Motion in Space: Learn It 2

Components of the Acceleration Vector

We can combine concepts from arc length and curvature with the acceleration vector to gain a deeper understanding of how this vector relates to motion. Recall that the unit tangent vector [latex]{\bf{T}}[/latex] and the unit normal vector [latex]{\bf{N}}[/latex] form an osculating plane at any point [latex]P[/latex] on the curve defined by [latex]{\bf{r}}(t)[/latex]. The acceleration vector [latex]{\bf{a}}(t)[/latex] always lies in this plane and can be written as a combination of these two unit vectors.

the plane of the acceleration vector

The acceleration vector [latex]{\bf{a}}(t)[/latex] of an object moving along a curve traced out by a twice-differentiable function [latex]{\bf{r}}(t)[/latex] lies in the plane formed by the unit tangent vector [latex]{\bf{T}}(t)[/latex] and the principal unit normal vector [latex]{\bf{N}}(t)[/latex].
[latex]\\[/latex]

The acceleration vector can be expressed as:

[latex]{\bf{a}}(t)=v'(t){\bf{T}}(t)+[v(t)]^{2}\kappa{\bf{N}}(t)[/latex]

where [latex]v(t)[/latex] is the speed of the object and [latex]\kappa[/latex] is the curvature of the curve traced out by [latex]{\bf{r}}(t)[/latex].

Proof

Because [latex]{\bf{v}}(t)={\bf{r}}'(t)[/latex] and [latex]{\bf{T}}(t)=\frac{{\bf{r}}'(t)}{\left\Vert{\bf{r}}'(t)\right\Vert}[/latex], we have [latex]{\bf{v}}(t)=\left\Vert{\bf{r}}'(t)\right\Vert{\bf{T}}(t)=v(t){\bf{T}}(t)[/latex]. Now we differentiate this equation:

[latex]{\bf{a}}(t)={\bf{v}}'(t)=\frac{d}{dt}(v(t){\bf{T}}(t))=v'(t){\bf{T}}(t)+v(t){\bf{T}}'(t).[/latex]

Since [latex]{\bf{N}}(t)=\frac{{\bf{T}}'(t)}{\left\Vert{\bf{T}}'(t)\right\Vert}[/latex], we know [latex]{\bf{T}}'(t)=\left\Vert{\bf{T}}'(t)\right\Vert{\bf{N}}(t)[/latex], so

[latex]{\bf{a}}(t)=v'(t){\bf{T}}(t)+v(t)\left\Vert{\bf{T}}'(t)\right\Vert{\bf{N}}(t).[/latex]

A formula for curvature is [latex]\kappa=\frac{\left\Vert{\bf{T}}'(t)\right\Vert}{\left\Vert{\bf{r}}'(t)\right\Vert}[/latex], so [latex]\left\Vert{\bf{T}}'(t)\right\Vert=\kappa\left\Vert{\bf{r}}'(t)\right\Vert=\kappa v(t)[/latex]. This gives [latex]{\bf{a}}(t)=v'(t){\bf{T}}(t)+\kappa(v(t))^{2}{\bf{N}}(t)[/latex].

[latex]_\blacksquare[/latex]

The coefficients of [latex]{\bf{T}}(t)[/latex] and [latex]{\bf{N}}(t)[/latex] in this formula have special names. We call [latex]v'(t)[/latex] the tangential component of acceleration and write it as [latex]a_{\bf{T}}[/latex]. We call [latex][v(t)]^{2}\kappa[/latex] the normal component of acceleration and write it as [latex]a_{\bf{N}}[/latex].

tangential and normal components of acceleration

Let [latex]{\bf{r}}(t)[/latex] be a vector-valued function that denotes the position of an object as a function of time. Then [latex]{\bf{a}}(t)={\bf{r}}''(t)[/latex] is the acceleration vector. The tangential and normal components of acceleration are:

  • Tangential component: [latex]a_{\bf{T}}={\bf{a}}\cdot{\bf{T}}=\frac{{\bf{v}}\cdot{\bf{a}}}{\left\Vert{\bf{v}}\right\Vert}[/latex]
  • Normal component: [latex]a_{\bf{N}}={\bf{a}}\cdot{\bf{N}}=\frac{\left\Vert{\bf{v}}\times{\bf{a}}\right\Vert}{\left\Vert{\bf{v}}\right\Vert}=\sqrt{{\left\Vert{\bf{a}}\right\Vert}^{2}-a^{2}_{\bf{T}}}[/latex]

These components are related by:

[latex]{\bf{a}}(t)=a_{\bf{T}}{\bf{T}}(t)+a_{\bf{N}}{\bf{N}}(t)[/latex]

where [latex]{\bf{T}}(t)[/latex] is the unit tangent vector and [latex]{\bf{N}}(t)[/latex] is the unit normal vector to the curve.

Understanding Centripetal Acceleration

The normal component of acceleration is also called the centripetal component of acceleration or the radial component of acceleration. To understand this concept, imagine traveling in a car on a circular track at constant speed. The acceleration vector points toward the center of the track at all times. As a rider in the car, you feel a pull toward the outside of the track because you are constantly turning. This sensation acts in the opposite direction of centripetal acceleration.

The same holds true for noncircular paths. Your body tends to travel in a straight line and resists the force from acceleration that pushes it toward the side. Notice in Figure 1 that at point [latex]B[/latex], the acceleration vector points backward. This happens because the car is decelerating as it goes into the curve.

This figure has a curve representing the path of a car. The curve decreases and increases. There are two circles along the path The first circle has point A where the curve meets the circle. At point A there are three vectors. The first vector is asubt and is tangent to the curve at A. The second vector is asubr and is orthogonal to vector asubt. In between these vectors is vector a. The second circle has point B where the curve meets the circle. At point A there are three vectors. The first vector is asubt and is tangent to the curve at A. The second vector is asubr and is orthogonal to vector asubt. In between these vectors is vector a.
Figure 1. The tangential and normal components of acceleration can be used to describe the acceleration vector.

The tangential and normal unit vectors at any point on the curve provide a frame of reference at that point. Think of the components this way:

  • Tangential component ([latex]a_{\bf{T}}[/latex]): Measures how quickly the speed is changing (speeding up or slowing down)
  • Normal component ([latex]a_{\bf{N}}[/latex]): Measures how quickly the direction is changing (how sharply you’re turning)

The tangential and normal components of acceleration are simply the projections of the acceleration vector onto [latex]{\bf{T}}[/latex] and [latex]{\bf{N}}[/latex], respectively.

A particle moves in a path defined by the vector-valued function [latex]{\bf{r}}(t)=t^{2}{\bf{i}}+(2t-3){\bf{j}}+(3t^{2}-3t){\bf{k}}[/latex], where [latex]t[/latex] measures time in seconds and distance is measured in feet.