How do you calculate the motion of a ball rolling on a rotating table?

[Chestermiller]

We can move on now. Everything is cleared till that far.

We next focus on the dynamics. This starts with the free body diagram for member AB:

View attachment 1588073847854.png
The above FBD shows the forces acting on strut AB. The tension F/4 at pin A is the portion of the imposted load F/2 that acts on strut AB. The tension R at pin A represents the contact force that strut DA imposes at A to keep vertex A from moving in the y direction. The tension T at pin B represents the contact force that strut CB imposes at B to pull vertex B back.

Next, we write the Newton's 2nd law force- and moment balances on strut AB. The force balances on AB in the x- and y-directions, respectively (based on the free body diagram) are:
$$\frac{F}{4}-T=ma_{cmx}=m\frac{a_{Ax}}{2}\tag{1}$$
$$-R=mA_{cmy}=-m\frac{a_{Ax}}{2}\tag{2}$$where use has been made here of some of the kinematic relations in post #29 to express the acceleration components of the center of mass of strut AB in terms of the x-direction acceleration of vertex A relative to the center of mass of the overall structure.

A clockwise moment balance on strut AB about its center of mass gives:
$$R\frac{b}{2\sqrt{2}}-\frac{F}{4}\frac{b}{2\sqrt{2}}-T\frac{b}{2\sqrt{2}}=I\frac{d^2 \theta}{dt^2}\tag{3}$$where I is the moment of inertial of the strut about its center of mass: $$I=m\frac{b^2}{12}\tag{4}$$If we combine Eqns. 3 and 4, we obtain:
$$R-\frac{F}{4}-T=\frac{m}{3}\frac{b}{\sqrt{2}}\frac{d^2 \theta}{dt^2}\tag{5}$$But, from post #29, $$\frac{b}{\sqrt{2}}\frac{d^2 \theta}{dt^2}=-a_{Ax}$$Therefore, $$\frac{F}{4}+T-R=\frac{m}{3}a_{Ax}\tag{6}$$
If we now combine Eqns. 1,2, and 6 to eliminate T and R, we obtain: $$a_{Ax}=\frac{3}{8}\frac{F}{m}$$Based on this and the results of Part 1 loading, we find that the absolute acceleration of vertex A is $\frac{F}{4m}+\frac{3}{8}\frac{F}{m}=\frac{5}{8}\frac{F}{m}$ and that the absolute acceleration of vertex C is $\frac{F}{4m}-\frac{3}{8}\frac{F}{m}=-\frac{1}{8}\frac{F}{m}$

 

[etotheipi]

@Chestermiller I had a question about your method. In the second part of your analysis you are using the fact that the behaviour of the system in the centre of mass frame is equivalent to having two opposite forces of F/2 on either side of the truss structure, causing it to pantograph.

I then recognized that the loading on the structure could be resolved into two superimposable parts:

1. F/2 at A in the positive x-direction and F/2 at C in the positive x-direction
2. F/2 at A in the positive x-direction and F/2 at C in the negative x-direction

I was wondering why it is possible to decompose the loading in this way? Superimposing all of those forces in their specified position indeed results in the same loading as in the problem statement, but why is it valid to only consider two of those forces at a time?

Essentially, why is the loading in the centre of mass frame only F/2 on each side? Since I'd presumed that in the accelerating centre of mass frame a fictitious F would act in the negative $x$ direction and the real F would act in the positive $x$ direction. Thanks!

 

[Adesh]

Thank you @Chestermiller

 

[wrobel]

I had a question about your method. In the second part of your analysis you are using the fact that the behaviour of the system in the centre of mass frame is equivalent to having two opposite forces of F/2 on either side of the truss structure, causing it to pantograph.

Yes indeed, I also have the same question. (Actually I have a lot of questions) Why is it F/2 that was applied not F/10 for example?

 

[Chestermiller]

@Chestermiller I had a question about your method. In the second part of your analysis you are using the fact that the behaviour of the system in the centre of mass frame is equivalent to having two opposite forces of F/2 on either side of the truss structure, causing it to pantograph.

I was wondering why it is possible to decompose the loading in this way? Superimposing all of those forces in their specified position indeed results in the same loading as in the problem statement, but why is it valid to only consider two of those forces at a time?

Essentially, why is the loading in the centre of mass frame only F/2 on each side? Since I'd presumed that in the accelerating centre of mass frame a fictitious F would act in the negative $x$ direction and the real F would act in the positive $x$ direction. Thanks!

