Problem with understanding angular momentum

[mohamed_a]

I have a problem in understanding angular momentum equation (mrv), especially the part where radius is involved.

imagine an elastic collision occurred between sphere of mass (M) attached to a string forming a circle of radius (R) and moving with velocity (V) and another stationary sphere having the same variables but with lower case.
let M=m
mrv = mRV
rv = RV
if R = 2r
v = 2V

how is that possible although using kinetic energy gives the following:
0.5mv^2 = 0.5MV^2
v = V

Radius should be omitted from first equation since the speed is the same in two spheres and this doesn't mean that they will REVOLVE with same radial velocity since W = V/R so the sphere attached to ahorter rope will revolve faster.

 

[BvU]

Hello @mohamed_a ,

$\qquad$!​

You write

mrv = mRV

But that is not the conservation of angular momentum around one point of reference !
You choose the center of one circle for one anguar momentum and the center of the other circle for the other. These two centers can not be one and the same point if the radii differ.

$\$

 

[mohamed_a]

Hello @mohamed_a ,

$\qquad$!​

You write

But that is not the conservation of angular momentum around one point of reference !
You choose the center of one circle for one anguar momentum and the center of the other circle for the other. These two centers can not be one and the same point if the radii differ.

$\$

Does that mean kinetic energy approach is the best in this case and angular momentum approach describes the difference in speed between parts of the same rod (stick) attached to one pivot (the part near the pivot rotates slower)?

 

[BvU]

Does that mean kinetic energy approach is the best in this case

For a collision that is fully elastic the conservation of kinetic energy holds.

and angular momentum approach describes the difference in speed between parts of the same rod (stick) attached to one pivot (the part near the pivot rotates slower)?

I do not understand your question. Thus far there were no sticks in the thread.
Can you give a complete description of what you have in mind ?

 

[mohamed_a]

For a collision that is fully elastic the conservation of kinetic energy holds.
I do not understand your question. Thus far there were no sticks in the thread.
Can you give a complete description of what you have in mind ?

this is a separate topic i am saying if there was a stick attached to a pivot the angular momentum conservation describes how the part near the pivot moves slower than that farther away .Right?

And why would the angular momentum conservation work in that way (mvr is conserved) in a single frame of reference, why would velocity increase if the object came closer to the center. Shouldn't the velocity remain the same and thus the angular velocity would be the only variable increasing?

 

[jbriggs444]

this is a separate topic i am saying if there was a stick attached to a pivot the angular momentum conservation describes how the part near the pivot moves slower than that farther away .Right?

No.

Angular momentum conservation says that the stick will keep rotating at a constant rate as long as the pivot is frictionless and you do not exert a force on the stick anywhere else.

Angular momentum conservation also says that if you have a massive doughnut around the far end of the stick and reel it in (carefully not applying any torque as you do so), the stick will rotate more rapidly so that angular momentum still remains constant.

 

[PeroK]

if there was a stick attached to a pivot the angular momentum conservation describes how the part near the pivot moves slower than that farther away .Right?

That's a constraint of the system that the rod is rigid. If the rod is not held together, then it won't necessarily rotate with a single angular velocity.

And why would the angular momentum conservation work in that way (mvr is conserved) in a single frame of reference, why would velocity increase if the object came closer to the center. Shouldn't the velocity remain the same and thus the angular velocity would be the only variable increasing?

There is no conservation law for velocity. There is conservation of linear momentum, angular momentum and, in some cases, kinetic energy.

 

[mohamed_a]

No.

Angular momentum conservation says that the stick will keep rotating at a constant rate as long as the pivot is frictionless and you do not exert a force on the stick anywhere else.

Angular momentum conservation also says that if you have a massive doughnut around the far end of the stick and reel it in (carefully not applying any torque as you do so), the stick will rotate more rapidly so that angular momentum still remains constant.

I mean where does the extra velocity which the ,now stick with doughnut, gain come from? What i understand uptil now is that the doughnut at the far end is revolving with (V) and this velocity is larger than the velocity of the stick material in the middle of the stick. Now, when i move the doughnut towards the middle ,for example, the mass inside the dougnut is revolving with the same (V) and it resists the change to a lower velocity by pushing the middle of stick to (V), but why does the speed increase to (2V) according to conservation of angular momentum?

 

[jbriggs444]

I mean where does the extra velocity which the ,now stick with doughnut, gain come from?

