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

[babaliaris]

$T = m[\frac{|V(t)|^2}{L} + gcos(θ(t))]$

And If I'm correct, this is not a conservative force since it depends on the velocity which depends on time, so, if I was using the principle of conservation of mechanical energy (potential energy in highest point + kinetic energy in highest point = potential energy at the lowest point + kinetic energy at the lowest point) it would be wrong.

 

[kuruman]

$T = m[\frac{|V(t)|^2}{L} + gcos(θ(t))]$

And If I'm correct, this is not a conservative force since it depends on the velocity which depends on time, so, if I was using the principle of conservation of mechanical energy (potential energy in highest point + kinetic energy in highest point = potential energy at the lowest point + kinetic energy at the lowest point) it would be wrong.

It doesn't matter if the tension is conservative or not. It is always perpendicular to the path and $dW=\mathbf{T}\cdot d\mathbf{s}=0$. This means that the tension does zero work on the mass. It changes the direction of the velocity $\mathbf{V}$ but not the speed $V$. Use energy conservation to find an expression for $mV^2$ in terms of $\theta$ and put in the equation.

 

[vanhees71]

It's easier with Lagrange I, i.e., you don't introduce the angle as a generalized parameter but work with (plane) Cartesian coordinates and imposing the constraint with a Lagrange parameter. The corresponding Lagrangian reads (setting $\vec{x}=(x,y)$)
$$L=\frac{m}{2} \dot{\vec{x}}^2 -mg y +\frac{\lambda}{2} (\vec{x}^2-L^2).$$
The Euler-Lagrange equations read
$$m \ddot{x} = \lambda x, \quad m \ddot{y}=-mg + \lambda y.$$
From this it follows
$$\vec{T}=m \ddot{x}+m g \vec{e}_y = \lambda \vec{x}.$$
To get $\lambda$ note that
$$m x \ddot{x} + m (\ddot{y}+g) y = \lambda(x^2+y^2)=\lambda L^2 \; \Rightarrow \; \lambda = \frac{m(\vec{x} \cdot \ddot{\vec{x}}+g y)}{L^2}.$$
Now you introduce $\theta$ via $\vec{x}=L(\sin \theta,-\cos \theta)$. Plugging this into the formula for $\lambda$ you find
$$\lambda=-\frac{m}{L} (L \dot{\theta}^2 + g \cos \theta).$$
Thus
$$\vec{T}=-m(L \dot{\theta}^2 + g \cos \theta) \begin{pmatrix}\sin \theta \\ -\cos \theta \end{pmatrix}$$
and
$$|\vec{T}|=m(L \dot{\theta}^2 + g \cos \theta).$$

 

[babaliaris]

It doesn't matter if the tension is conservative or not. It is always perpendicular to the path and $dW=\mathbf{T}\cdot d\mathbf{s}=0$. This means that the tension does zero work on the mass. It changes the direction of the velocity $\mathbf{V}$ but not the speed $V$. Use energy conservation to find an expression for $mV^2$ in terms of $\theta$ and put in the equation.

I understand what you are saying, thanks for this note.

I'm not sure about the potential energy when the pendulum is at its maximum angle, because I don't know what the height is there. I saw on the internet people taking the height = L when the angle of the pendulum is 90 degrees but is this correct? I don't know if the pendulum can reach that high. Maybe it does not matter in terms of energy, since they could reach there if I calmy v(0) = 0 leaving it to fall while holding it at 90 degrees.

It's easier with Lagrange I, i.e., you don't introduce the angle as a generalized parameter but work with (plane) Cartesian coordinates and imposing the constraint with a Lagrange parameter. The corresponding Lagrangian reads (setting $\vec{x}=(x,y)$)
$$L=\frac{m}{2} \dot{\vec{x}}^2 -mg y +\frac{\lambda}{2} (\vec{x}^2-L^2).$$
The Euler-Lagrange equations read
$$m \ddot{x} = \lambda x, \quad m \ddot{y}=-mg + \lambda y.$$
From this it follows
$$\vec{T}=m \ddot{x}+m g \vec{e}_y = \lambda \vec{x}.$$
To get $\lambda$ note that
$$m x \ddot{x} + m (\ddot{y}+g) y = \lambda(x^2+y^2)=\lambda L^2 \; \Rightarrow \; \lambda = \frac{m(\vec{x} \cdot \ddot{\vec{x}}+g y)}{L^2}.$$
Now you introduce $\theta$ via $\vec{x}=L(\sin \theta,-\cos \theta)$. Plugging this into the formula for $\lambda$ you find
$$\lambda=-\frac{m}{L} (L \dot{\theta}^2 + g \cos \theta).$$
Thus
$$\vec{T}=-m(L \dot{\theta}^2 + g \cos \theta) \begin{pmatrix}\sin \theta \\ -\cos \theta \end{pmatrix}$$
and
$$|\vec{T}|=m(L \dot{\theta}^2 + g \cos \theta).$$

I don't understand this math :P

 

[kuruman]

I understand what you are saying, thanks for this note.

I'm not sure about the potential energy when the pendulum is at its maximum angle, because I don't know what the height is there. I saw on the internet people taking the height = L when the angle of the pendulum is 90 degrees but is this correct? I don't know if the pendulum can reach that high. Maybe it does not matter in terms of energy, since they could reach there if I calmy v(0) = 0 leaving it to fall while holding it at 90 degrees.

I don't understand this math :P

Gravitational potential energy near the surface of the Earth is $U=mgy$ where $y$ is the vertical distance from an origin of your choice to the point where you want to find the potential energy. If you choose the lowest point of the motion as the origin, then what do you think is the vertical distance of the mass from the lowest point? Look at the picture that you yourself posted.

If you have not seen Lagrangians, you will not be able to understand the math in post #38.