The loading from Part 1 exactly cancels the pseudo body force resulting from the frame of reference acceleration. The Part 1 loading pushes from behind with a force of F/2 at C and pulls from the front with a force of F/2 at A. The reason I chose to resolve the superposition in the way that I did was because the Part 1 loading simply results in a rigid body translation of the truss and the Part 2 loading would simply result in pantographing.

 

[etotheipi]

The loading from Part 1 exactly cancels the pseudo body force resulting from the frame of reference acceleration. The Part 1 loading pushes from behind with a force of F/2 at C and pulls from the front with a force of F/2 at A. The reason I chose to resolve the superposition in the way that I did was because the Part 1 loading simply results in a rigid body translation of the truss and the Part 2 loading would simply result in pantographing.

I guess I'm just slightly confused about this method of superposition. I have never come across such an approach before. What allows us to break it down into a part 1 and a part 2 in the first place?

To my mind, we could either analyse this in the lab frame in which there is a sole force F acting at C, and zero external force at A. Or, we could transform into the centre of mass frame in which we still have a force F acting at C, in addition to a fictitious body force which essentially reduces to four forces of magnitude F/4 acting through the centres of mass of each of the rods, in the negative $x$ direction. In either of those frames, you could do force and torque analysis assuming you also draw the internal constraint forces between each of the rods on your free body diagram.

So then I'm not sure why it's permitted to first add on extra external forces at C, and secondly to consider two different scenarios in each of which only half of the forces act on the truss structure. Because neither part 1 nor part 2 represent the state of the system.

Thanks for your patience!

 

[Chestermiller]

I guess I'm just slightly confused about this method of superposition. I have never come across such an approach before. What allows us to break it down into a part 1 and a part 2 in the first place?

In my judgment, the equations for the two parts would be linearly superimposable to the overall equations.

To my mind, we could either analyse this in the lab frame in which there is a sole force F acting at C, and zero external force at A. Or, we could transform into the centre of mass frame in which we still have a force F acting at C, in addition to a fictitious body force which essentially reduces to four forces of magnitude F/4 acting through the centres of mass of each of the rods, in the negative $x$ direction. In either of those frames, you could do force and torque analysis assuming you also draw the internal constraint forces between each of the rods on your free body diagram.

So then I'm not sure why it's permitted to first add on extra external forces at C, and secondly to consider two different scenarios in each of which only half of the forces act on the truss structure. Because neither part 1 nor part 2 represent the state of the system.

Thanks for your patience!

Sorry, I don't think I can explain it any better than I did. However, I find it hard to believe that the analysis I presented could come up with exactly the correct answer if there were a fundamental flaw somewhere.

 

[etotheipi]

In my judgment, the equations for the two parts would be linearly superimposable to the overall equations.

This sort of makes sense to me though I don't really have a concrete picture of why it works. It appears to be a clever algebraic trick. It must be a valid method, though.

Just out of interest, do you know of any books or online resources which explain the general idea behind the superposition approach you took? It's an interesting idea.

 

[Chestermiller]

This sort of makes sense to me though I don't really have a concrete picture of why it works. It appears to be a clever algebraic trick. It must be a valid method, though.

Just out of interest, do you know of any books or online resources which explain the general idea behind the superposition approach you took? It's an interesting idea.

No. It's just something that popped into my head for this problem. I spent a lot of time thinking about this problem over several days before I arrived at this.

 

[Chestermiller]

I guess I'm just slightly confused about this method of superposition. I have never come across such an approach before. What allows us to break it down into a part 1 and a part 2 in the first place?

To my mind, we could either analyse this in the lab frame in which there is a sole force F acting at C, and zero external force at A. Or, we could transform into the centre of mass frame in which we still have a force F acting at C, in addition to a fictitious body force which essentially reduces to four forces of magnitude F/4 acting through the centres of mass of each of the rods, in the negative $x$ direction. In either of those frames, you could do force and torque analysis assuming you also draw the internal constraint forces between each of the rods on your free body diagram.

So then I'm not sure why it's permitted to first add on extra external forces at C, and secondly to consider two different scenarios in each of which only half of the forces act on the truss structure. Because neither part 1 nor part 2 represent the state of the system.

Thanks for your patience!

I thought about this some more. Try visualizing this problem: Suppose you had external forces of +F/2 in the positive x-direction at C and A. Would this guarantee that the truss translated as a rigid body? What would be the internal reaction forces at the ends of the struts? Suppose that you then looked at this from the center of mass frame of reference. It would be the same as having a gravitational body force acceleration g = F/(4m) acting in the negative x-direction, with the imposed forces at C and A acting to just balance the gravitation. OK so far?