Suppose that you have a rock on a string. If you reel in the string, that takes force. That force does work. That work increases the energy of the rock. The rock moves faster. That is an increase in speed.

Or, you can think of it like this. You pull in the rock. It acquires a bit of inward radial velocity. In one quarter of a revolution, that bit of inward radial velocity will have become a bit of tangential velocity. Because the direction you previously thought of as "inward radial" has become "forward tangential".

You might wonder whether the math works out.

The rock originally had tangential velocity $v_t$. Accordingly, it had kinetic energy $\frac{1}{2}{v_t}^2$. Suppose that you gave the rock radial velocity $v_r$. All other things being equal, this would have kinetic energy $\frac{1}{2}{v_r}^2$. If we combine the velocities with the Pythagorean theorem we get ${v_\text{tot}}^2 = {v_t}^2 + {v_r}^2$. One can see that the energies will add as they should. Whatever work you do that increases the rock's radial velocity, the resulting energy increase adds consistently to the rock's total velocity.

 

[PeroK]

I mean where does the extra velocity which the ,now stick with doughnut, gain come from? What i understand uptil now is that the doughnut at the far end is revolving with (V) and this velocity is larger than the velocity of the stick material in the middle of the stick. Now, when i move the doughnut towards the middle ,for example, the mass inside the dougnut is revolving with the same (V) and it resists the change to a lower velocity by pushing the middle of stick to (V), but why does the speed increase to (2V) according to conservation of angular momentum?

If you imagine a single particle in a circular path and you exert an additional force towards the centre, then it cannot remain in a circular path. The additional force pulls it into a spiral. Hence the additional force towards the centre is no longer perpendicular to the motion and increases the speed of the particle.

Something similar happens for a planet in an elliptical orbit. As the planet gets nearer the star, the gravitational force speeds the planet up; and slows it down again as it moves away.

If you want to move the particle to a smaller circular orbit without increasing its speed, you need a further force to oppose the direction of motion and prevent the particle speeding up as it moves inwards.

This can be understood by conservation of angular momentum, or by analysing the specific forces using polar coordinates.

 

[mohamed_a]

If you imagine a single particle in a circular path and you exert an additional force towards the centre, then it cannot remain in a circular path. The additional force pulls it into a spiral. Hence the additional force towards the centre is no longer perpendicular to the motion and increases the speed of the particle.

Something similar happens for a planet in an elliptical orbit. As the planet gets nearer the star, the gravitational force speeds the planet up; and slows it down again as it moves away.

If you want to move the particle to a smaller circular orbit without increasing its speed, you need a further force to oppose the direction of motion and prevent the particle speeding up as it moves inwards.

This can be understood by conservation of angular momentum, or by analysing the specific forces using polar coordinates.

You got my exact problem if you could refer me to a textbook or source that deeply analyzes this issue especially the part of analysing using polar coordinates, I would be very grateful .

 

[mohamed_a]

Suppose that you have a rock on a string. If you reel in the string, that takes force. That force does work. That work increases the energy of the rock. The rock moves faster. That is an increase in speed.

Or, you can think of it like this. You pull in the rock. It acquires a bit of inward radial velocity. In one quarter of a revolution, that bit of inward radial velocity will have become a bit of tangential velocity. Because the direction you previously thought of as "inward radial" has become "forward tangential".

You might wonder whether the math works out.

The rock originally had tangential velocity $v_t$. Accordingly, it had kinetic energy $\frac{1}{2}{v_t}^2$. Suppose that you gave the rock radial velocity $v_r$. All other things being equal, this would have kinetic energy $\frac{1}{2}{v_r}^2$. If we combine the velocities with the Pythagorean theorem we get ${v_\text{tot}}^2 = {v_t}^2 + {v_r}^2$. One can see that the energies will add as they should. Whatever work you do that increases the rock's radial velocity, the resulting energy increase adds consistently to the rock's total velocity.

I like the simple mathematic representation. If you know some sources that discuss physics topics, in general, (including this issue) deeply and in a way similar to your answer, then please recommend it.

 

[PeroK]

You got my exact problem if you could refer me to a textbook or source that deeply analyzes this issue especially the part of analysing using polar coordinates, I would be very grateful .

Most textbooks will derive the conservation of angular momentum in general from Newton's laws, then use that in any particular dynamics problem.

