Analyzing an Angular Impulse Problem

[PiEpsilon]

Homework Statement

A pair of $10cm$ radius, $2kg$ disc flywheels with frictionless bearings are spinning at $1000 rads\space s^-1$ and are supported at a distance $d = 15cm$ on either side of a universal bearing by a small diameter bar $AB$ of mass $M = 1kg$ and length $4d$, as shown in Fig.16-21.

If a ball of mass $m = 10g$ is dropped from a height $h = 5cm$ onto the tip $A$ of the bar and rebounds upwards, give the components of resulting angular momentum $\vec L$ of the flywheels and sketch the motion of the tip of the bar as seen from $+x$ direction. Also, give the angular velocity $\Omega_n$ of this motion and radius $r$ of the circle described by the tip once the circular motion is attained.

Relevant Equations

-

What we know:
The ball is dropped at the tip A with some speed $v_0$ and rebounds with speed $v$. This collision produces an angular impulse, changing the angular momentum of the bar with the flywheels.

Solution inspired by an answer provided by @TSny in the similar question.

Angular impulse is:
$m(v + v_0)\cdot 2d = L_y$
By conservation of energy:
$\frac 12m v_0^2 + \frac12 I_x\omega_0^2 = \frac12 mv^2 + \frac12 I_x\omega_0^2 + \frac12 I_y\omega_y^2 \implies m(v_0^2-v^2) = L_y^2/I_y$

Therefore:
$4m^2d^2(v+v_0)^2 = I_ym(v_0^2-v^2)\implies 4md^2(v+v_0) = I_y(v_o-v)$

Solving for $v$ yields:
$$v = \biggr(\frac{I_y-4md^2}{I_y+4md^2}\biggl)v_0\approx(1-8md^2/I_y)\cdot v_0$$

(approximation ignores higher order terms)
​

$I_y = 2\cdot (\frac14 M_{0}R_{0}^2 + M_{0}d^2) + \frac1{12}M(4d)^2 = (\frac43 M + 2M_{0})d^2 + \frac12 M_{0}R_{0}^2 = 0.13 [kg\cdot m^2]$
where $M_0$ and $R_0$ is the mass and radius of the disc respectively.

By the work energy theorem:
$mgh = \frac12 mv_0^2\implies v_0=\sqrt{2gh}\approx 0.99\frac{m}{s}$

Therefore:
$L_y = 2d\cdot mv_0(\frac{I_y-4md^2}{I_y+4md^2}+1) = 2* 0.15 *0.01*\sqrt{2*9.81*0.05}\biggl(\frac{0.13 - 4*0.01*0.15^2}{0.13 + 4*0.01*0.15^2}+1)\biggr)\approx 0.0059 Js^{-1}$

$L_x = I_x\omega_0 = 2\cdot (\frac12M_0R_0^2)\omega_0 = 20Js^{-1}$

And

$\vec L = (20\hat e_x + 0.0059\hat e_y)Js^{-1}$

Problems with the exercise:

The answer of $L_y$ differs by a factor of 2 (should be twice as big).

I assume that $\Omega_n$ is the angular velocity (pointing in + z-direction) as the bar rotates and aligns with angular momentum vector $\vec L$.
The angle swept is: $\lvert\Delta\vec L\rvert/\lvert\vec L\rvert=\lvert L_y\rvert/\lvert\vec L\rvert=\sin\theta$
However in what time does it happen?
I've been trying to use the nutation equation $\vec\tau = \vec\Omega \times \vec L$ too, but I was unable to come up with anything clever.

I am also unable to visualize the tip describing a circular motion. The tip should dip a little bit to conserve angular momentum in the z-direction and rotate about that axis to align itself with the final angular momentum vector $\vec L$. Considering nutation quickly lead to the mess, that seemed like an overkill for this type of a problem.

 

[TSny]

Your work looks correct to me. I don't see how they could get an answer for $L_y$ that is twice your value.

Once the collision is over, the body rotates without any applied torque. If you've studied Euler's equations for rotating bodies, then I think you can use them to help answer the question about the circular motion of point $A$. I'm not sure why the statement of the problem includes the phrase "once the circular motion is attained". It seems to me that the circular motion of point $A$ begins immediately after the collision.

Did they provide numerical answers for $\Omega_n$ and $r$?

 

[PiEpsilon]

Your work looks correct to me. I don't see how they could get an answer for $L_y$ that is twice your value.