 

[etotheipi]

It would be the same as having a gravitational body force acceleration g = F/(4m) acting in the negative x-direction, with the imposed forces at C and A acting to just balance the gravitation. OK so far?

That checks out, yes. OK so far!

 

[Chestermiller]

That checks out, yes. OK so far!

OK. Just so you're comfortable with this, what do you get for the reaction forces at the ends of the struts?

So now you have a truss situated in a gravitational field, with identical forces applied at its two ends to just hold it in equilibrium. This reminds me of a somewhat analogous situation of a mass in a gravitational field hanging in equilibrium at the end of a spring. You displace the mass vertically and then release it. Does the subsequent simple harmonic motion depend on g?

 

[etotheipi]

OK. Just so you're comfortable with this, what do you get for the reaction forces at the ends of the struts?

Suppose the truss is rigid. If you consider any single strut in the truss, there can be no $x$ component of reaction force at either hinge A or hinge C, due to reflection symmetry across the $x$ axis. The external force acting in the positive $x$ direction on each strut is is of magnitude F/4, which means that the $x$ component at B/D must also be zero for the correct $x$ acceleration of each strut. The $y$ components at each end of each strut must also be opposite forces. Torques about either A or C for the struts on either side of the $x$ axis mean these components have to be zero too. About the CM of each strut there is now a non-zero torque, which is a contradiction. So the truss is not rigid in this stage.

So now you have a truss situated in a gravitational field, with identical forces applied at its two ends to just hold it in equilibrium. This reminds me of a somewhat analogous situation of a mass in a gravitational field hanging in equilibrium at the end of a spring. You displace the mass vertically and then release it. Does the subsequent simple harmonic motion depend on g?

The motion is independent of $g$.

 

[Chestermiller]

Suppose the truss is rigid. If you consider any single strut in the truss, there can be no $x$ component of reaction force at either hinge A or hinge C, due to reflection symmetry across the $x$ axis. The external force acting in the positive $x$ direction on each strut is is of magnitude F/4, which means that the $x$ component at B/D must also be zero for the correct $x$ acceleration of each strut. The $y$ components at each end of each strut must also be opposite forces. Torques about either A or C for the struts on either side of the $x$ axis mean these components have to be zero too. About the CM of each strut there is now a non-zero torque, which is a contradiction. So the truss is not rigid in this stage.

This is correct with regard to the x-components, but not with regard to the y components. For strut AB, the y component of the reaction force imposed by strut BC at B is in the + y direction, and the y component of the reaction force imposed by strut DA at A is in the - y direction. These y direction forces are equal in magnitude. The couple created by these y-component reaction forces is equal and opposite to the couple associated with the x-gravitational force (at the center of mass) and the x imposed force at A. The x-forces are of magnitude F/4, and the magnitudes of the y reaction forces are F/8 (because of the larger moment arm). So there actually is a zero torque on the strut, and the truss does remain rigid.

 

[etotheipi]

These y direction forces are equal in magnitude. The couple created by these y-component reaction forces is equal and opposite to the couple associated with the x-gravitational force (at the center of mass) and the x imposed force at A. The

You're right, I concluded that the torques of the $y$ components canceled but of course they are both in the same sense.

 

[Chestermiller]

You're right, I concluded that the torques of the $y$ components canceled but of course they are both in the same sense.

I've decided that I'm done screwing around with this problem. I'm going to solve it only using the laboratory (fixed) frame of reference, and without using the resolution of the problem into two separate problems that I've done so far. To do this, I'm going to focus on the top half of the truss, and do force and moment balances of struts AB and CD.

First the kinematics. There are only two parameters that characterize the kinematics of this problem. This is the angle $\theta$ and the location of the center of mass of the overall rhombus. Using the analysis and notation in post #29, the acceleration components of the center of mass AB are:

$a_C+a_{Ax}/2$ in the x direction
and
$-a_{Ax}/2$ in the y direction

And for strut BC, they are
$a_C-a_{Ax}/2$ in the x-direction
and
$-a_{Ax}/2$ in the y direction$where$a_C## is the x-direction acceleration of the center of mass of the overall truss.

The figure below shows the free body diagram for the two struts comprising the upper half of the truss:

Based on this, the force- and moment balances on strut AB are:
$$\frac{F}{2}-T=m\left(a_C+\frac{a_{Ax}}{2}\right)$$
$$R_B-R_A=-m\frac{a_{Ax}}{2}$$
$$(R_B+R_A)\frac{b}{2\sqrt{2}}-\left(\frac{F}{2}+T\right)\frac{b}{2\sqrt{2}}=m\frac{b^2}{12}\frac{d^2\theta}{dt^2}$$
And for strut BC, they are:
$$T=m\left(a_C-\frac{a_{Ax}}{2}\right)$$
$$R_B+R_C=m\frac{a_{Ax}}{2}$$
$$-(R_C-R_B-T)\frac{b}{2\sqrt{2}}=-m\frac{b^2}{12}\frac{d^2\theta}{dt^2}$$

OK so far?

