Freely hinged rods on a table - Linear velocities of CM after the impulse

[PiEpsilon]

Homework Statement:

Two uniform, equal stiff rods AB and AC are freely hinged at A and placed on a smooth horizontal table with AC perpendicular to AB, as shown in Fig. 16-2.
A horizontal blow is delivered perpendicular to AC at C. Find the ratio of the resulting linear velocities of the centers of mass of the rods Vac/Vab, immediately after this impulse.

Relevant Equations:

None

We know that impulse is
$$\vec J = \vec F \Delta t = \Delta \vec p$$
Let $l, m$ be the length of single rod and its mass respectively.

Analyzing torques and forces on each rod separately we have:
Rod $AC$:
$$F\Delta t +N_x\Delta t = mV_{ac,x} \space\space\text{ eq. }(1)$$
$$F\Delta t\cdot \frac l2 - N_x\Delta t\cdot \frac l2 = \frac{ml^2 \omega_{ac}}{12}\implies F\Delta t - N_x\Delta t = \frac{ml\omega_{ac}}{6}\space\space\text{ eq. }(2)$$
Rod $AB$:
$$ -N_x \Delta t = mV_{ab,x} \space\space\text{ eq. } (3)$$
$$\text{where } N_x \text{ is the force acting from the hinge connection}$$


Adding $(1)$ to $-1*(2)$, we have the following relation:
$$2N_x \Delta t = mV_{ac,x} - \frac{ml\omega_{ac}}{6} \implies 2V_{ab,x} + V_{ac,x} = \frac{ml\omega_{ac}}{6} \space\space\text{ eq. }(4)$$

Analyzing the situation from the CM view left me with no meaningful result.
Work energy consideration seemed also to lead nowhere.

Next logical step would be to represent right side of equation $(4)$ as some other relation of velocities of $V_{ab,x}$ and $V_{ac,x}$ which I am unable to do.


I have left vertical direction, as vertical velocities of rods must be equal in magnitude but opposite in direction.*

*when the system of rods is considered, the force acting is horizontal.
Since $\vec F = 2m\vec A_{cm} \implies \vec V_{cm}$ is horizontal and:
$2m\vec V_{cm} = m\vec V_{ac} + m\vec V_{ab} \implies V_{cm,y} = 0 \iff V_{ac,y} = -V_{ab,y}$

#
The horizontal positive $x$ direction is taken as pointing to the right (from B to A)
The vertical positive $y$ direction is taken as pointing up (from C to A)

 

[TSny]

Is there a kinematical relationship that relates $V_{ac,x}, V_{ab,x}$, and $\omega_{ac}$?

 

[PiEpsilon]

In general, we can relate linear velocity with angular velocity by the following relationship:
$$\vec v = \vec \omega \times \ \vec r$$
From there we could get linear velocity at the end of the rod , but it doesn't seem to be much of use.

 

I'd like to suggest another approach for you to think about (although you should definitely try to finish your method too). The blow delivers a horizontal impulse $I$ to the system; what is then the velocity of the centre of mass of the entire system right after the blow?

Transform into the centre of mass frame of the entire system (also, where exactly is the centre of mass?); if in this frame the CoM of rod AC moves at $\mathbf{v}$, then the CoM of AB must move at $-\mathbf{v}$ (why?). What was the angular impulse of the blow about the centre of mass of the entire system, and what is the final angular momentum about the centre of mass of the entire system; how are these related?

After you found $\mathbf{v}$, you can add back the velocity of the centre of mass frame to deduce the initial velocities of the CoMs of both rods.

 

[PiEpsilon]

The CM of the figure, before the blow, is located at $\vec R_{cm} = -\frac l4 \hat i - \frac l4 \hat j$ , measured from the hinge A.
We know that after the impulse the figure CM moves with some horizontal velocity $\vec V_{cm}$ and rotates around it with some angular velocity $\omega_{cm}\hat k$

In frame of reference centered at the at CM moving with velocity $\vec V_{cm}:$
$$\vec F \Delta t = \Delta P_{cm} = 2m\vec V_{cm}$$
$$\vec \tau \Delta t = F \Delta t \cdot \frac{3l}{4} \hat k=\Delta \vec L = I_{cm}\vec\omega_{cm}$$
$$\vec v'_{ab} + \vec v'_{ac} = 0 $$
From these we can get:
$$\frac32 ml V_{cm} = I_{cm}\omega_{cm}$$

I am not sure I can see the solution from here.

 

Careful with the final angular momentum $\Delta \mathbf{L}$; since the rods are freely hinged they don't form a rigid body, and we can't define an angular velocity vector for the configuration as a whole. So $I_{\text{cm}} \boldsymbol{\omega}_{\text{cm}}$ is not right.

Try and think about what @TSny said in post #2, in order to relate the angular velocity of $AC$ to the velocities of the CoMs of both rods. [Hint: find two expressions for the velocity of the hinge, and equate them!]

 

[TSny]

In general, we can relate linear velocity with angular velocity by the following relationship:
$$\vec v = \vec \omega \times \ \vec r$$
From there we could get linear velocity at the end of the rod , but it doesn't seem to be much of use.

So far, in your first post, you have 3 equations involving 4 unknowns: $N_x$, $V_{ab,x}$, $V_{ac,x}$, and $\omega_{ac}$. A fourth equation is obtainable by considering the kinematic constraint imposed by the fact that the rods are connected. How does the velocity of the right end of rod AB compare to the velocity of the top end of rod AC? Just after the collision can you express the x-component of the velocity of the right end of rod AB in terms of $V_{ab,x}$? Can you express the x-component of the velocity of the top end of rod AC in terms of $V_{ac, x}$ and $\omega_{ac}$?