Once the collision is over, the body rotates without any applied torque. If you've studied Euler's equations for rotating bodies, then I think you can use them to help answer the question about the circular motion of point $A$. I'm not sure why the statement of the problem includes the phrase "once the circular motion is attained". It seems to me that the circular motion of point $A$ begins immediately after the collision.

Did they provide numerical answers for $\Omega_n$ and $r$?

Oh right! The Euler's equations should do it.

The answers for these quantities are:
$\Omega_n = 154s^{-1}$
and
$r = 1.77 \times 10^{-2}cm$

I am not sure why angular velocity has no angle measurement units.

 

[TSny]

Oh right! The Euler's equations should do it.

The answers for these quantities are:
$\Omega_n = 154s^{-1}$
and
$r = 1.77 \times 10^{-2}cm$

I am not sure why angular velocity has no angle measurement units.

I get their value for $\Omega_n$. I get half their value for $r$ due to our $L_y$ being half of theirs.

 

[PiEpsilon]

From the Euler equations:
$I_x\dot\omega_x = (I_y - I_z)\omega_y\omega_z$
$I_y\dot\omega_y = (I_z - I_x)\omega_z\omega_x$
$I_z\dot\omega_z = (I_x - I_y)\omega_x\omega_y$

Since $I_y = I_z$, we have:
$\omega_x = const.$
$\dot\omega_y = (1 - I_x/I_y)\omega_z\omega_x$
$\dot\omega_z = (I_x/I_y - 1)\omega_x\omega_y$

Let $(1-I_x/I_y)\equiv A = const.$

Then:
$\omega_x = const.$
$\dot\omega_y = A\omega_z\omega_x \implies \ddot \omega_y = A\omega_x\dot \omega_z$
$\dot\omega_z = -A\omega_x\omega_y \implies \ddot \omega_z = -A\omega_x\dot \omega_y$

Therefore:
$\ddot \omega_z +A^2\omega_x^2\omega_z = 0$
$\ddot \omega_y + A^2\omega_x^2\omega_y = 0$

The solution to this second order differential equation:

The characteristic equation has the roots:
$r = \pm i A\omega_x$

Let $A\omega_x \equiv B$
Complex solution leads to general equations:
$\omega_y(t) = c_1\cos(Bt) + c_2\sin(Bt)$
$\omega_z(t) = c_3\cos(Bt) + c_4\sin(Bt)$

For t = 0
$c_1 = \omega_{y0}$
$c_3 = \omega_{z0}$

Taking time derivative:
$\dot\omega_y(t) = -B\omega_{y0}\sin(Bt) + Bc_2\cos(Bt)$
$\dot\omega_z(t) = -B\omega_{z0}\sin(Bt) + Bc_4\cos(Bt)$

For t = 0
$c_2 = \dot\omega_{yo}/B$
$c_4 = \dot\omega_{zo}/B$

Therefore the equations of motion are:
$\omega_y(t) = \omega_{y0}\cos(Bt) + (\dot\omega_{yo}/B)\sin(Bt)$
$\omega_z(t) = \omega_{z0}\cos(Bt) + (\dot\omega_{zo}/B)\sin(Bt)$

where $B = (1-I_x/I_y)\omega_x$

The question is how do I get the values for {$\omega_{yo},\omega_{zo},\dot\omega_{yo},\dot\omega_{zo}$}?

 

[TSny]

From the nature of the collision at $t = 0$, what is the value of $\omega_z(0)$?

Form Euler's equations, what can you conclude about $\dot \omega_y(0)$?

 

[PiEpsilon]

Let's see...
Initially the figure rotates about the x-axis. During the collision torque acts in the x-y plane.
Angular momentum about the z-axis is then zero at all times. However the figure clearly rotates about that axis as it aligns itself with angular momentum vector $\vec L$
At time $t = 0$, i.e. immediately after the collision, it is however reasonable to assume that $\omega_z(0)=0$

If above holds then:

$\omega_{z0} = 0$
and
$\dot\omega_y(0) = 0$ (because $\dot\omega_y(t) = (1-I_x/I_y)\omega_x\omega_z(t)$)

So the equations become:

$\omega_y(t) = \omega_{y0}\cos(Bt)$
$\omega_z(t) = (\dot\omega_{z0}/B)\sin(Bt)$

I cannot get my head around (!) the two remaining angular constants.

 

[TSny]

Let's see...
Initially the figure rotates about the x-axis. During the collision torque acts in the x-y plane.
Angular momentum about the z-axis is then zero at all times.