 

[etotheipi]

$a_C+a_{Ax}/2$ in the x direction
and
$-a_{Ax}/2$ in the y direction

And this is because in the CM frame where $(x,y) = (\frac{b}{2}\cos{\theta}, \frac{b}{2}\sin{\theta})$, we also have $(\ddot{x}, \ddot{y}) = (-\frac{b}{2}\cos{\theta} (\dot{\theta})^2 - \frac{b}{2}\sin{\theta}\ddot{\theta}, \frac{b}{2}\cos{\theta}\ddot{\theta} - \frac{b}{2}(\dot{\theta})^2 \sin{\theta})$. And initially with $\theta = \frac{\pi}{4}$ that does mean that the initial $y$ acceleration of the CM of AB is $-\frac{a_{Ax}}{2}$. The torque and force analyses on the struts look good.

I spent a lot of time thinking about your first method yesterday and it's really clever. I came to the conclusion that so long as you can assume that $\vec{a}_C$ is a linear function of all of the applied forces, then you can apply the superposition principle.

 

[Chestermiller]

And this is because in the CM frame where $(x,y) = (\frac{b}{2}\cos{\theta}, \frac{b}{2}\sin{\theta})$, we also have $(\ddot{x}, \ddot{y}) = (-\frac{b}{2}\cos{\theta} (\dot{\theta})^2 - \frac{b}{2}\sin{\theta}\ddot{\theta}, \frac{b}{2}\cos{\theta}\ddot{\theta} - \frac{b}{2}(\dot{\theta})^2 \sin{\theta})$. And initially with $\theta = \frac{\pi}{4}$ that does mean that the initial $y$ acceleration of the CM of AB is $-\frac{a_{Ax}}{2}$. The torque and force analyses on the struts look good.

OK. Continuing with the analysis, if we add the force- and moment balances for the two struts together, we obtain the following:
$$\frac{F}{2}=2ma_C\tag{1}$$
$$2R_B+R_C-R_A=0\tag{2}$$
$$2R_B+R_A-R_C-\frac{F}{2}=0\tag{3}$$
From Eqn. 1, if follows that $$a_C=\frac{F}{4m}\tag{4a}$$From Eqns. 2 and 3, it follows that $$R_B=\frac{F}{8}\tag{4b}$$
Eqns. 4 are identical to the relationships obtained previously for Part 1 loading. So adding the force and moment balance equations for the two struts is found to deliver the exact same equations as for Part 1 loading.

From Eqn. 2, it follows that $$\frac{(R_A-R_C)}{2}=R_B=\frac{F}{8}\tag{5}$$
Now, with no loss of generality, we can then also write:
$$R_A=R+R_B\tag{6a}$$
$$R_C=R-R_B\tag{6b}$$ where $$R=\frac{(R_A+R_C)}{2}$$
Next, subtracting the force and moment balances of the two struts and combining these with Eqns. 6 yields:
$$\frac{F}{2}-2T=ma_{Ax}\tag{7a}$$
$$2R=ma_{Ax}\tag{7b}$$
$$\left(2R-2T-\frac{F}{2}\right)=2m\frac{b^2}{12}\frac{d^2\theta}{dt^2}\tag{7c}$$

Eqns. 7 are identical to the relationships obtained previously for Part 2 loading. So subtracting the force and moment balance equations for the two trusses is found to deliver the exact same equations as for Part 2 loading.

 

[etotheipi]

Eqns. 7 are identical to the relationships obtained previously for Part 2 loading. So subtracting the force and moment balance equations for the two trusses is found to deliver the exact same equations as for Part 2 loading.

Amazing. You make it look easy! Do you get double points for solving it twice?

It'll take me a little while to put all of the pieces together but I could follow through your reasoning. Thanks!

 

[JD_PM]

My 2 cents worth contribution to this beautiful thread.

Spoiler: Extended version of wrobel's solution of problem 2

Please note that I used the same approach that @wrobel at #21 and, to be fair, I checked it before getting the answer by myself.

I modified Chestermiller's nice diagram so that it can be applied to this approach

Where $O$ is at rest relative to the lab frame.

We first note that we have a time-dependent external force (exerted on A) which has no potential energy associated to it due to the fact that the force is not conservative.

