Where is the centripetal force acting on mass M?

[PMANNAS]

TL;DR Summary

I cannot understand how a centripetal force component can be achieved via the push and pull forces acting on mass M when the rigid triangle is angularly accelerating.

Look at the image below where I've drawn a rigid triangle, with thin lightweight rods and a fixed mass M attached as shown. There is a constant torque (2Fr) applied at a fixed pivot point P bisecting the top rod. Let's assume M is at rest and then a torque is added about pivot point P which will angularly accelerate the triangle and M in a circular path. Lets also assume the triangle is on a frictionless flat table so there is no gravity involved.

1. Am I correct that forces F1 and F2 will act on mass M, with a resultant net tangential force (as shown in image below) when the triangle is at rest?
2. I am assuming that the forces F1 and F2 will remain unchanged when the triangle is angularly accelerating . I've been informed they may change but that doesn't make any sense to me because the torque forces F are constant. There is a push force F1 on M via the left arm and a pull force F2 on M via the right arm. When I add those 2 forces , the resultant force is always tangential to the circular path but where is the centripetal force to keep mass M moving in a circle?

 

[BvU]

Hello @PMANNAS ,

$\qquad$ !
​

1. Am I correct that forces F1 and F2 will act on mass M, with a resultant net tangential force (as shown in image below) when the triangle is at rest?

Yes. At $t = 0$ when the forces start acting.

2. I am assuming that the forces F1 and F2 will remain unchanged when the triangle is angularly accelerating

Which means that the trajectory of $M$ is a straight line along $F_3$, right ?

$\$

 

[berkeman]

TL;DR Summary: I cannot understand how a centripetal force component can be achieved via the push and pull forces acting on mass M when the rigid triangle is angularly accelerating.

When I add those 2 forces , the resultant force is always tangential to the circular path but where is the centripetal force to keep mass M moving in a circle?

Please write the equations for the angular acceleration due to those forces and the centripital acceleration pulling that mass toward the center of rotation as a function of the angular velocity. If the two forces are constant as the assembly spins up, there will be a tangential acceleration resulting in an angular acceleration in addition to the centripital force which depends on the angular velocity. Can you see that?

 

[Juanda]

TL;DR Summary: I cannot understand how a centripetal force component can be achieved via the push and pull forces acting on mass M when the rigid triangle is angularly accelerating.

Look at the image below where I've drawn a rigid triangle, with thin lightweight rods and a fixed mass M attached as shown. There is a constant torque (2Fr) applied at a fixed pivot point P bisecting the top rod. Let's assume M is at rest and then a torque is added about pivot point P which will angularly accelerate the triangle and M in a circular path. Lets also assume the triangle is on a frictionless flat table so there is no gravity involved.

1. Am I correct that forces F1 and F2 will act on mass M, with a resultant net tangential force (as shown in image below) when the triangle is at rest?
2. I am assuming that the forces F1 and F2 will remain unchanged when the triangle is angularly accelerating . I've been informed they may change but that doesn't make any sense to me because the torque forces F are constant. There is a push force F1 on M via the left arm and a pull force F2 on M via the right arm. When I add those 2 forces , the resultant force is always tangential to the circular path but where is the centripetal force to keep mass M moving in a circle?

View attachment 328142

I believe you are missing the force done by the support so the forces at the rods $F_1$ and $F_2$ don't have the values you stated.

I can't get my hands on it now but I'll give it a shot once I'm back.

Basically, as @berkeman said, the problem can be solved using the equation for angular acceleration because we know it will be a circular motion. Then, using the equations for the forces you can get the necessary reaction at the support to obtain such motion.
Finally, from the external forces and the acceleration of the system, the internal forces at the rods can be obtained by analyzing the dynamics of this "mechanism".

 

[PMANNAS]

Hello @PMANNAS ,

$\qquad$ !
​

Yes. At $t = 0$ when the forces start acting.Which means that the trajectory of $M$ is a straight line along $F_3$, right ?

