Pendulum Tension Force -- How to calculate the full vector?

[babaliaris]

Hello! I'm trying to understand how this pendulum works. I found this video that explains how to calculate the T force from the rope.

He uses the preservation of kinetic and potential energy in order to find the magnitude of the velocity and then using Newton's second law, he calculates the T force.

My question is, if I understand correctly he only calculated the T in the direction of the y axis, not the x. So this is not the full answer.
In other words $T_{y_{0}}=3mg$.

I want to know the full vector $T = T_{x_{0}} + T_{y_{0}}$ or $T(x,y)$

Can I follow the same idea and use ΣF=ma on the x-axis? The gravity force is zero there, but what should I use for a? Maybe the tangent acceleration $a_{T} = \frac{dV}{dt}$ (where V is the magnitude of the velocity)?

 

[kuruman]

Newton's second law says $\mathbf{F}_{\text{net}}=m\mathbf{a}$. There are two forces acting on the pendulum, gravity and tension. When the pendulum passes through the vertical position, they do not have a horizontal component.

Question 1: What is the horizontal component of the net force?
Question 2: According to Newton's second law, what should the horizontal component of the acceleration be?

 

[babaliaris]

Question 1: What is the horizontal component of the net force?

You are right, when it passes through the vertical position the horizontal component of T is zero. At that point the horizontal component of the net force should also be zero because gravity doesn't have one as well, in fact, gravity is always zero on the x-axis along the whole motion. So $T_{x_{0}} + W_{x_{0}} = 0$

Question 2: According to Newton's second law, what should the horizontal component of the acceleration be?

$Σf = ma <= > -mgy_{0} + T1x_{0} + T2y_{0} = m(a1x_{0} + a2y_{0})$ so $a1x_{0} = \frac{T1}{m}x_{0}$

And if I'm not mistaken, |T| = 3mg since T is always constant in magnitude, only its direction changes.

 

[kuruman]

You are right, when it passes through the vertical position the horizontal component of T is zero. At that point the horizontal component of the net force should also be zero because gravity doesn't have one as well, in fact, gravity is always zero on the x-axis along the whole motion. So $T_{x_{0}} + W_{x_{0}} = 0$$Σf = ma <= > -mgy_{0} + T1x_{0} + T2y_{0} = m(a1x_{0} + a2y_{0})$ so $a1x_{0} = \frac{T1}{m}x_{0}$

And if I'm not mistaken, |T| = 3mg since T is always constant in magnitude, only its direction changes.

I agree that
$F_{\text{net,x}}=T_{x_0}+W_{x_0}=0.$
Newton says
$F_{\text{net,x}}=ma_x.$
Can you put the two equations together to find the horizontal acceleration?

 

[babaliaris]

Can you put the two equations together to find the horizontal acceleration?

It should be 0 right? $a_{x} = \frac{F_{net,x}}{m} = 0$ We are talking about when the m passes through the vertical position right?

 

[kuruman]

Right and right.

 

[babaliaris]

Now, what happens at a random point in time? Can you give me an idea on how to approach this problem in order to find out T as a vector? I believe the magnitude of T does not change, only it's direction, that's what my senses are telling me. If this is true, then from the calculations of my original post, |T| should be equal to |T| = 3mg. So I just need to find a way to describe that vector as a function of θ(t) which will change T's direction.

 

[kuruman]

Well, you senses are deceiving you here. Draw a free body diagram of the mass when it is at angle $\theta$ with respect to the vertical moving with speed $v$. Calculate the tension as a function of angle and $v$.

 

[babaliaris]

I'll try it and report back when I'm ready!

 

[babaliaris]

This is as far as I reached. I don't know what to do now since I'm not sure what a1, and a2 are equal to.

PS: I also attached a Pendulum Analysis.pdf file in case you can't read the pictures.

But these equations don't stand in this problem. The magnitude of V can't be Lω(t) since this would be true if it was in a circular motion, but in this problem, the pendulum oscillates in an arc motion.

 

[kuruman]

Diagram 2 is where you start. You should not substitute $x(t)$ in anything. This is how to proceed.

1. Consider perpendicular axes along the rope, radial or r-direction, and perpendicular to it, tangential or t-direction.
2. Write the components of the two forces in the r-direction.
3. Write Newton's second law in the r-direction. That should give you the tension in the rope when the pendulum is at angle $\theta$ and the pendulum mass has speed $v.$

 

[babaliaris]

 

[kuruman]

Please execute step 2 in post #11 and type your answers.

In the r-direction
T_r =
W_r =
a_r =

Also note that there is only one angle θ. All other pertinent angles are either θ or 90°- θ. There is no need for subscripts because cos(90°- θ) = sinθ and sin(90°- θ) =cosθ.

 

[babaliaris]

I can not understand what you mean by the r direction. Maybe I'm missing some mathematical knowledge?

 

[kuruman]

The r-direction is in the direction of the rope. See the diagram below. Vectors are red, axes are black.