At this point we think of the definition of generalized force (see for instance Classical Mechanics by Gregory, Chapter 12, page 337, definition 12.6)

$$Q_j = \sum_i \vec F_i^S \cdot \frac{\partial \vec r_i}{\partial q_j}$$

Where $\vec F_i^S$ is the specified force at a given point and $\vec r_i$ is the distance from $O$ to the given point.

Let's go step by step

1) Compute generalized forces

The generalized forces in this problem are:

$$Q_x = \vec F_A^S \cdot \frac{\partial \vec r_A}{\partial x}+\vec F_B^S \cdot \frac{\partial \vec r_B}{\partial x}+\vec F_C^S \cdot \frac{\partial \vec r_C}{\partial x}+\vec F_D^S \cdot \frac{\partial \vec r_D}{\partial x}$$

$$Q_{\theta} = \vec F_A^S \cdot \frac{\partial \vec r_A}{\partial \theta}+\vec F_B^S \cdot \frac{\partial \vec r_B}{\partial \theta}+\vec F_C^S \cdot \frac{\partial \vec r_C}{\partial \theta}+\vec F_D^S \cdot \frac{\partial \vec r_D}{\partial \theta}$$

Before computing anything we note that we've only been specified a force at A. Thus the rest are zero and we end up with:

$$Q_x = \vec F_A^S \cdot \frac{\partial \vec r_A}{\partial x}=F$$

$$Q_{\theta} = \vec F_A^S \cdot \frac{\partial \vec r_A}{\partial \theta}=-Fb \sin\theta$$

2) Determine the kinetic energy $(T)$ of the system

Note that we are going to work with two frames here: the lab frame (external observer) and the rhombic frame $XYS$. König's theorem asserts that the kinetic energy of a system is the sum of two terms: the kinetic energy associated to the movement of the center of mass with respect to the lab frame (let me label it $T_1$) and the kinetic energy associated to the movement of particles making up the system with respect of the center of mass of the system (let me label it $T_2$).

- Let's determine $T_1$: the center of mass (wrt the lab frame) is located at a distance OS, which is precisely our definition for the general coordinate x. Thus we get

$$T_1 = \frac 1 2 (4m) \dot x^2$$

- Let's determine $T_2$: Let's first get the coordinates of the center of mass of the $AB$ rod

$$\Big( \frac b 2 \cos \theta, \ \frac b 2 \sin \theta \Big)$$

Then the velocity coordinates are

$$\Big( -\frac b 2 \sin \theta \dot \theta, \ \frac b 2 \cos \theta \dot \theta \Big)$$

Or in the more usual form: $\vec v_{T_2}=-\frac b 2 \sin \theta \dot \theta \hat x + \frac b 2 \cos \theta \dot \theta \hat y$.

The nice thing here is that, although neither the coordinates nor the velocity coordinates of the other 3 rod's center of masses are the same as the ones for the $AB$ rod, the squared modulus of the velocity ($|\vec v_{T_2}|^2$) is.

$$|\vec v_{T_2}|^2=\frac{b^2}{4} \dot \theta^2$$

Besides, the rods possesses moment of inertia with respect to their center of mass ($\frac{mb^2}{12}$ each), which means that we have to take into account the kinetic energy associated to the moment of inertia of all 4 rods.

Thus $T_2$ is given by

$$T_2=4 \Big( \frac 1 2 m \frac{b^2}{4} \dot \theta^2 + \frac{1}{24} mb^2 \dot \theta^2 \Big)=\frac{2}{3} mb^2 \dot \theta^2$$

Thus the kinetic energy $(T)$ of the system is

$$T = 2m \dot x^2 + \frac{2}{3} mb^2 \dot \theta^2$$

3) Compute the equations of motion related to each generalized coordinate

The Lagrange equations for a general standard system are:

$$\frac{d}{dt} \Big( \frac{\partial T}{\partial \dot q_j} \Big) - \frac{\partial T}{\partial q_j} = Q_j$$

- For $x$ we get

$$4m \ddot x = F$$

- For $\theta$ we get

$$\frac 4 3 mb \ddot \theta = -F \sin \theta$$

4) Get $a_c$

Based on the diagram we see that

$$a_c=\frac{d^2}{dt^2} OC, \ \ \ \ \theta=\frac{\pi}{4},\ \ \ \ OC=x-b \cos\theta$$

Thus, using the equations of motion we've got, we end up with

$$a_c= \ddot x + b\cos \theta \ddot \theta^2 = \frac{F}{4m}-\frac{3F}{4m} \sin \theta \cos \theta=-\frac{F}{8m}$$

 

[Adesh]

My 2 cents worth contribution to this beautiful thread.

