How High Must a Rod Fall to Hit a Table Elastically at Both Ends?

[sergiokapone]

Homework Statement

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.

Homework 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)

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.

 

[haruspex]

Energy conservation law mgh=I_{CM}\dfrac{\omega^2}{2} (1)

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.

 

[sergiokapone]

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.

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

 

[sergiokapone]

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.

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

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.

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.

 

[haruspex]

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

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

 

[haruspex]

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

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.

 

[sergiokapone]

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

I agree with you, then I made a mistake.

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.

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.

 

[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.

 

[sergiokapone]

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.)

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?

 

[haruspex]

m\vec{u}=m\vec{v}+\vec{J}

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.

 

[sergiokapone]

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.

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?

 

[haruspex]

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?

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?

 

[sergiokapone]

Yes, but be careful with signs. Which way are you measuring ω?

The clockwise direction.

There's also energy conservation through the bounce, i.e. relating u, v and ω.

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

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?

(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.

 

[haruspex]

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

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.)

(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?

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.

 

[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}

 

[haruspex]

If we accept the hypothesis that the center of mass stops at zero level, then:

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 $\Delta h = vt-\frac{gt^2}{2}$

During this time t, rod rotated by an angle \pi-2\phi.

No it's even simpler than that. Think again.

 

[sergiokapone]

above the ground

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

just put Δh = 0 in ...

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

No it's even simpler than that. Think again.

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

 

[haruspex]

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

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

 

[sergiokapone]

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

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

 

[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)

 

[haruspex]

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 agree with all the above.

$I=\dfrac{ml^2}{12}$

The length of the rod is 2l.

$mgh=\dfrac{mu^2}{2}+mglcos\phi$ (zero of potential is in the groung)

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.

 

[sergiokapone]

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.

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 $h=\dfrac{\phi l}{24 cos\phi}\dfrac{\left(3cos^2\phi+1\right)^2}{1- 3cos^2\phi}$

 

[TSny]

First eqn: $(u-v)=\dfrac{\omega l}{3cos\phi}$ (1)

I think there might be a sign problem in the above equation. Are you assuming $v$ is a positive or negative number?

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

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 $2l$.

 

[sergiokapone]

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 $2l$.

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

I think there might be a sign problem in the above equation. Are you assuming $v$ is a positive or negative number?

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

 

[sergiokapone]

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