For example, acceleration in plane polar coordinates is given by:
$$\ddot{\mathbf r} = (\ddot r - r\dot \theta^2)\mathbf{\hat r} + (r\ddot \theta + 2\dot r \dot \theta)\mathbf{\hat \theta}$$That's an entirely mathematical derivation based solely on the definition of polar coordinates.

We now apply Newton's law $\mathbf F = m\mathbf a$ for a central force $\mathbf F = F \mathbf{\hat r}$ using polar coordinates above and we see that $a_{\theta} = 0$ and this gives:
$$ m(r\ddot \theta + 2\dot r \dot \theta) = 0$$Which leads to the the conservation of angular momentum:
$$\frac{d}{dt}(mr^2\dot \theta) = m(r^2\ddot \theta + 2r \dot r \dot \theta) = 0$$$$\Rightarrow \ mr^2\dot \theta = \text{constant}$$And, in particular, in the problem I described, as $r$ decreases, so $\dot \theta$ must increase in magnitude.

 

[mohamed_a]

Most textbooks will derive the conservation of angular momentum in general from Newton's laws, then use that in any particular dynamics problem.

For example, acceleration in plane polar coordinates is given by:
$$\ddot{\mathbf r} = (\ddot r - r\dot \theta^2)\mathbf{\hat r} + (r\ddot \theta + 2\dot r \dot \theta)\mathbf{\hat \theta}$$That's an entirely mathematical derivation based solely on the definition of polar coordinates.

We now apply Newton's law $\mathbf F = m\mathbf a$ for a central force $\mathbf F = F \mathbf{\hat r}$ using polar coordinates above and we see that $a_{\theta} = 0$ and this gives:
$$ m(r\ddot \theta + 2\dot r \dot \theta) = 0$$Which leads to the the conservation of angular momentum:
$$\frac{d}{dt}(mr^2\dot \theta) = m(r^2\ddot \theta + 2r \dot r \dot \theta) = 0$$$$\Rightarrow \ mr^2\dot \theta = \text{constant}$$And, in particular, in the problem I described, as $r$ decreases, so $\dot \theta$ must increase in magnitude.

The problem is the textbooks i use either give a very shallow explanation or much deeper explanation (with tensors and highly complex definitions) .I have a free access to springerlink and onlinewileylibrary, so i would like if you recommend a good book (something like: Differential Equations with Applications and Historical Notes) from these publisher

 

[PeroK]

The problem is the textbooks i use either give a very shallow explanation or much deeper explanation (with tensors and highly complex definitions) .I have a free access to springerlink and onlinewileylibrary, so i would like if you recommend a good book (something like: Differential Equations with Applications and Historical Notes) from these publisher

What are you looking for exactly?

 

[mohamed_a]

What are you looking for exactly?

I am looking for a book that connects differential equations and vector analysis with classical (and if possible quantum) physics and explains the every step followed to get the required physical quantity (how torque and angular momentum equations were slowly developed as an example)

 

[PeroK]

I am looking for a book that connects differential equations and vector analysis with classical (and if possible quantum) physics and explains the every step followed to get the required physical quantity (how torque and angular momentum equations were slowly developed as an example)

I've shown the derivation of the conservation of angular momentum for a particle in a central force above. More generally, we can see that:
$$\frac{d}{dt}(mr^2\dot \theta) = mr(r\ddot \theta + 2 \dot r \dot \theta) = mra_{\theta} = rF_{\theta} = \tau$$Where $\tau \equiv rF_{\theta}$ is the torque. And, that's the relationship between torque and angular momentum for a single particle in two dimensions.

The next step, which all introductary texts on classical mechanics take, is to extend that to a rigid body, which requires the concept of moment of inertia:

https://en.wikipedia.org/wiki/Moment_of_inertia

Then, the angular equivalent of Newton's second law becomes:$$\boldsymbol {\tau} = I\boldsymbol {\alpha}$$ where $I$ is the moment of inertia and $\boldsymbol {\alpha}$ is the angular acceleration.

Quantum mechanics is beyond the scope of this classical physics thread. That's another subject entirely.