Spoiler: Extended version of wrobel's solution of problem 2

Please note that I used the same approach that @wrobel at #21 and, to be fair, I checked it before getting the answer by myself.

I modified Chestermiller's nice diagram so that it can be applied to this approach

View attachment 261815
Where $O$ is at rest relative to the lab frame.

We first note that we have a time-dependent external force (exerted on A) which has no potential energy associated to it due to the fact that the force is not conservative.

At this point we think of the definition of generalized force (see for instance Classical Mechanics by Gregory, Chapter 12, page 337, definition 12.6)

$$Q_j = \sum_i \vec F_i^S \cdot \frac{\partial \vec r_i}{\partial q_j}$$

Where $\vec F_i^S$ is the specified force at a given point and $\vec r_i$ is the distance from $O$ to the given point.

Let's go step by step

1) Compute generalized forces

The generalized forces in this problem are:

$$Q_x = \vec F_A^S \cdot \frac{\partial \vec r_A}{\partial x}+\vec F_B^S \cdot \frac{\partial \vec r_B}{\partial x}+\vec F_C^S \cdot \frac{\partial \vec r_C}{\partial x}+\vec F_D^S \cdot \frac{\partial \vec r_D}{\partial x}$$

$$Q_{\alpha} = \vec F_A^S \cdot \frac{\partial \vec r_A}{\partial \theta}+\vec F_B^S \cdot \frac{\partial \vec r_B}{\partial \theta}+\vec F_C^S \cdot \frac{\partial \vec r_C}{\partial \theta}+\vec F_D^S \cdot \frac{\partial \vec r_D}{\partial \theta}$$

Before computing anything we note that we've only been specified a force at A. Thus the rest are zero and we end up with:

$$Q_x = \vec F_A^S \cdot \frac{\partial \vec r_A}{\partial x}=F$$

$$Q_{\alpha} = \vec F_A^S \cdot \frac{\partial \vec r_A}{\partial \theta}=-Fb \sin\alpha$$

2) Determine the kinetic energy $(T)$ of the system

Note that we are going to work with two frames here: the lab frame (external observer) and the rhombic frame $XYS$. König's theorem asserts that the kinetic energy of a system is the sum of two terms: the kinetic energy associated to the movement of the center of mass with respect to the lab frame (let me label it $T_1$) and the kinetic energy associated to the movement of particles making up the system with respect of the center of mass of the system (let me label it $T_2$).

- Let's determine $T_1$: the center of mass (wrt the lab frame) is located at a distance OS, which is precisely our definition for the general coordinate x. Thus we get

$$T_1 = \frac 1 2 (4m) \dot x^2$$

- Let's determine $T_2$: Let's first get the coordinates of the center of mass of the $AB$ rod

$$\Big( \frac b 2 \cos \theta, \ \frac b 2 \sin \theta \Big)$$

Then the velocity coordinates are

$$\Big( -\frac b 2 \sin \theta \dot \theta, \ \frac b 2 \cos \theta \dot \theta \Big)$$

Or in the more usual form: $\vec v_{T_2}=-\frac b 2 \sin \theta \dot \theta \hat x + \frac b 2 \cos \theta \dot \theta \hat y$.

The nice thing here is that, although neither the coordinates nor the velocity coordinates of the other 3 rod's center of masses are the same as the ones for the $AB$ rod, the squared modulus of the velocity ($|\vec v_{T_2}|^2$) is.

$$|\vec v_{T_2}|^2=\frac{b^2}{4} \dot \theta^2$$

Besides, the rods possesses moment of inertia with respect to their center of mass ($\frac{mb^2}{12}$ each), which means that we have to take into account the kinetic energy associated to the moment of inertia of all 4 rods.

Thus $T_2$ is given by

$$T_2=4 \Big( \frac 1 2 m \frac{b^2}{4} \dot \theta^2 + \frac{1}{24} mb^2 \dot \theta^2 \Big)=\frac{2}{3} mb^2 \dot \theta^2$$

Thus the kinetic energy $(T)$ of the system is

$$T = 2m \dot x^2 + \frac{2}{3} mb^2 \dot \theta^2$$

3) Compute the equations of motion related to each generalized coordinate

The Lagrange equations for a general standard system are:

$$\frac{d}{dt} \Big( \frac{\partial T}{\partial \dot q_j} \Big) - \frac{\partial T}{\partial q_j} = Q_j$$

