Physics with Calculus/Mechanics/Rotational Motion

Centripetal Acceleration

Some notational issues

<---! Some of this discussion might belong in another place in the book!--->

1. It is often convenient to write vectors in component form, without unit vectors, for example: ${\vec {r}}(t)=\left[\,(x(t),y(t),z(t)\,\right]$
2. The ${\vec {r}}$ is unique in that it's components are not \left[\,r_x, r_y,r_z\,\right] </math>
3. There exists an irreparable ambiguity regarding differential's of vectors. It is universally recognized that $\Delta r$ represents the differential of the magnitude, $\Delta r\equiv \Delta \left|{\vec {r}}\,\right|$. On the other hand, in uniform circular motion, $|\,v\,|$ is a constant, so it might be more convenient (on board or paper) to write $\Delta v$ as $\left|\Delta {\vec {v}}\,\right|$. So, should we write such non-standard notation into text?

Rigorous discussion of acceleration along a curved path

Simply stated, the acceleration of a particle consists of two parts:

${\vec {a}}={\frac {\rm {dv}}{\rm {dt}}}{\hat {\tau }}+{\frac {v^{2}}{R}}{\hat {\eta }}$

where ${\hat {\eta }}$ and ${\hat {\tau }}$ are unit vectors (called eta-hat and tau-hat); ${\hat {\tau }}$ points tangential to the path, and ${\hat {\eta }}$ points to the center of curvature (perpendicular to ${\hat {\tau }}$). The distance to the center of curvature is R. It is common to use the symbol ${\rm {d\ell }}$ to denote a small change in position. Using component notation for vectors:

${\rm {d{\vec {\ell }}\equiv {\rm {d{\vec {r}}=\left[\,{\rm {dx,{\rm {dy,{\rm {dz\,}}}}}}\right]}}}}$.

Some authorsrefer to this as ${\rm {ds}}$. In both electromagnetics (Ampere's Law) and general relativity (proper time) the scalar magnitude of this vector is important:

${\rm {d\ell ={\sqrt {dx^{2}+dy^{2}+dz^{2}}}}}$

It is convenient to use dots to denote differential with respect to time:

${\vec {v}}={\frac {d{\vec {x}}}{dt}}\equiv {\dot {\vec {x}}}$

${\vec {a}}={\frac {d{\vec {v}}}{dt}}\equiv {\dot {\vec {v}}}$

Note that $v\equiv |{\vec {v}}|={\frac {d\ell }{dt}}$. Next we define the tangent unit vector ${\hat {\tau }}$, as a unit vector parallel to the velocity:

${\hat {\tau }}={\vec {v}}/v$

We now show that the differential of ${\hat {\tau }}$ is normal (perpendicular) to ${\hat {\tau }}$. We begin by defining the ${\vec {\eta }}$ vector as follows:

${\vec {\eta }}\equiv {\frac {d{\hat {\tau }}}{d\ell }}$

Since (by definition of unit vector) the derivative of ${\hat {\tau }}\cdot {\hat {\tau }}$ vanishes, we have:

${\frac {d({\hat {\tau }}\cdot {\hat {\tau }})}{d\ell }}=0=2{\vec {\eta }}\cdot {\hat {\tau }}$

What makes ${\hat {\eta }}$ and ${\hat {\tau }}$ interesting is that they depend only on the path taken by the particle through space, and not on the rate (speed) at which the particle takes this path. Define the magnitude of ${\vec {\eta }}$ to be $\kappa$, so that:

$\kappa {\hat {\eta }}={\vec {\eta }}$.

Thus, we can take the derivative of velocity, ${\vec {v}}={\hat {\tau }}({\rm {d\ell /dt)}}$, as the derivative of the product of two terms:

${\vec {a}}={\frac {d{\vec {v}}}{dt}}={\frac {d{\hat {\tau }}}{dt}}{\frac {d\ell }{dt}}+{\hat {\tau }}{\frac {d^{2}\ell }{dt^{2}}}=\kappa \left({\frac {d\ell }{dt}}\right)^{2}{\hat {\eta }}+{\frac {d^{2}\ell }{dt^{2}}}{\hat {\tau }}$.

This means first that a particle never accelerates in any direction except the normal or the tangent (or of course a combination of both). That is, it never accelerates at all in the ${\hat {\tau }}\times {\hat {\eta }}$ direction. Furthermore, the acceleration in the tangent direction changes speed and only speed, whereas acceleration in the normal direction changes direction and only direction. To discern the meaning of $\kappa$, we consider a particle making uniform circular motion about a circle of radius R. In x-y space,

${\vec {r}}=[R\cos \omega t,R\sin \omega t]$

It is easily shown that $\kappa$ (called the curvature) is given by,

${\vec {r}}=\kappa ={\frac {1}{R}}$(this problem is unsolved)

Discussion temp

Memorize this. It is known as centripetal acceleration. If you remember nothing else from this section, remember that formula. Notice that the centripetal acceleration is directed toward the center of the circle, not out. You might beg to differ, having gone on a merry-go-round or carousel because you feel a definite force outward. Many people like to say "that is not a real force, it is simply inertia." You might say, "yes, but I know I feel a force." And indeed you do feel a force, and it is a real force in a spinning reference frame. Say you are a particle moving around on a circle, then that means there is some force pulling you toward the center, call it F.

$F=m{\frac {v^{2}}{R}}$.

But to you, you don't see yourself moving, so you say the sum of the forces is zero, and write out

$F_{total}=F-m{\frac {v^{2}}{R}}=0$.

Lo and behold, on the left hand side is a mysterious force, the centrifugal force, and it is pointing the opposite direction, radially outward! In a rotating frame, you have to add this extra centrifugal force term (along with another one, the Coriolis force) to make Newton's second law true.

Rotational Kinematics

If you know linear kinematics, rotational kinematics are a breeze. We use exactly the same arguments to construct almost identical formulas.

For rotational motion, we first assume that everything is in a plane, and that everything moves in a circle. This is a rather boring situation, so we will generalize it in a bit. The angle between some reference point (usually the x axis) and the particle. The angle can have any real value, and it will never jump from $2\pi$ to 0 or anything funny like that. The derivative of angle is angular velocity $\omega$ and the derivative of that is angular acceleration $\alpha$. Of course, by the very definition of angle,

$\theta ={\frac {s}{R}}$

where s is the arclength. Thus we have

$\omega ={\frac {v}{R}}$

and

$\alpha ={\frac {a}{R}}$.

All the normal kinematic equations hold, replacing a with $\alpha$, v with $\omega$, and x with $\theta$.

Rotational Dynamics

We can describe the position of a particle with a vector. If we add two displacement vectors, we get the total displacement, and it does not matter what order we add them in. However, there is no such vector quantity for angles, because rotations do not commute, meaning that it matters what order you add them in. For example rotating a book 90 degrees around the horizontal then vertical axis is not the same as rotating 90 degrees around the vertical then horizontal axis.

It can be shown that very small rotations do commute, so it is possible to define a vector for the rate of change of the angle. While the angle is not a vector, the rate of change of an angle is.

We can define the angular velocity, $\omega =(d\alpha /dt,d\beta /dt,d\gamma /dt)$ where $\alpha ,\beta ,\gamma$ are the angles to the x, y, and z axes respectively. The angular velocity of an object has magnitude equal to the speed of a particle going around the axis of rotation at unit radius, the direction of the axis, and sense according to the right hand rule.