$\$

Yes , that was what I assumed but it looks that I may be wrong according to other member's posts.

 

[PMANNAS]

Please write the equations for the angular acceleration due to those forces and the centripital acceleration pulling that mass toward the center of rotation as a function of the angular velocity. If the two forces are constant as the assembly spins up, there will be a tangential acceleration resulting in an angular acceleration in addition to the centripital force which depends on the angular velocity. Can you see that?

Yes , I can see that a 'Centripetal Force' can be theoretically worked out but I cannot see how it could be physically achieved.

 

[PMANNAS]

I believe you are missing the force done by the support so the forces at the rods $F_1$ and $F_2$ don't have the values you stated.

I can't get my hands on it now but I'll give it a shot once I'm back.

Basically, as @berkeman said, the problem can be solved using the equation for angular acceleration because we know it will be a circular motion. Then, using the equations for the forces you can get the necessary reaction at the support to obtain such motion.
Finally, from the external forces and the acceleration of the system, the internal forces at the rods can be obtained by analyzing the dynamics of this "mechanism".

I'm more interested in how that centripetal force can physically be achieved . It must be via the 2 arms connecting to mass M . You've mentioned that I am missing forces done by the support , so are you able to draw some force diagrams (no need for maths) to just show me how that centripetal force could be created as the triangle angularly accelerates?

 

[berkeman]

so you able to draw some force diagrams (no need for maths) to just show me how that centripetal force could be created?

If the triangle is a rigid piece of material, then the centripital force (pull on the mass toward the center) is transmitted through that solid triangular material.

If the triangle is made up of "rigid" rods, the central pulling force is a bending force on the lower two rods. If they are not very rigid, they will bow out as part of pulling the mass in toward the center.

 

[PMANNAS]

If the triangle is a rigid piece of material, then the centripital force (pull on the mass toward the center) is transmitted through that solid triangular material.

If the triangle is made up of "rigid" rods, the central pulling force is a bending force on the lower two rods. If they are not very rigid, they will bow out as part of pulling the mass in toward the center.

Ahhh! Now I think I can visualise it . So the centripetal forces via the pivot arms will start pulling on lower arms increasing the strain tension which means there will be pull forces acting on mass M by both arms. When those forces are added together we will get a centripetal and tangential force acting on mass M as it angularly accelerates.

On reflection , won't the 'bowing' out strain tension acting on mass M via both arms cancel out the tangential force? And if there is no tangential force , then no angular acceleration.

 

[PeroK]

Yes , I can see that a 'Centripetal Force' can be theoretically worked out but I cannot see how it could be physically achieved.

Initially, as the mass is at rest, there is an instantaneous acceleration but zero initial velocity. There is, therefore, no centripetal component of the force on the mass. You can calculate $F_1, F_2$ and the external force at the pivot from that. You can also calculate the angular acceleration - which must be constant over time, as the external torque about the pivot is constant.

As the mass speeds up, there must be a centripetal force on the mass, which must come from an imbalance in $F_1$ and $F_2$, as those are the only forces acting on the mass. The angular acceleration gives you the speed of the mass at any time and from that you can calculate the changing forces over time.

 

[PMANNAS]

Initially, as the mass is at rest, there is an instantaneous acceleration but zero initial velocity. There is, therefore, no centripetal component of the force on the mass. You can calculate $F_1, F_2$ and the external force at the pivot from that. You can also calculate the angular acceleration - which must be constant over time, as the external torque about the pivot is constant.

As the mass speeds up, there must be a centripetal force on the mass, which must come from an imbalance in $F_1$ and $F_2$, as those are the only forces acting on the mass. The angular acceleration gives you the speed of the mass at any time and from that you can calculate the changing forces over time.

