Physics with Calculus/Mechanics/Rotational Motion

Centripetal Acceleration

We will attempt to develop some sort of understanding of a particle moving in any given motion. Since any given small motion can be thought of as an infinitessimal screw motion, and we already know about linear motion, this will mainly focus on rotational motion.

The position of the particle, of mass m, as a function of time is

$\overrightarrow{x}(t)=(x(t),y(t),z(t))$.

It's velocity and acceleration we define as

$\overrightarrow{v}=\frac{d\overrightarrow{x}}{dt}\equiv {\overrightarrow{x}}^{\prime }$

$\overrightarrow{a}=\frac{d\overrightarrow{v}}{dt}\equiv {\overrightarrow{v}}^{\prime }$

where the primes indicate differentiation with respect to time. It is possible to write position not only as a function of time, but as other things as well. We can in fact write x as a function of u where u is a sufficiently nice function of t. Of course there are restrictions, such as u must have the same range as t. Never the less, the most important parameterization is writing x in terms of arclength, s. In practice we will never find something as a function of arclength, but I assure you, it is possible. Arclength is, of course, the "length" of the curve. We could be very mathematical here and rigorously define arclength and say when it exists, and prove everything, but that is not necessary since we are only looking for the results. For notation, a dot above something will indicate differentiation with respect to arclength, two dots twice differentiating, etc.

$\overrightarrow{T}=\dot{\overrightarrow{x}}$

T is tangent to the curve, because as two points on x come closer and closer together, the vector between them, which has the same direction as T, becomes closer and closer to the tangent. T is also a unit vector, meaning it has magnitude 1 because as two points on x come closer and closer together, the distance between becomes closer and closer to the arclength between them. Since T is a unit vector, we write $\hat{T}$.

$\frac{d\hat{T}}{ds}=\overrightarrow{N}$

N is perpendicular to T because

$\frac{d\hat{T}\cdot \hat{T}}{ds}=\frac{d1}{ds}$

$2\overrightarrow{N}\cdot \hat{T}=0$

We call the magnitude of N $\kappa$ and the direction $\hat{N}$.

Thus, we can write velocity as

$\hat{T}\frac{ds}{dt}$

because s' is simply the arclength per second, which is the definition of speed. Acceleration becomes

$\overrightarrow{a}=\frac{d\hat{T}}{dt}\frac{ds}{dt}+\hat{T}\frac{d^{2}s}{d{t}^2}=\kappa {\left(\frac{ds}{dt}\right)}^2\hat{N}+\frac{d^{2}s}{d{t}^2}\hat{T}$.

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 $\hat{T}\times \hat{N}$ direction. Furthermore, the acceleration in the tangent direction changes speed and only speed, whereas acceleration in the normal direction changes direction and only direction. Thus, perhaps somewhat counterintuitively, something in circular orbit feels only a force directed toward the center of the circle. You can see this when you swing something around on a rope -- there is not tangental force or else the rope would bend.

Let's examine the coefficient in front of the N term a little bit. It turns out $\kappa$ is $1/\rho$ where $\rho$ is the radius of curvature. It has units of length, and it is the radius of the circle which most closely approximates the curve locally (you can try to prove this if you want or try out a few examples). This means that for a particle moving in a circle, the normal acceleration it must feel is

$\frac{||v|{|}^2}{r}$.

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 oppostite 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 they 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.