All the mathematical knowledge you need is how to resolve a vector into components along two perpendicular directions.

 

[babaliaris]

But this is what I did, didn't I?

Then I calculated Wy and Wx where Wy is in the opposite r-direction as you describe.
T is always parallel with the r-direction which means it has only one component, the other one is zero.

 

[kuruman]

OK, if that's what you did, then the next step is to use what you did and write expressions for the component of the net force in the r-direction.

$F_{\text{net,r}} =~$ _______________

and the acceleration in the r-direction

$a_{\text{r}} =~$ _______________

Fill in the blanks. If you insist on posting hand-drawn diagrams, make sure the above are clearly shown as indicated above. Also, your expressions on the right-hand side must have no symbols other than $\theta$ and its trig functions, the mass $m$, the length $L$, the acceleration of gravity $g$, the speed $v$ and tension $T$.

Please do not use $\theta_1$, $\theta_2$ and $\theta_3$. The trig functions of all these angles can be reduced to functions of angle $\theta$ as defined in post #15. The goal is to find the dependence of the tension on this angle $\theta$.

 

[babaliaris]

do not use $\theta_1$, $\theta_2$ and $\theta_3$. The trig functions of all these angles can be reduced to functions of angle $\theta$ as defined in post #15. The goal is to find the dependence of the tension on this angle $\theta$.

Yes I know how to reduce it to just Θ and how to use the odd and even properties of sine and cosine to get rid off the 90 degrees as well.

The problem now is acceleration. This is the only thing (for now) that I don't know what it's equal to. I don't even know how to draw it on the diagram. Should I just draw it in a random angle, and just analyse it's components using sine and cosine as functions of Θ?

 

[PeroK]

Should I just draw it in a random angle, and just analyse it's components using sine and cosine as functions of Θ?

Yes, exactly!

 

[kuruman]

Yes I know how to reduce it to just Θ and how to use the odd and even properties of sine and cosine to get rid off the 90 degrees as well.

Then do it and post the results.

The problem now is acceleration. This is the only thing (for now) that I don't know what it's equal to. I don't even know how to draw it on the diagram. Should I just draw it in a random angle, and just analyse it's components using sine and cosine as functions of Θ?

The acceleration would not be a problem if you did what I asked you to do, namely,

1. Item 2 in post ##1 or
2. fill in the blanks in post #13 or
3. fill in the blanks in post #17

In general you would have to write two equations for Newton's second law, one in the r-direction and one in the t-direction (see post #15). For what you want to do here, the r-direction is of value and the t-direction is not. I want to help you but if you choose not to take my advice, I will have nothing more to say because I have said it three times already. If you don't understand or don't know how to follow my advice, just say so and I will help you along.

 

[Lnewqban]

...
The problem now is acceleration. This is the only thing (for now) that I don't know what it's equal to. I don't even know how to draw it on the diagram...

 

[Lnewqban]

@babaliaris
Regarding previous animation:
Please, note that at each instant of the oscillation, the shown acceleration (red vector) can be visualized as the vectorial addition of the centripetal and tangential accelerations.

As you can see, the value of the centripetal acceleration (tension in your cord) depends on the tangential velocity, therefore, it is zero when the pendulum momentarily stops, and maximum at the lowest point.

 

[kuruman]

@babaliaris
Regarding previous animation:
Please, note that at each instant of the oscillation, the shown acceleration (red vector) can be visualized as the vectorial addition of the centripetal and tangential accelerations.

As you can see, the value of the centripetal acceleration (tension in your cord) depends on the tangential velocity, therefore, it is zero when the pendulum momentarily stops, and maximum at the lowest point.

This is a nice animation. The animator also shows the conventional counterclockwise direction of increasing angle $\theta.$ Maybe I'm greedy but it would be nice if there were a way to show the angular velocity and angular acceleration vectors as well.

 

[babaliaris]

I think I figured it out (If I don't have any mistakes).

But I think I have. I calculated T and $tan(θ)sin(θ)-cos(θ)$ gives a negative number.
But T should always point towards the positive y-axis. Something feels wrong...

Can you tell me my mistake? Thank you!

 

[kuruman]

OK, you chose "y" instead of "r" for the direction along the rope. That's fine. Let's look at $T_y$. The first occurrence of the tension in your most recent work is here

First of all, the variables in this equation are scalar quantities, not vectors, so you should be putting the arrows over them. This equation is correct and is the one I wanted you to start from. So you have

$W_y+T=ma_y$

Now farther up in work you have equation (4) $W_y=mg\cos\theta$. That is incorrect. Do you see why? Hint: The trig function is correct. Also, you will need to find an expression for $a_y$ in terms of the speed $v$. Hint: the mass is describing a circle as it moves.

 

[kuruman]

Also, you cannot replace $|\vec a|$ in the top equation with $g$ and get the bottom equation. That is absolutely wrong because if the acceleration of the mass is $g$ that means it is in free fall which it isn't.

 

[babaliaris]

Is this correct about the magnitude of the acceleration?