For the y-components, you have deduced that $V_{ac, y} = -V_{ab, y}$. By considering a possible vertical force $N_y$ acting on rod AC (and corresponding reaction force on rod AB), you should be able to go further and deduce the individual values for $V_{ac, y}$ and $V_{ab, y}$.

 

[PiEpsilon]

Since the rods are hinged, it must be that the velocity of right end of rod AB is equal to velocity of top end of rod AC.
The only horizontal force acting on rod AB is $-N_{x}$, therefore it's horizontal velocity is given by eq. (3)

The x-component of velocity of rod top AC is $V_{ac,x} - \omega_{ac}\cdot \frac l2$
Moreover, considering vertical direction we have:
$$-N_y\Delta t \cdot \frac l2 = \frac {ml^2\omega_{ab}}{12}\implies \frac{ml\omega_{ab}}{6}= -N_y \Delta t$$
and
$$-N_y\Delta t = - mV_{ac,y} = mV_{ab,y}$$

Considering the velocities at the hinge:
$$ V_{ab,x} = V_{ac,x} - \omega_{ac}\cdot \frac l2 $$
and
$$ V_{ac,y} = V_{ab,y} - \omega_{ab}\cdot\frac l2 $$

Therefore we have:
$$l\omega_{ac} = 2V_{ac,x} - 2V_{ab,x}$$
$$l\omega_{ac} = 6V_{ac,x} + 12V_{ab,x}$$

and

$$l\omega_{ab} = 6V_{ab,y}$$
$$l\omega_{ab} = 2V_{ab,y} - 2V_{ac,y}$$
So $4V_{ab,y} = - 2V_{ac,y}$ and $V_{ab,y} = - V_{ac,y} \iff V_{ab,y} = V_{ac,y} = 0$

Finally the ratio is:
$$V_{ac}/V_{ab}=\frac{-7}2$$

Thank you!

Considering all those rotations my head itself was spinning and I did not consider the crucial kinematics restriction.
I am still not certain how to approach this problem in CM frame. The angular momentum before the impulse is 0. After it we must consider $\vec L = \Sigma m_i\vec r_i\times\vec v_i$ measured from CM.

 

[PiEpsilon]

Here is my reasoning regarding calculations from CM frame, although it seems I am making some mistake.

Angular momentum in CM frame can be described as rotation of two rods of equal moment of inertia, but different angular velocities.
By parallel axis theorem moment of inertia of the rod about CM is equal to $I_{rod\space center} + mD^2$ where $D^2 = 2\cdot ( \frac l4)^2$
I.e.
$$\Delta L = (ml^2/12 + ml^2/8)(\omega_{ac} + \omega_{ab})$$
From previous result:
$$\frac32mlV_{cm} = \Delta L$$
Or
$$\frac54(\omega_{ac} + \omega_{ab})\cdot l = V_{cm}$$
Velocity of the rod at the hinge is given by $D\cdot \omega_{ac}$ , which by constraint, must be equal to $D\cdot \omega_{ab}$
Therefore:
$\omega_{ab} = \omega_{ac}$

And:
$\frac52\omega_{ac}\cdot l = V_{cm}$

The AC rod CM velocity $v'_{ac}$ is equal to $l\omega_{ac}/\sqrt8$
So $v'_{ac} = \frac5{\sqrt2}V_{cm}= -v'_{ab}$

Finally,
$v_{ac} = V_{cm} + v'_{ac}$
And
$v_{ab} = V_{cm} - v'_{ac}$

which gives invalid answer

 

[TSny]

Your work in post #8 looks good to me.

Finally the ratio is:
$$V_{ac}/V_{ab}=\frac{-7}2$$

It's a little odd for the question to ask for the ratio of two velocities since velocities are vector quantities. $-\frac{7}2$ is the ratio of the x-components of the velocities. (Maybe I'm being too picky here.)

 

[PiEpsilon]

Your work in post #8 looks good to me.


It's a little odd for the question to ask for the ratio of two velocities since velocities are vector quantities. $-\frac{7}2$ is the ratio of the x-components of the velocities. (Maybe I'm being too picky here.)

I completely agree! Thanks again for your help.

 

I'm going to be a bit busy the coming days so won't be around to give tutorial-style help. Given you've already solved it, I'll just sketch my solution

Argue as before that the velocities of the CoMs of the rods have no vertical components. Denoting the velocities of the CoMs of the rods in the centre of mass frame $\mathbf{v}_{ac} := \mathbf{v}$ and $\mathbf{v}_{ab} = -\mathbf{v}$ (and writing $v := |\mathbf{v}|$) and denoting the angular velocity of $AC$ by $\boldsymbol{\omega}$, then the total angular momentum about the centre of mass is$$L_z = \left[ \frac{mlv}{4} \right] + \left[ \frac{mlv}{4} + \frac{ml^2 \omega}{12} \right] = \frac{ml}{2} \left( v + \frac{l\omega}{6} \right)$$which must equal the angular impulse $J = (3l/4)I$. The kinematic constraint is $-v = v - \frac{l \omega}{2} \implies \omega = \frac{4v}{l}$ from which it follows that\begin{align*}

\frac{3l}{4}I = \frac{ml}{2} \left( v + \frac{l\omega}{6} \right) \, \implies \, \frac{3I}{2m} = v + \frac{l\omega}{6} = \frac{5v}{3} \, \implies \, v = \frac{9I}{10m}


\end{align*}adding back the centre of mass velocity $I/2m$ to transform back to the velocities in the lab frame tells you\begin{align*}
V_{ac} = \frac{I}{2m} + \frac{9I}{10m} = \frac{7I}{5m} \\

V_{ab} = \frac{I}{2m} - \frac{9I}{10m} = \frac{-2I}{5m}
\end{align*}