Yes, we can take the torque to be in the $y$ direction.

However the figure clearly rotates about that axis as it aligns itself with angular momentum vector $\vec L$

I'm not sure what you mean by: "...as it aligns itself with angular momentum vector $\vec L$".

At time $t = 0$, i.e. immediately after the collision, it is however reasonable to assume that $\omega_z(0)=0$

Yes

$\omega_y(t) = \omega_{y0}\cos(Bt)$
$\omega_z(t) = (\dot\omega_{z0}/B)\sin(Bt)$

I cannot get my head around (!) the two remaining angular constants.

Consider the Euler equation for $\dot \omega_z$. This should allow you to relate $\dot \omega_{z0}$ to $\omega_{y0}$.

In your first post, you found values for $L_{y0}$ and $I_y$. So, you can find $\omega_{y0}$.

 

[PiEpsilon]

From the Euler's equations:
$\dot\omega_{z0} = -A\omega_x\omega_{y0}$

(I will keep denoting $\omega_{y0}\equiv\omega_y (0)$ although I should have left it unchanged for ease of reading)

$\omega_{y0} = L_y/I_y = 0.0059/0.13 Js^{-1}kg^{-1}m^{-2}\approx 0.0454s^{-1}$Therefore
$\dot\omega_{z0} = -A\omega_x L_y/I_y$

So the specific solutions to these differential equations are:
$$\omega_y(t) = (L_y/I_y)\cos(Bt)=\omega_{y0}\cos(Bt)$$

and​

$$\omega_z(t) = -\frac{(A\omega_x L_y)}{(BI_y)}\sin(Bt)=-\omega_{y0} \sin(Bt) $$with $(1-I_x/I_y) = (1-\frac{0.02}{0.13})\approx 0.846$
we have
$A \approx 0.846$
$B \approx 846 \frac{rad}{s}$

and:
$$\omega_y(t) = 0.0454\cos(846t) [_{rad}/_{s}]$$

and​

$$\omega_z(t) = -0.0454\sin(846t) [_{rad}/_{s}] $$

What really do the equations tell me?
As seen from rotating body frame of reference (due to the nature of Euler's equations), the components of angular velocity $\omega_y(t)$ and $\omega_z(t)$ traces the circle about the x-axis in the counterclockwise direction (as seen from origin towards + x-axis).

From the geometry and because the bar tip aligns with angular momentum*, we conclude that:
$(\sqrt{L_y^2+L_z^2})/L_x = r/2d\implies r = 2*15cm*\sqrt{0.13^2*0.0454^2}/20 \approx 0.885\times10^{-2}cm$

$\biggl(\sqrt{L_y^2+L_z^2})=\sqrt{I_y^2\omega_y^2+I_z^2\omega_z^2}=\sqrt{I_y^2(\omega_y^2+\omega_z^2)}\biggr)$​

Angular velocity of this motion is $\Omega_n = B = 846[_{rad}/_{s}]$

which clearly disagrees with the answer given.

* I assume the rod tip aligns with angular momentum (i.e. is parallel to it), yet I am unable to prove it beyond reasonable doubt (inertia tensor is diagonal, so angular velocity components are parallel to angular momentum components?)

 

[TSny]

Your results so far look good. Note that $\Omega_n$ is the angular velocity that the vectors $\vec L$ and $\vec \omega$ precess about the x-axis of the body frame as measured in the body frame. So, we don't expect this to equal the angular velocity of point $A$ in the lab frame. You can see from your results that the angle between $\vec L$ and the rod remains constant. So, the rod will not try to align with the angular momentum. Also, you can easily show that $\vec L$, $\vec \omega$, and the rod always lie in the same plane.

The tricky part is going from these results in the body frame to visualizing what's happening in the lab frame. In the lab frame $\vec L$ is a fixed vector (conservation of angular momentum). In the lab frame, the x-axis of the body (i.e, the rod) and the instantaneous angular velocity vector $\vec \omega$ precess about the fixed vector $\vec L$ such that $\vec L$, $\vec \omega$, and the rod always lie in the same plane. One way to visualize it is to use the concepts of the "body cone" and the "space cone" and to imagine the body cone as rolling without slipping on the space cone. But I don't know if you have studied this. Doing a quick web search I found https://www.researchgate.net/publication/231043528_Inertial_rotation_of_a_rigid_body which you can download. I have not read it, but glancing at it, it looks pretty good. Section 5 of this paper states the results that are relevant to your problem.

