0
$\begingroup$

I was reading 1.11.2 in Introduction to Mechanics by Kleppner and Kolenkow when they started discussing the derivative of the r unit vector with respect to time. I understand that a change in r would be a new vector $r(\theta + \delta\theta) - r(\theta)$. But how would we incorporate time into it? What would the rate of change of the unit radius vector look like?

$\endgroup$
1
  • 1
    $\begingroup$ I do not understand why this question has been asked as in section 1.11.2 on page 29 the equation $\frac{d \bf\hat r}{dt}=\dot \theta \bf\hat \theta$ is derived. $\endgroup$
    – Farcher
    Commented Nov 2, 2022 at 23:24

4 Answers 4

1
$\begingroup$

The unit vector $V(t)$ has coordinates $(\cos(u(t)),\sin(u(t))$. We have thus $$ V' =dV/dt = (-\sin(u(t))u', \cos(u(t)) u')= u'(-\sin(u(t)), \cos(u(t))) $$ the modulus of $V'$ is thus $u'$ and $V'$ is orthogonal to $V$.

$\endgroup$
1
$\begingroup$

If you have a unit vector that stays a unit vector while it changes in time, then it is changing direction. This is a rotation.

In that case the derivative would be the angular velocity. How fast it's changing direction, and what direction it's changing. By convention that gets written as a vector perpendicular to both the unit vector and to the direction of change, according to the right-hand rule.

$\endgroup$
1
$\begingroup$

It represents the instantaneous velocity vector directed along the tangent to the moving path. It is the sum of resolved radial ( same direction as position vector) and ( the perpendicular) circumferential components of velocity.

The changes occur whether or not the initial position vector is of unit length or not.

$\endgroup$
1
$\begingroup$

Often in physics problems, how each variable depends on all the other variables is left to be implicit. Although this makes it more convenient to write things it can obscure certain details, making the problem more difficult.

A general (time-dependent) position vector in polar coordinates is written as follows $$\vec r(r(t),\theta(t)),$$ i.e. it depends on the radius and angle which can both be functions of time. We restrict ourselves to motion on the unit circle so we can set $r(t)=1$. Now our function only depends on $\theta$: $$\vec r(\theta(t))$$ The derivative of this function with respect to time can be calculated by applying the vector version of the chain rule. \begin{align} \frac d{dt}\vec r(\theta(t))&=\theta'(t)\frac d{d\theta}\vec r(\theta(t))\\ &=\theta'(t)\begin{pmatrix}\frac{d{x}}{d{\theta}}\\\frac{d{y}}{d{\theta}}\end{pmatrix} \end{align} After knowing this explicit definition you might be still confused how you can take the derivative of a vector which is constrained to lie on the unit circle. The answer is that it is still possible as long as this derivative (read: the velocity) is perpendicular to $\vec r$ itself. This can be shown using the following proof. Assume $\vec r$ can depend on $r$ and $\theta$ and then impose $\frac d {dt}(\vec r\cdot \vec r)=0$, which is equivalent to saying the radius is constant. Working out this expression gives $\vec r\cdot \vec v=0$, which means $\vec v$ has to be perpendicular to $\vec r$ in order to have a constant radius.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.