Components of the Acceleration Vector

Learning Outcomes

  • Explain the tangential and normal components of acceleration.

We can combine some of the concepts discussed in Arc Length and Curvature with the acceleration vector to gain a deeper understanding of how this vector relates to motion in the plane and in space. 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 a vector-valued function [latex]{\bf{r}}\,(t)[/latex]. The following theorem shows that the acceleration vector [latex]{\bf{a}}\,(t)[/latex] lies in the osculating plane and can be written as a linear combination of the unit tangent and the unit normal vectors.

The Plane of the Acceleration Vector Theorem

The acceleration vector [latex]{\bf{a}}\,(t)[/latex] of an object moving along 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] to [latex]C[/latex]. Furthermore,

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

 

Here, [latex]v\,(t)[/latex] is the speed of the object and [latex]\kappa[/latex] is the curvature of [latex]C[/latex] 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] are referred to as the tangential component of acceleration and the normal component of acceleration, respectively. We write [latex]a_{\bf{T}}[/latex] to denote the tangential component and [latex]a_{\bf{N}}[/latex] to denote the normal component.

Tangential and Normal Components of Acceleration Theorem

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 [latex]a_{\bf{T}}[/latex] and [latex]a_{\bf{N}}[/latex] are given by the formulas

[latex]\large{a_{\bf{T}}={\bf{a}}\cdot{\bf{T}}=\frac{{\bf{v}}\cdot{\bf{a}}}{\left\Vert{\bf{v}}\right\Vert}}[/latex].

and

[latex]\large{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 the formula

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

 

Here [latex]{\bf{T}}\,(t)[/latex] is the unit tangent vector to the curve defined by [latex]{\bf{r}}\,(t)[/latex], and [latex]{\bf{N}}\,(t)[/latex] is the unit normal vector to the curve defined by [latex]{\bf{r}}\,(t)[/latex].

The normal component of acceleration is also called the centripetal component of acceleration or sometimes the radial component of acceleration. To understand centripetal acceleration, suppose you are traveling in a car on a circular track at a constant speed. Then, as we saw earlier, 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. The reason is that your body tends to travel in a straight line and resists the force resulting from acceleration that push it toward the side. Note that at point [latex]B[/latex] in Figure 4 the acceleration vector is pointing backward. This is 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 given point on the curve provide a frame of reference at that point. The tangential and normal components of acceleration are the projections of the acceleration vector onto [latex]{\bf{T}}[/latex] and [latex]{\bf{N}}[/latex], respectively.

Example: Finding Components of Acceleration

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.

try it

An object moves in a path defined by the vector-valued function [latex]{\bf{r}}\,(t)=4t\,{\bf{i}}+t^{2}\,{\bf{j}}[/latex], where [latex]t[/latex] measures time in seconds.

  1. Find [latex]a_{\bf{T}}[/latex] and [latex]a_{\bf{N}}[/latex] as functions of [latex]t[/latex].
  2. Find [latex]a_{\bf{T}}[/latex] and [latex]a_{\bf{N}}[/latex] at time [latex]t=-3[/latex].

Watch the following video to see the worked solution to the above Try It