of the Earth-Rod, then you can write the law of conservation.

$m\vec{u}=m\vec{v}+\vec{J}$
$\vec{J}$ - momentum gained Earth.

Did you mean it?

10. Mar 20, 2013

### haruspex

That's right. I actually meant J to be the impulse from the Earth, making it -J in the equation above, but it doesn't matter as long as we're consistent. If you also write the one for angular moment you can eliminate J between them to obtain a useful equation.

11. Mar 20, 2013

### sergiokapone

Ok. Let us consider the momentum conservation law ($y$- axis downward). Then
$mu=-mv_{up}+\Delta p$

The momentum conservation law:
$I_{CM}\omega=\Delta p l cos \phi$

Energy concervation law (before impact) $mgh=\dfrac{mu^2}{2}$
Am I right?

12. Mar 20, 2013

### haruspex

Yes, but be careful with signs. Which way are you measuring ω?
There's also energy conservation through the bounce, i.e. relating u, v and ω.
Next, if time t elapses between bounces, what equations can you write for the vertical height of (a) the centre of the rod and (b) the right-hand end of the rod at the second bounce?

13. Mar 20, 2013

### sergiokapone

The clockwise direction.

$\dfrac{mu^2}{2}=\dfrac{I_{CM}\omega^2}{2}+\dfrac{mv^2}{2}$

(a)
$\Delta h = vt-\frac{gt^2}{2}$, $\Delta h$ vertical height abter first bounce.

(b)
???

I always thought that the rod moves as shown. First, gliding over the surface of the left end so that the center of mass is in the same vertical. Then start upward motion, with $\vec v$. Am I right?

Or the center of mass of the rod stops at zeroі level, and then begins to rise up? I do not understand this point.

14. Mar 20, 2013

### haruspex

That's certainly the generic equation, but I defined t as the time to the second bounce. We know what Δh is at the time of second bounce, yes? (Remember, the angle of contact is the same as in the first bounce, just reflected.)
Yes, the centre of mass stays in the same vertical line. Asking for the height of the end may have put you on the wrong track. What I'm looking for is the relationship between t, ϕ and ω.

15. Mar 20, 2013

### sergiokapone

If we accept the hypothesis that the center of mass stops at zero level, then:
$\Delta h = v\frac{t}{2}-g\frac{t^2}{8}$, if t - is the time to the second bounce.
During this time t, rod rotated by an angle $\pi-2\phi$. Then $t=\frac{\pi-2\phi}{\omega}$

16. Mar 20, 2013

### haruspex

No, it doesn't stop at zero level; it must move up. But it does return to zero level at time of second bounce. Its height above the ground is l sin(ϕ) in both cases. Just put Δh = 0 in $\Delta h = vt-\frac{gt^2}{2}$
No it's even simpler than that. Think again.

17. Mar 20, 2013

### sergiokapone

Ok, your zero's level is in the ground.

$t=\frac{2v}{g}$

Now I get that $t=\frac{\pi+2\phi}{\omega}$ and no other.

18. Mar 20, 2013

### haruspex

It's rotating clockwise, so the angle reduces from ϕ to zero and continues on to -ϕ. What angle has it turned through?

19. Mar 20, 2013

### sergiokapone

I am being stupied!!!!! Of course $2\phi$

20. Mar 20, 2013

### sergiokapone

So, I have
$m(u-v)=\dfrac{I\omega}{lcos\phi}$
$\dfrac{mu^2}{2}=\dfrac{I\omega^2}{2}+\dfrac{mv^2}{2}$
$\omega=\dfrac{g\phi}{v}$
$I=\dfrac{ml^2}{12}$
$mgh=\dfrac{mu^2}{2}+mglcos\phi$ (zero of potential is in the groung)

21. Mar 20, 2013

### haruspex

I agree with all the above.
The length of the rod is 2l.
h is the distance it falls to first contact, and u is the speed acquired in doing so.
$mgh=\dfrac{mu^2}{2}$
So you now have the right number of equations for your unknowns. Go for it.

22. Mar 21, 2013

### sergiokapone

OK.
First eqn: $(u-v)=\dfrac{\omega l}{3cos\phi}$ (1)
Second eqn: $u^2-v^2=\dfrac{\omega^2 l^2}{3}$ (2)
Or
$(u-v)(u+v)=\dfrac{\omega^2 l^2}{3}$
$(u-v)(u+v)=(u-v)\omega l cos\phi$

$(u+v)=\omega l cos\phi$ (3)

Let's add (1) and (3) we get $2u=\omega l \dfrac{1+3cos^2\phi}{3cos\phi}$ (1')
Substract (3) and (1) we get $2v=\omega l \dfrac{3cos^2\phi-1}{3cos\phi}$ (2')
Eliminate a v from the equation (1') from (2')
Finally, we get
$h=\dfrac{\phi l}{12 cos\phi}\dfrac{\left(3cos^2\phi+1\right)^2}{3cos^2\phi-1}$

I think there it is a bug somewhere. The answer in the book of problems $h=\dfrac{\phi l}{24 cos\phi}\dfrac{\left(3cos^2\phi+1\right)^2}{1- 3cos^2\phi}$

Last edited: Mar 21, 2013
23. Mar 21, 2013

### TSny

I think there might be a sign problem in the above equation. Are you assuming $v$ is a positive or negative number?
I believe your factor of 12 in the denominator is correct and the factor of 24 in the book's answer is incorrect. The book's answer would be correct if the length of the rod is $l$ rather than $2l$.

24. Mar 21, 2013

### sergiokapone

Yes, I've simplified, and then forgot about it.

This value must be positive, but negative. Thank you. Now, with signs and with the answer all right.

25. Mar 21, 2013

### sergiokapone

Thank you haruspex! Thank you TSny. The problem is solved.