- For $x$ we get

$$4m \ddot x = F$$

- For $\theta$ we get

$$\frac 4 3 mb \ddot \theta = -F \sin \theta$$

4) Get $a_c$

Based on the diagram we see that

$$a_c=\frac{d^2}{dt^2} OC, \ \ \ \ \theta=\frac{\pi}{4},\ \ \ \ OC=x-b \cos\theta$$

Thus, using the equations of motion we've got, we end up with

$$a_c= \ddot x + b\cos \theta \ddot \theta^2 = \frac{F}{4m}-\frac{3F}{4m} \sin \theta \cos \theta=-\frac{F}{8m}$$

Although, I couldn’t understand your solution (due to my lack of knowledge about Lagrangian Mechanics) but then also it was a beautiful solution, very well organised and clear.

 

[etotheipi]

My 2 cents worth contribution to this beautiful thread.

Are your cents made out of gold?

 

[TSny]

Work-energy approach to problem 2

Discussions by @wrobel and others in the thread https://www.physicsforums.com/threads/acceleration-of-a-vertex.672835/ got me thinking about an energy approach to the problem.

The picture shows the applied forces in the center-of-mass reference frame. Each rod experiences an inertial force of magnitude F/4 as shown. The figure is shown for an arbitrary value of $\theta$.

We will obtain the acceleration of joint C using the principle that the rate of change of kinetic energy of the system equals the rate at which the applied forces do work. The four inertial forces do zero net work in the center-of-mass frame since the horizontal displacements of the centers of rods AB and AD are to the right while the horizontal displacements of the other two rods are the same amount to the left. So, we only need to consider the work done by the applied force F at joint A.

It is not hard to show that the total kinetic energy of the system in the center-of-mass frame is $T = \frac{2}{3}mL^2 \dot \theta^2$, where $m$ and $L$ are the mass and length of one rod. Also, the position of joint A is $x_A = L \cos \theta$. The rate of change of $T$ equals the rate that $F$ does work:

$\dot T = F \dot x_A$

So, $\frac{4}{3}mL^2 \dot \theta \ddot \theta = F\left(-L\sin \theta \, \dot \theta \right)$

Thus, $\ddot \theta = - \large \frac{3F}{4mL}\sin \theta \,\,\,\,$ [“pendulum” equation of motion for $\theta$].

The position of joint C is $x_C = -L \cos \theta$. Therefore,

$\ddot x_C = L \sin \theta \, \ddot\theta \,\,$ whenever $\dot \theta = 0$, as when $F$ is first applied. Substitute for $\ddot \theta$ to obtain

$\ddot x_C = -\frac{3F}{4m}\sin^2 \theta\,\,$ when $F$ is first applied.

This gives the acceleration of joint C relative to the center-of-mass frame. Relative to the lab frame, the acceleration of the center of mass is $\frac{F}{4m}$ to the right. So, the acceleration of joint C relative to the lab is

$\ddot x_{C/ \rm lab} =\frac{F}{4m} - \frac{3F}{4m}\sin^2 \theta$

or,

$\boxed{\ddot x_{C/ \rm lab} = \frac{F}{4m}\left(1-3 \sin^2 \theta \right)} \,\,\,$ when $F$ is first applied.

For $\theta = 45^o$, this gives $\ddot x_{C/ \rm lab} = - \large \frac{F}{8m}$.

The value of $\theta$ for which $\ddot x_{C/ \rm lab} = 0$ when $F$ is applied is $\theta = \arcsin (1/ \sqrt{3}) \approx 35^o$.

 

[wrobel]

I vote for this solution by TSny ,at least this is only solution I understand

 

[Chestermiller]

I vote for this solution by TSny ,at least this is only solution I understand

I'm amazed that you don't understand the solution I presented in post's #51 and 53; it is just straightforward freshman physics using force- and moment balances.

 

[Spinnor]

My 2 cents worth contribution to this beautiful thread.

Spoiler: Extended version of wrobel's solution of problem 2

Please note that I used the same approach that @wrobel at #21 and, to be fair, I checked it before getting the answer by myself.

I modified Chestermiller's nice diagram so that it can be applied to this approach

View attachment 261815
Where $O$ is at rest relative to the lab frame.

We first note that we have a time-dependent external force (exerted on A) which has no potential energy associated to it due to the fact that the force is not conservative.

At this point we think of the definition of generalized force (see for instance Classical Mechanics by Gregory, Chapter 12, page 337, definition 12.6)

$$Q_j = \sum_i \vec F_i^S \cdot \frac{\partial \vec r_i}{\partial q_j}$$

Where $\vec F_i^S$ is the specified force at a given point and $\vec r_i$ is the distance from $O$ to the given point.