 

[vanhees71]

Take as the most simple non-trivial example a system of two interacting particles within Newtonian mechanics. From the symmetry properties of Newtonian spacetime the particles must interact via a central interaction force that is derivable from a potential, i.e.,
$$V=V(|\vec{x}_1-\vec{x}_2|)=V(r), \quad \vec{r}=\vec{x}_1-\vec{x}_2.$$
From this you get the forces acting on the particles
$$\vec{F}_1=-\vec{\nabla}_1 V=-\frac{\vec{r}}{r} V'(r),$$
$$\vec{F}_2=+\frac{\vec{r}}{r} V'(r)=-\vec{F}_1.$$
The equations of motion read
$$m_1 \ddot{\vec{x}}_1=-\frac{\vec{r}}{r} V'(r), \quad m_2 \ddot{\vec{x}}_2 = +\frac{\vec{r}}{r} V'(r).$$
The total angular momentum is given by
$$\vec{J}=m_1 \vec{x}_1 \times \dot{\vec{x}}_1+m_2 \vec{x}_2 \times \dot{\vec{x}}_2.$$
Taking the time derivative gives
$$\dot{\vec{J}} = m_1 \vec{x}_1 \times \ddot{\vec{x}}_1 + m_2 \vec{x}_2 \times \ddot{\vec{x}}_2.$$
With the equations of motion this becomes
$$\dot{\vec{J}}=\vec{x}_1 \times \vec{F}_1 + \vec{x}_2 \times \vec{F}_2 = (\vec{x}_1-\vec{x}_2) \times \vec{F}_1 = -\vec{r} \times \frac{\vec{r}}{r} V'(r)=0.$$
That shows that the total angular momentum of a closed system of two particles always is conserved.

The same can be derived for arbitrary closed systems of point particles. The conservation of angular momentum follows from the symmetry of Galilei-Newton spacetime under rotations (isotropy of space), which restricts the allowed forms for the interaction forces, such that angular momentum is always conserved (Noether's theorem).

 

[mohamed_a]

I've shown the derivation of the conservation of angular momentum for a particle in a central force above. More generally, we can see that:
$$\frac{d}{dt}(mr^2\dot \theta) = mr(r\ddot \theta + 2 \dot r \dot \theta) = mra_{\theta} = rF_{\theta} = \tau$$Where $\tau \equiv rF_{\theta}$ is the torque. And, that's the relationship between torque and angular momentum for a single particle in two dimensions.

The next step, which all introductary texts on classical mechanics take, is to extend that to a rigid body, which requires the concept of moment of inertia:

https://en.wikipedia.org/wiki/Moment_of_inertia

Then, the angular equivalent of Newton's second law becomes:$$\boldsymbol {\tau} = I\boldsymbol {\alpha}$$ where $I$ is the moment of inertia and $\boldsymbol {\alpha}$ is the angular acceleration.

Quantum mechanics is beyond the scope of this classical physics thread. That's another subject entirely.

but if $$a_{\theta } = 0$$ then:
$$r\dot{\theta}\hat{\theta } = constant = C$$
$$r \omega\hat{\theta} = C$$
$$v\hat{\theta} = C$$
so v will remain constant. this is th problem where does the extra r (radius) comes from?

 

[mohamed_a]

Take as the most simple non-trivial example a system of two interacting particles within Newtonian mechanics. From the symmetry properties of Newtonian spacetime the particles must interact via a central interaction force that is derivable from a potential, i.e.,
$$V=V(|\vec{x}_1-\vec{x}_2|)=V(r), \quad \vec{r}=\vec{x}_1-\vec{x}_2.$$
From this you get the forces acting on the particles
$$\vec{F}_1=-\vec{\nabla}_1 V=-\frac{\vec{r}}{r} V'(r),$$
$$\vec{F}_2=+\frac{\vec{r}}{r} V'(r)=-\vec{F}_1.$$
The equations of motion read
$$m_1 \ddot{\vec{x}}_1=-\frac{\vec{r}}{r} V'(r), \quad m_2 \ddot{\vec{x}}_2 = +\frac{\vec{r}}{r} V'(r).$$
The total angular momentum is given by
$$\vec{J}=m_1 \vec{x}_1 \times \dot{\vec{x}}_1+m_2 \vec{x}_2 \times \dot{\vec{x}}_2.$$
Taking the time derivative gives
$$\dot{\vec{J}} = m_1 \vec{x}_1 \times \ddot{\vec{x}}_1 + m_2 \vec{x}_2 \times \ddot{\vec{x}}_2.$$
With the equations of motion this becomes
$$\dot{\vec{J}}=\vec{x}_1 \times \vec{F}_1 + \vec{x}_2 \times \vec{F}_2 = (\vec{x}_1-\vec{x}_2) \times \vec{F}_1 = -\vec{r} \times \frac{\vec{r}}{r} V'(r)=0.$$
That shows that the total angular momentum of a closed system of two particles always is conserved.