Yes , I understand that there must be an imbalance in the forces F1 and F2 but I had problems visualising how that could physically happen. The pivot arms are rotating and there must be centripetal forces pulling in the rods (and would bend them if they weren't rigid). So the arms would tend to bow out and introduce strain tension and forces acting on M in the opposite directions to F1 and F2. The net force acting on M must end up with a tangential and centripetal force . But why would the bowing in one arm be different to the other? If the bowing was the same, then wouldn't there be no net tangential force?

 

[PeroK]

But why would the bowing in one arm be different to the other? If the bowing was the same, then wouldn't there be no net tangential force?

There is an asymmetry between the two arms created by the direction of the angular acceleration.

 

[PMANNAS]

There is an asymmetry between the two arms created by the direction of the angular acceleration.

I'm not sure how there can be an asymmetry but obviously it is happening. Is it because one arm (ie. the left one) is being bowed inwards (concave) while the other arm ( the right one) outwards?

 

[PeroK]

I'm not sure how there can be an asymmetry but obviously it is happening. Is it because one arm (ie. the left one) is being bowed inwards (concave) while the other arm ( the right one) outwards?

The direction of motion defines two different roles for the arms: one pushing (with a centrifugal component) and the other pulling (with a centripetal component). That is an asymmetry.

 

[PMANNAS]

Maybe I could split the problem into a net eccentric force acting on each arm like the images below.

Look at the 1st image further below:
This could be the yellow net eccentric force applied on the left arm which, according to 'Chasles Theorem', causes a clockwise couple about the COM (ie. M in this case if the rods are lightweight). The moment arm would be 'y' and I suspect it would cause an inward bowing of the left arm. This would cause a strain force SL to be applied on M .

The 2nd image further below would show the net eccentric force on the right arm and an anti-clockwise couple about M , but the moment arm 'x' is larger so I suspect the inward bowing would be larger , therefore the strain tension force SR acting on M will be larger.

So looking at the net effect on M , we have strain tension forces SL and SR that could theoretically cause the centripetal and tangential forces.

Is this a feasible explanation or have I got it completely and utterly wrong? Obviously , because I've defined the rods as being rigid there wouldn't be any deformation which I think complicates the visualisation when trying to solve problems like these.

 

[Juanda]

I'm more interested in how that centripetal force can physically be achieved . It must be via the 2 arms connecting to mass M . You've mentioned that I am missing forces done by the support , so are you able to draw some force diagrams (no need for maths) to just show me how that centripetal force could be created as the triangle angularly accelerates?

I think the math is what really will give you the proper insight but here you have the diagram without equations.

In red you can see the external forces on the system. In green, the acceleration of the mass is due to those external forces. This is assuming the rods are massless and the rotational speed is already greater than 0.

Realize there are the two forces you mentioned in your picture but also the reactions at the articulated support. The reactions at the support are typically omitted because, since we know the system will be rotating around that point, it is then possible to find the description of the angular position of the mass using just the equation for moments. However, the reactions at the support are a necessary part of the system to make the movement possible.

Regarding how the centripetal acceleration is transmitted to the mass, it is easier to imagine the system being static and changing the accelerations by the equivalent inertial forces by multiplying them with the mass and changing its sign.
I said no equations would be used but some things are just easier to see in equations than explained so... $\sum \vec{F}=m \vec{a}\rightarrow \sum \vec{F}-m \vec{a}=0$

From the previous picture, we can imagine how both rods will be working under tension but the left rod will have a greater tension than the one at the right. That imbalance makes it accelerate counterclockwise. On the other hand, the centripetal component from the tension in both rods is what keeps the mass from flying away in the radial direction.

The tangential component of the acceleration is caused by the angular acceleration while the normal component is caused by the centripetal acceleration which is caused by the rotational speed. So as time goes by and the angular speed increases, the normal component will get bigger and bigger.