Let's go step by step

1) Compute generalized forces

The generalized forces in this problem are:

$$Q_x = \vec F_A^S \cdot \frac{\partial \vec r_A}{\partial x}+\vec F_B^S \cdot \frac{\partial \vec r_B}{\partial x}+\vec F_C^S \cdot \frac{\partial \vec r_C}{\partial x}+\vec F_D^S \cdot \frac{\partial \vec r_D}{\partial x}$$

$$Q_{\theta} = \vec F_A^S \cdot \frac{\partial \vec r_A}{\partial \theta}+\vec F_B^S \cdot \frac{\partial \vec r_B}{\partial \theta}+\vec F_C^S \cdot \frac{\partial \vec r_C}{\partial \theta}+\vec F_D^S \cdot \frac{\partial \vec r_D}{\partial \theta}$$

Before computing anything we note that we've only been specified a force at A. Thus the rest are zero and we end up with:

$$Q_x = \vec F_A^S \cdot \frac{\partial \vec r_A}{\partial x}=F$$

$$Q_{\theta} = \vec F_A^S \cdot \frac{\partial \vec r_A}{\partial \theta}=-Fb \sin\theta$$

2) Determine the kinetic energy $(T)$ of the system

Note that we are going to work with two frames here: the lab frame (external observer) and the rhombic frame $XYS$. König's theorem asserts that the kinetic energy of a system is the sum of two terms: the kinetic energy associated to the movement of the center of mass with respect to the lab frame (let me label it $T_1$) and the kinetic energy associated to the movement of particles making up the system with respect of the center of mass of the system (let me label it $T_2$).

- Let's determine $T_1$: the center of mass (wrt the lab frame) is located at a distance OS, which is precisely our definition for the general coordinate x. Thus we get

$$T_1 = \frac 1 2 (4m) \dot x^2$$

- Let's determine $T_2$: Let's first get the coordinates of the center of mass of the $AB$ rod

$$\Big( \frac b 2 \cos \theta, \ \frac b 2 \sin \theta \Big)$$

Then the velocity coordinates are

$$\Big( -\frac b 2 \sin \theta \dot \theta, \ \frac b 2 \cos \theta \dot \theta \Big)$$

Or in the more usual form: $\vec v_{T_2}=-\frac b 2 \sin \theta \dot \theta \hat x + \frac b 2 \cos \theta \dot \theta \hat y$.

The nice thing here is that, although neither the coordinates nor the velocity coordinates of the other 3 rod's center of masses are the same as the ones for the $AB$ rod, the squared modulus of the velocity ($|\vec v_{T_2}|^2$) is.

$$|\vec v_{T_2}|^2=\frac{b^2}{4} \dot \theta^2$$

Besides, the rods possesses moment of inertia with respect to their center of mass ($\frac{mb^2}{12}$ each), which means that we have to take into account the kinetic energy associated to the moment of inertia of all 4 rods.

Thus $T_2$ is given by

$$T_2=4 \Big( \frac 1 2 m \frac{b^2}{4} \dot \theta^2 + \frac{1}{24} mb^2 \dot \theta^2 \Big)=\frac{2}{3} mb^2 \dot \theta^2$$

Thus the kinetic energy $(T)$ of the system is

$$T = 2m \dot x^2 + \frac{2}{3} mb^2 \dot \theta^2$$

3) Compute the equations of motion related to each generalized coordinate

The Lagrange equations for a general standard system are:

$$\frac{d}{dt} \Big( \frac{\partial T}{\partial \dot q_j} \Big) - \frac{\partial T}{\partial q_j} = Q_j$$

- For $x$ we get

$$4m \ddot x = F$$

- For $\theta$ we get

$$\frac 4 3 mb \ddot \theta = -F \sin \theta$$

4) Get $a_c$

Based on the diagram we see that

$$a_c=\frac{d^2}{dt^2} OC, \ \ \ \ \theta=\frac{\pi}{4},\ \ \ \ OC=x-b \cos\theta$$

Thus, using the equations of motion we've got, we end up with

$$a_c= \ddot x + b\cos \theta \ddot \theta^2 = \frac{F}{4m}-\frac{3F}{4m} \sin \theta \cos \theta=-\frac{F}{8m}$$

You get the same answer for theta = 0 and 90 degrees, that result does not agree with my experiment but the results of post #58 do?

 

[TSny]

I vote for this solution by TSny ,at least this is only solution I understand

Thank you, but both of @Chestermiller's solutions look correct to me.

 

[wrobel]

Yes you are right