The same can be derived for arbitrary closed systems of point particles. The conservation of angular momentum follows from the symmetry of Galilei-Newton spacetime under rotations (isotropy of space), which restricts the allowed forms for the interaction forces, such that angular momentum is always conserved (Noether's theorem).

The problem is not with conservation of angular momentum its about why angular momnetum has r in it? why $$ \vec{J}=m_1 \vec{x}_1 \times \dot{\vec{x}}_1 $$ has $$\vec{x}_1$$? Is it something related to the definition or terminology?

 

[PeroK]

but if $$a_{\theta } = 0$$ then:
$$r\dot{\theta}\hat{\theta } = constant = C$$

If $$a_{\theta } = 0$$ then:$$r^2\dot \theta = \text{constant}$$

 

[PeroK]

The problem is not with conservation of angular momentum its about why angular momnetum has r in it? why $$ \vec{J}=m_1 \vec{x}_1 \times \dot{\vec{x}}_1 $$ has $$\vec{x}_1$$? Is it something related to the definition or terminology?

The post by @vanhees71 is well beyond your level of understanding at this stage.

 

[mohamed_a]

If $$a_{\theta } = 0$$ then:$$r^2\dot \theta = \text{constant}$$

but according to velocity : $$ v = \dot{r}\hat{r} + r\theta \hat{\theta} $$, so if there is no acceleration the part in theta direction, must remain constant where did the multiplied r come from?

 

[PeroK]

but according to velocity : $$ v = \dot{r}\hat{r} + r\theta \hat{\theta} $$, so if there is no acceleration the part in theta direction, must remain constant where did the multiplied r come from?

I showed you the acceleration in plane polar coordinates in post #13 above:

For example, acceleration in plane polar coordinates is given by:
$$\ddot{\mathbf r} = (\ddot r - r\dot \theta^2)\mathbf{\hat r} + (r\ddot \theta + 2\dot r \dot \theta)\mathbf{\hat \theta}$$That's an entirely mathematical derivation based solely on the definition of polar coordinates.

You can find a derivation of that here, for example:

https://ocw.mit.edu/courses/aeronau...fall-2009/lecture-notes/MIT16_07F09_Lec05.pdf

If you think that:$$\ddot{\mathbf r} = \ddot r \mathbf{\hat r} + r\ddot \theta \boldsymbol{ \hat \theta}$$then you won't be the first or last student to make that mistake.

 

[mohamed_a]

Thank you for holding up with me uptill now, i found the solution in this link :

https://ocw.mit.edu/courses/aeronau...fall-2009/lecture-notes/MIT16_07F09_Lec15.pdf

,it's a link to another MIT lecture (it is an amazing source, that's i what i needed), the problem is i forgot the power rule of functions (so trivial)

this discussion has been very useful to me and I am greatful to you so much. May god guide and bliss you.

if you know another source like MIT lectures then please recommend it to me

 

[vanhees71]

The post by @vanhees71 is well beyond your level of understanding at this stage.

It's much simpler than the calculation in polar coordinates ;-)) and much more generally valid at the same time.

 

[mohamed_a]

It's much simpler than the calculation in polar coordinates ;-)) and much more generally valid at the same time.

The problem wasn't proving angular momentum conservation but it was the reason why angular momentum had $$r^2$$ and not $$r$$ only but turns out it is a part of the derivation where I missed the power rule of function

 

[vanhees71]

I guess you refer to the special case of single particle moving on a circle with constant angular velocity $\vec{\omega}$. Take the 3-axis as the rotation axis. Then the trajectory of the particle is described by
$$\vec{x}(t)=r \begin{pmatrix} \cos(\omega t) \\ \sin(\omega t) \\0 \end{pmatrix}.$$
The velocity is
$$\dot{\vec{x}}(t)=r\omega \begin{pmatrix} -\sin(\omega t) \\ \cos (\omega t) \\ 0 \end{pmatrix}.$$
And the angular momentum with the center point in the origin (i.e. the center of the circle) is
$$\vec{L}=m \vec{x} \times \dot{\vec{x}}=r^2 \omega \begin{pmatrix}0\\0\\1 \end{pmatrix}.$$
That's of course equivalent to the above representation in cylindrical coordinates.