To see how it deforms we would need to analyze the internal stress of each of the bars. It can be done somewhat easily by assuming each of the corners of the triangle is articulated although it's a tedious process and I won't be doing it now. In that case, the lateral rods will be increasing in length due to the tension although in different magnitudes. The rod at the base will bow (like a smiley face) due to the bending moments present in it. Also, its left half will be working in compression while the right half will be working in tension so you could expect some changes in length due to that. However, the deformation due to bending is typically much greater than the one caused by normal forces so it is ignored very often.

 

[Lnewqban]

Yes , I understand that there must be an imbalance in the forces F1 and F2 but I had problems visualising how that could physically happen. The pivot arms are rotating and there must be centripetal forces pulling in the rods (and would bend them if they weren't rigid). So the arms would tend to bow out and introduce strain tension and forces acting on M in the opposite directions to F1 and F2. The net force acting on M must end up with a tangential and centripetal force . But why would the bowing in one arm be different to the other? If the bowing was the same, then wouldn't there be no net tangential force?

For your example, it is recommendable to consider the members as articulated at each node.
That way, any complicating bending moment is eliminated from the analysis, as you have an armature, formed by members that can only “feel” compression or tension stresses.

The magnitude of F1 can only be equal to the magnitude of F2 when there is no angular acceleration.
Just imaging those two rigid members as springs.
Also, when you swing a stone attached to a string around a horizontal trajectory, your hand must describe a horizontal circular trajectory around the center of rotation in order to transfer energy to the stone.

Can you see that there is certain rotational or angular velocity for which the force F1 becomes zero?
For that particular case, we could eliminate that left arm.
As the angular velocity becomes huge eventually, the direction of represented F1 must reverse, even reaching the point of breaking the arms due to excessive internal tension (the right member will break first).

 

[Juanda]

For your example, it is recommendable to consider the members as articulated at each node.
That way, any complicating bending moment is eliminated from the analysis, as you have an armature, formed by members that can only “feel” compression or tension stresses.

Watch out, that is only true if the forces are applied at the articulated points. Here is a nice and quick video about it.
The rod at the base actually will bend. See my post at #16 for the details.

 

[PMANNAS]

I think the math is what really will give you the proper insight but here you have the diagram without equations.

In red you can see the external forces on the system. In green, the acceleration of the mass is due to those external forces. This is assuming the rods are massless and the rotational speed is already greater than 0.
View attachment 328184

Realize there are the two forces you mentioned in your picture but also the reactions at the articulated support. The reactions at the support are typically omitted because, since we know the system will be rotating around that point, it is then possible to find the description of the angular position of the mass using just the equation for moments. However, the reactions at the support are a necessary part of the system to make the movement possible.

Regarding how the centripetal acceleration is transmitted to the mass, it is easier to imagine the system being static and changing the accelerations by the equivalent inertial forces by multiplying them with the mass and changing its sign.
I said no equations would be used but some things are just easier to see in equations than explained so... $\sum \vec{F}=m \vec{a}\rightarrow \sum \vec{F}-m \vec{a}=0$
View attachment 328185

From the previous picture, we can imagine how both rods will be working under tension but the left rod will have a greater tension than the one at the right. That imbalance makes it accelerate counterclockwise. On the other hand, the centripetal component from the tension in both rods is what keeps the mass from flying away in the radial direction.

The tangential component of the acceleration is caused by the angular acceleration while the normal component is caused by the centripetal acceleration which is caused by the rotational speed. So as time goes by and the angular speed increases, the normal component will get bigger and bigger.

To see how it deforms we would need to analyze the internal stress of each of the bars. It can be done somewhat easily by assuming each of the corners of the triangle is articulated although it's a tedious process and I won't be doing it now. In that case, the lateral rods will be increasing in length due to the tension although in different magnitudes. The rod at the base will bow (like a smiley face) due to the bending moments present in it. Also, its left half will be working in compression while the right half will be working in tension so you could expect some changes in length due to that. However, the deformation due to bending is typically much greater than the one caused by normal forces so it is ignored very often.