 

[PiEpsilon]

Summing everything up:

Initially the bar with the flywheels rotates about the x-axis. Initial angular velocity and angular momentum are collinear and point along that axis too.

After the ball is dropped it produces an angular impulse that changes the angular momentum.
Reasoning as in the first post, we conclude that the final angular momentum vector $\vec L$ is in the x-y plane.
Since no torques are acting on a system, conservation of angular momentum implies that this vector does not change both in length and direction in the inertial frame of reference (i.e. is 'fixed' in space).

Moreover we are interested what exactly happens after the impact (what type of motion is produced after this incident).
One of the ways to analyze this situation is solving Euler's equations. Doing so, as seen in further posts, we get an answer for the motion of the rod. However due to the nature of the Euler's equations, the result is valid as seen from the reference frame 'attached' to the rotating body. In this frame the angular velocity vector precesses about the x'-axis with angular frequency $B$
As measured in the body frame (all primed quantities are measured w.r.t. body frame):
$$\vec \omega'=
\begin{pmatrix}
\omega_x'\\
\omega_{y0'}\cos(Bt)\\
-\omega_{y0'}\sin(Bt)\\
\end{pmatrix}
$$
and
$$\vec L' =
\begin{pmatrix}
I_x\omega_x'\\
I_y\omega_{y0'}\cos(Bt)\\
-I_y\omega_{y0'}\sin(Bt)\\
\end{pmatrix}
$$Notice that $\vec L' = I_x\omega_x'\hat {i'} + I_y\omega_y'\hat {j'} + I_z\omega_z'\hat {k'} = I_x\omega_x'\hat {i'} + I_y(\omega_y'\hat {j'} + \omega_z'\hat {k'}) = (I_x - I_y)\omega_x'\hat {i'} + I_y\vec \omega'$
I.e. the angular momentum vector lies in the plane spanned by the x'-axis and angular velocity vector, so the vector must rotate as does the angular velocity vector

Moreover, the tip of the rod $A$ lies along $\hat{i'}$ at all times. Since $\hat{i'}\cdot \vec L' = const.$ and $\hat{i'}\cdot \vec \omega' = const.$, the angle between these vectors and the rod's end does not change.

Summarizing, in the body frame the x'-axis is fixed in space, while the angular velocity $\vec\omega'$ and angular momentum $\vec L'$ are rotating about that axis with the same angular frequency $B$ with the constant angle between them. At any time, all three (the x'-axis and vectors) lie on the same plane.
Also at $t=0$, i.e. shortly after the impulse, it must be that all primed quantities are equal to the unprimed ones, because the axis in both frames coincides at that time. (so $\vec L = \vec L', \vec\omega= \vec\omega'$, etc. for $t=0$)

We would like now to see, what happens in an inertial frame of reference.
Knowing that angular momentum $\vec L$ is fixed in space in that frame, we conclude that x'-axis and $\vec\omega$ must rotate around it.

Note that $\vec\omega' = \vec L'/I_y + B\hat{i'}$
or for $t=0,\space \vec\omega = \vec L/I_y + B\hat{i'}$

We know that relation between inertial frame and rotational frame is:

$\biggl(\frac{d}{dt}\biggr)_{lab} = \biggl(\biggl(\frac{d}{dt}\biggr)_{rot.} + \vec\omega \times\biggr)$​

Therefore (for t = 0):

$$\biggl(\frac{d\hat{i'}}{dt}\biggr)_{lab} = \biggl(\biggl(\frac{d\hat{i'}}{dt}\biggr)_{rot.} + \vec\omega \times \hat {i'}\biggr)=\biggl(\vec L/I_y + B\hat{i'}\biggr)\times \hat{i'}$$
And
$$\frac{d\hat{i'}}{dt} = (L/I_y) \hat L \times \hat {i'} $$

which is the same form as
$\vec v = \vec\Omega \times \vec r$
and tells us that the rods end $A$ rotates around $\vec L$ with the angular frequency $\Omega_n = L/I_y \approx 153.85 s^{-1}$

At the same time (t=0), the rod coincides with the x-axis and rotates around $\vec L$. So for small angles:
$L_y/L_x \approx r/2d \implies r \approx 0.885\times10^{-2}cm$

 

[TSny]

That all looks good to me.

 

[PiEpsilon]

Thank you for checking this! I appreciate your help.