 

[mohamed_a]

I guess you refer to the special case of single particle moving on a circle with constant angular velocity $\vec{\omega}$. Take the 3-axis as the rotation axis. Then the trajectory of the particle is described by
$$\vec{x}(t)=r \begin{pmatrix} \cos(\omega t) \\ \sin(\omega t) \\0 \end{pmatrix}.$$
The velocity is
$$\dot{\vec{x}}(t)=r\omega \begin{pmatrix} -\sin(\omega t) \\ \cos (\omega t) \\ 0 \end{pmatrix}.$$
And the angular momentum with the center point in the origin (i.e. the center of the circle) is
$$\vec{L}=m \vec{x} \times \dot{\vec{x}}=r^2 \omega \begin{pmatrix}0\\0\\1 \end{pmatrix}.$$
That's of course equivalent to the above representation in cylindrical coordinates.

It is a nice representation. What I meant is in PeroK answer it was proven that $$mr^2\dot{\theta} = mvr = constant$$ in a single point around a center, and since momentum is a physical quantity that is reserved under no force exertion so angular momentum = mvr. My problem was why angular momentum is not mv or $$m\dot{x} $$ only. I know that angular momentum is moment of momentum and that's why it is equal to $$p. r$$ but that didn't mean sense to me I had to find out why the radius was crucial to be present in $$ p. r = mvr = mx. \dot{x} $$

 

[vanhees71]

You should use vectors always. It can only lead to confusion, if you try to learn mechanics without vectors. The correct definitions are: $\vec{x}=\vec{x}(t)$ is a position vector, referring to an inertial frame of reference (with an "origin" O and $\vec{x}=\overrightarrow{OP}$, where $P$ is the position of the particle at time $t$. Then you define velocity and acceleration, which are time derivatives of the position vector and thus also vectors:
$$\vec{v}=\dot{\vec{x}}=\frac{\mathrm{d}}{\mathrm{d} t} \vec{x}, \quad \vec{a}=\dot{\vec{v}}=\ddot{\vec{x}}.$$
Then you define the momentum as
$$\vec{p}=m \vec{v}.$$
Since the mass, $m$, is a scalar quantity and $\vec{v}$ is a vector, $\vec{p}$ is also a vector.

Finally the angular momentum (wrt. the chosen origin $O$ of the reference frame) is defined by
$$\vec{L}=\vec{x} \times \vec{p}=m \vec{x} \times \vec{v}$$
and thus also a vector.

Now you can of course use any kind of coordinates to specify your position vector and all the vectors derived from it. The most simple choice are Cartesian coordinates. They are defined by choosing an arbitrary (right-handed) system of three unit vectors which are pair-wise perpendicular to each other, i.e., these three vectors fulfill the condition $\vec{e}_1 \times \vec{e}_2=\vec{e}_3$. Then you can write
$$\vec{x}=x_j \vec{e}_j,$$
where the sum over repeated indices is understood (Einstein summation convention). It's also convenient to associate the Cartesian components in terms of column vectors,
$$(x_j)=\begin{pmatrix}x_1 \\ x_2 \\ x_3 \end{pmatrix}.$$
Since the $\vec{e}_j$ are time-independent the time derivatives of the vectors are mapped uniquely to the time derivatives of the Cartesian components, e.g.,
$$\vec{v} = \dot{\vec{x}} = \frac{\mathrm{d}}{\mathrm{d} t}(x_j \vec{e}_j) = \vec{e}_j \frac{\mathrm{d}}{\mathrm{d} t} x_j = e_j \dot{x}_j$$
and thus
$$(v_j)=(\dot{x}_j)=\begin{pmatrix} \dot{x}_1 \\ \dot{x}_2 \\ \dot{x}_3 \end{pmatrix}.$$
Before you go further and introduce more general coordinates like the polar coordinates used above, you should get familiar with this most simple formalism for Cartesian coordinates first!

 

[PeroK]

You should use vectors always. It can only lead to confusion, if you try to learn mechanics without vectors.

Particularly when studying angular momentum, which requires motion in at least two dimensions.