Tangential and normal acceleration
Question
The electron again
The electron is joined by a tiny rubber blob of length and width each equal to $0.1$. Usual motion:
$$\mathbf f(t)=\langle \cos(t)(2-\cos(4t)),\sin(t)(2-\cos(4t)),\sin(4t)\rangle$$
Headwind (resp. tiny monkey) causes rubber blob to change length (resp. width) in proportion to the acceleration $a$ along the path (resp. the acceleration $b$ normal to the path).
$$\Delta(\textrm{length})=-0.01a\qquad \Delta(\textrm{width})=0.01b$$
- What are the length and width of the rubber blob at time $t$?
- More basic question: how can we describe the acceleration along the path or perpendicular to the path? What does it mean?
Examples
What is acceleration along and perpendicular to these paths?
- $(t^{45},0,0)$
- $(t^2-1,t^3-t,0)$
- $(\cos(t),\sin(t),t)$
Examples
First: $(t^{45},0,0), -1\leq t\leq 1$ loop
Examples
First: $(t^{45},0,0)$
Motion on a string!
The acceleration in the direction of motion is just the vector $\langle 1980 t^{43},0,0\rangle$. It always points in the direction of the motion (including negative scalars, so could point in opposite direction).
How is the sign change at $t=0$ reflected in the motion?
Examples
Second: $(t^2-1,t(t^2-1),0), -2\leq t\leq 2$ loop
acceleration
tangential component
normal component
Examples
Second: $(t^2-1,t(t^2-1),0)$, observations
- What we call "turning" looks like increased magnitude of normal component of acceleration.
- Tangential component of acceleration can be very negative for a while before things slow down!
- What else?
Examples
Second: $(t^2-1,t(t^2-1),0)$
To calculate: find the unit tangent and normal vectors, then calculate components of acceleration.
- Unit tangent: $\mathbf T(t)=\frac{1}{\sqrt{9t^4-2t^2+1}}\langle 2t,3t^2-1,0\rangle$
- Unit normal: $\mathbf N(t)=\frac{1}{\sqrt{9t^4-2t^2+1}}\langle 1-3t^2,2t,0\rangle$
- Acceleration: $\mathbf a(t)=\langle 2,6t,0\rangle$
- The components of $\mathbf a(t)$ in the directions of $\mathbf T(t)$ and $\mathbf N(t)$:
- $(\mathbf a(t)\cdot\mathbf T(t))\mathbf T(t)=\frac{18t^3-2t}{9t^4-2t^2+1}\langle 2t,3t^2-1,0\rangle$
- $(\mathbf a(t)\cdot\mathbf N(t))\mathbf N(t)=\frac{6t^2+2}{9t^4-2t^2+1}\langle 1-3t^2,2t,0\rangle$
- Which component dominates for large $t$? How about for $t$ near $0$? Consistent with observations?
Examples
Do the third: the helix $(\cos(t),\sin(t),t)$
To calculate: find the unit tangent and normal vectors, then calculate components of acceleration.
- Unit tangent: $\mathbf T(t)=$
- Unit normal: $\mathbf N(t)=$
- Acceleration: $\mathbf a(t)=$
- The components of $\mathbf a(t)$ in the directions of $\mathbf T(t)$ and $\mathbf N(t)$:
- $(\mathbf a(t)\cdot\mathbf T(t))\mathbf T(t)=$
- $(\mathbf a(t)\cdot\mathbf N(t))\mathbf N(t)=$
Examples
Do the third: the helix $(\cos(t),\sin(t),t)$
To calculate: find the unit tangent and normal vectors, then calculate components of acceleration.
- Unit tangent: $\mathbf T(t)=\frac{1}{\sqrt{2}}\langle -\sin(t),\cos(t),1\rangle$
- Unit normal: $\mathbf N(t)=\langle -\cos(t),-\sin(t),0\rangle$
- Acceleration: $\mathbf a(t)=\langle -\cos(t),-\sin(t),0\rangle$
- The components of $\mathbf a(t)$ in the directions of $\mathbf T(t)$ and $\mathbf N(t)$:
- $(\mathbf a(t)\cdot\mathbf T(t))\mathbf T(t)=\mathbf 0$
- $(\mathbf a(t)\cdot\mathbf N(t))\mathbf N(t)=\langle -\cos(t),-\sin(t),0\rangle$
- Why does the acceleration have no tangential component?
- (More troubling?) Why does the normal component have no $z$-component? How can the particle be climbing??
General theory
The procedure is always the same: find unit tangent and unit normal, calculate components.
The book has a discussion of various formulas, deductions using curvature, the product rule, etc.
The upshot: given a path $\mathbf r(t)$ with unit tangent $\mathbf T(t)$ and unit normal $\mathbf N(t)$, we can write the acceleration $\mathbf a=a_T\mathbf T+a_N\mathbf N$ where
$$a_T=\frac{\mathbf r'(t)\cdot\mathbf r''(t)}{|\mathbf r'(t)|}\qquad\quad a_N=\frac{|\mathbf r'(t)\times\mathbf r''(t)|}{|\mathbf r'(t)|}$$
Do one: the electron/rubber blob $$\mathbf f(t)=\langle \cos(t)(2-\cos(4t)),\sin(t)(2-\cos(4t)),\sin(4t)\rangle$$