I'll need to read this through a few times to make sure I understand. I'm still unsure why it's obvious that the left rod will have greater tension than the one on the right (although I tried using Chasles theorem in the previous post). I cannot understand why there is a force from the support along the base direction.

 

[A.T.]

I'm still unsure why it's obvious that the left rod will have greater tension than the one on the right

When you move your center of mass around, while sitting in a chair, isn't it obvious that the forces in the chair's legs will be different?

Forces in rigid structures adjust to whatever is needed to preserve their shape. If that requires applying a centripetal force to M (because otherwise the triangle would deform), then they will adjust to do that (if possible within material strength).

 

[BvU]

Yes , that was what I assumed but it looks that I may be wrong according to other member's posts.

Never too late to adjust an assumption: it's valid for $t=0$ and clearly wrong for $t>0$.
$M$ is constrained to follow a circular trajectory.
For the radial component of $F_3$ you have $F=mv^2/r$ and $v$ will have to follow from $\tau = I\alpha$.

Not much more to say without a problem statement.

$\$

 

[Juanda]

I'll need to read this through a few times to make sure I understand. I'm still unsure why it's obvious that the left rod will have greater tension than the one on the right (although I tried using Chasles theorem in the previous post). I cannot understand why there is a force from the support along the base direction.

I recommend first studying the external forces over the system and then moving on to the internal forces that keep it together resisting being torn apart by the external forces.

We know from boundary conditions that this thing will be traveling in circles. Then, the system can be described with just $\theta$.

Since it is traveling in circles, it is possible to use the Equation of Moments around the center of rotation. The reaction forces at the center of rotation are initially unknown but we will assume it exists and it has a horizontal and vertical component because that's what that kind of support does.
I will apply the equation of moments at the center of rotation. That way, the forces at the support can be ignored since they do not contribute (leverage arm is 0).
$\sum \tau =I\ddot{\theta}\rightarrow 2Fr=I\ddot{\theta}\rightarrow \left \{ I=mh^2 \right \}\rightarrow \ddot{\theta}=\frac{2Fr}{mh^2}$
So the angular acceleration $\ddot{\theta}$ is known. It can also be easily integrated with respect to time to obtain $\dot{\theta}$ and $\theta$.

Now if we make the equations for the horizontal forces it might seem like there are too many unknowns.
$\sum F =m\ddot{x}\rightarrow R_x=m\ddot{x}$

However, we actually know $\ddot{x}$ because we know $\theta$ and we said before that the system can be described with it.
$x=hcos(\theta)\rightarrow \dot{x}=-h\dot{\theta}sin(\theta)\rightarrow \ddot{x}=-h(\ddot{\theta}sin(\theta)+\dot{\theta}^2cos(\theta))$
So $R_x$, the horizontal reaction at the support, can be found.
$\sum F =m\ddot{x}\rightarrow R_x=m\ddot{x}\rightarrow R_x=m(-h(\ddot{\theta}sin(\theta)+\dot{\theta}^2cos(\theta)))$

A similar thing can be done to find the vertical component of the reaction $R_y$. With that, you should be able to now understand why I plotted the reactions at the support like this.
(External forces are shown in red, and acceleration of the mass is shown in green)

Then, to understand the internal forces, you'd need some background in statics. Even if this is a dynamic problem we can treat it as a static one by rearranging the terms.
$\sum \vec{F}=m \vec{a}\rightarrow \sum \vec{F}-m \vec{a}=0$

You could then analyze the internal forces at a given instant with this. Assume all the corners of the triangle are articulations.

I recommend you solve the problem for $\theta = \frac{\pi}{2}$ so the problem is not as tedious to solve because the triangle will be pointing upwards so the external forces are either horizontal or vertical. Also set $\dot{\theta} = 100 rad/s$ so you can see the effect from the centripetal acceleration.
See post #18 for a video on how to solve such a structure.