As you said the magnitude can not be just g. But also tan(θ) can give negative answers, so how is this a correct magnitude? Maybe, because my coordinate system moves with the ball the system flips when the ball is on the right-hand side of the diagram? So that the positive x-axis will point to the left? If you see θ as measured from the "world" coordinate system (the one that has (0,0) in the wall where the rope is attached) when θ is between 180 and 270 degrees (this is what I draw in my diagram) then tan(θ) is positive (because sin and cos are negative there, both of them), this explains why the magnitude is positive there. But when θ is between 270-360 degrees, tan(θ) < 0 so it seems like $a_{x}$ flips horizontally.

Also, you will need to find an expression for $a_y$ in terms of the speed $v$. Hint: the mass is describing a circle as it moves.

Well if my math for the acceleration is correct, then I can say $a = gtan(θ)cos(θ)x_{0} + gtan(θ)sin(θ)y_{0}$ then integrate to find the velocity. Then I can take the magnitude of the velocity and integrate that to find the path of the motion $\frac{ds}{dt}=|v|$

 

[kuruman]

You are off the track. Let's try something simple. Consider a mass going around a circle of radius $R$ at constant speed $V$. Is it accelerating? If "yes", what is its acceleration? If "no", why not?

 

[babaliaris]

Yes it is accelerating but the acceleration chages only the direction of the velocity keeping it's magnitude constant. It's acceleration I believe is $R \cdot \frac{dω(t)}{dt}$

Edit:
By the way, I noticed something in my way of thinking when you said that these are scalar quantities (in the $ΣF_{y}=ma_{y}$). I thought the cos and sine were giving the direction of these vectors, but the truth is that these components were described with a 0 < θ < 90, since if you see the angle between the components, it's 90 degrees. So for example $W_{x} = mgsin(θ)$ or $a_{x} = |a| cos(θ)$ the trig functions give 0<= cos(θ) <= 1, 0<= sin(θ) <= 1 (only positive numbers). So these trig functions control the length of the magnitude.

In other words, I had to add signs as well when writing the ΣF equations which I didn't do.

 

[kuruman]

Yes it is accelerating but the acceleration chages only the direction of the velocity keeping it's magnitude constant. It's acceleration I believe is $R \cdot \frac{dω(t)}{dt}$

If there is acceleration in the direction of the velocity, the mass will not move at constant speed which is the kind of motion we are considering. The component of the acceleration $R\frac{d\omea}{dt}$ is the tangential acceleration and changes the speed. Here the mass is assumed to go around at constant speed. So try again, what is its acceleration?

By the way, I noticed something in my way of thinking when you said that these are scalar quantities (in the $ΣF_{y}=ma_{y}$). I thought the cos and sine were giving the direction of these vectors, but the truth is that these components were described with a 0 < θ < 90, since if you see the angle between the components, it's 90 degrees. So for example $W_{x} = mgsin(θ)$ or $a_{x} = |a| cos(θ)$ the trig functions give 0<= cos(θ) <= 1, 0<= sin(θ) <= 1 (only positive numbers). So these trig functions control the length of the magnitude.

You have defined x and y axes. In your drawing (shown below) you have vectors $\vec T$ and $\vec W$. Which of these two (if any) has a negative component in either direction? What is this component?

 

[kuruman]

In other words, I had to add signs as well when writing the ΣF equations which I didn't do.

OK, then do it and write the correct equation for Newton's second law in the y (or r) direction. Be sure to have the correct expression for the acceleration in terms of the speed $V$.

 

[babaliaris]

I think I solved it. By the way, I wanted to find out θ(t).

 

[kuruman]

I think I solved it. By the way, I wanted to find out θ(t).

I don't think you solved it correctly. The tension depends on the speed $V$. Your expression for it does not. If you want to find θ(t), you have to solve the differential equation $$\frac{d^2\theta}{dt^2}+\frac{g}{L}\sin\theta =0.$$ You can only solve it approximately, e.g. use the small-angle approximation $\sin\theta \approx \theta$ which of course is invalid here because the pendulum starts at the horizontal position. However, you can find the angular speed as a function of angle, $\omega(\theta).$

 

[babaliaris]

Can anyone point out my mistake? Maybe the mistake is the way I describe acceleration, it's the only thing I'm not completely sure if it is correct.

Maybe the way that I said $a_{T} = |a|cos(θ)$ $a_{Ν} = |a|sin(θ)$ and then calculated $|a|$ from $ΣF_{T}=ma_{T}$ it's not correct.

And instead, maybe I should have written the $ΣF_{N} = ma_{N} = m \frac{|V|^{2}}{L}$ and use this one to find the tension.

But I would like to know what I did wrong.

I don't see any mistakes in my math, so it must be something in my "Physics way of thinking".

 

[kuruman]

And instead, maybe I should have written the $ΣFN=maN=\frac{m|V|^2}{L}$ and use this one to find the tension.

That’s exactly what you should have done and what I meant when I said relate the tension and the speed. What is that relation?