2.6: Tangential and Normal Components of Acceleration
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This section breaks down acceleration into two components called the tangential and normal components. Similar to how we break down all vectors into \hat{\textbf{i}}, \hat{\textbf{j}} , and \hat{\textbf{k}} components, we can do the same with acceleration. The addition of these two components will give us the overall acceleration.
Introduction
We're use to thinking about acceleration as the second derivative of position, and while that is one way to look at the overall acceleration, we can further break down acceleration into two components: tangential and normal acceleration. The tangential acceleration, denoted a_Tallows us to know how much of the acceleration acts in the direction of motion. The normal acceleration a_N is how much of the acceleration is orthogonal to the tangential acceleration.
Remember that vectors have magnitude AND direction. The tangential acceleration is a measure of the rate of change in the magnitude of the velocity vector, i.e. speed, and the normal acceleration are a measure of the rate of change of the direction of the velocity vector.
This approach to acceleration is particularly useful in physics applications, because we need to know how much of the total acceleration acts in any given direction. Think for example of designing brakes for a car or the engine of a rocket. Why might it be useful to separate acceleration into components?
Theoretical discussion with descriptive elaboration
We can find the tangential accelration by using Chain Rule to rewrite the velocity vector as follows:
\mathbf{v}=\dfrac{\mathrm{d\textbf{r}} }{\mathrm{d} t}=\dfrac{\mathrm{d\textbf{r}} }{\mathrm{d} s}\dfrac{\mathrm{d\textit{s}} }{\mathrm{d} t}=\textbf{T}\dfrac{\mathrm{d\mathit{s}} }{\mathrm{d} t} \nonumber
Now, since acceleration is simply the derivative of velocity, we find that:
\begin{align} \mathbf{a}&=\dfrac{\mathrm{d\mathbf{v}} }{\mathrm{d} t}\\ &=\dfrac{\mathrm{d} }{\mathrm{d} t}(\mathbf{T}\dfrac{\mathrm{d\mathit{s}} }{\mathrm{d} t}) \\ &=\dfrac{\mathrm{d} ^2\mathit{s}}{\mathrm{d} t^2}\mathbf{T}+\dfrac{\mathrm{d} s}{\mathrm{d} t}\dfrac{\mathrm{d} \mathbf{T}}{\mathrm{d} t} \\ & =\dfrac{\mathrm{d} ^2s}{\mathrm{d} t^2}\mathbf{T}+\dfrac{\mathrm{d} s}{\mathrm{d} t}\left ( \dfrac{\mathrm{d} \mathbf{T}}{\mathrm{d} s}\dfrac{\mathrm{d} s}{\mathrm{d} t} \right ) \\ & =\dfrac{\mathrm{d} ^2s}{\mathrm{d} t^2}\mathbf{T}+\dfrac{\mathrm{d} s}{\mathrm{d} t}\left ( \kappa \mathbf{N}\dfrac{\mathrm{d} s}{\mathrm{d} t} \right ) \\ & =\dfrac{\mathrm{d} ^2s}{\mathrm{d} t^2}\mathbf{T}+\kappa \left ( \dfrac{\mathrm{d} s}{\mathrm{d} t} \right )^2\mathbf{N} \end{align} \nonumber
\dfrac{\mathrm{d} \mathbf{T}}{\mathrm{d} s}=\kappa \mathbf{N} \nonumber
This, in turn, gives us the definition for acceleration by components.
If the acceleration vector is written as
\mathbf{a}=a_T\mathbf{T}+a_N\mathbf{N}, \nonumber
then
a_T=\dfrac{\mathrm{d} ^2s}{\mathrm{d} t^2}=\dfrac{\mathrm{d} }{\mathrm{d} t}|v| \ \text{ and } \ a_N=\kappa \left ( \dfrac{\mathrm{d} s}{\mathrm{d} t} \right )^2=\kappa|v|^2 \nonumber
To calculate the normal component of the accleration, use the following formula:
a_N=\sqrt{|a|^2-a_T^2} \label{Normal}
We can relate this back to a common physics principal-uniform circular motion. In uniform circulation motion, when the speed is not changing, there is no tangential acceleration, only normal accleration pointing towards the center of circle. Why do you think this is? Hint: look in the introduction section for the difference between the two components of acceleration.
Without finding T and N, write the accelration of the motion
\mathbf{r}(t)=(\cos t+t\sin t) \hat{ \textbf{i} } +(\sin t-t\cos t)\hat{ \textbf{j} } \nonumber for t>0.
To solve this problem, we must first find the particle's velocity.
\begin{align} \mathbf{v}&=\dfrac{\mathrm{d} \mathbf{r}}{\mathrm{d} t} \\ & =(-\sin t+\sin t+t\cos t) \hat{ \textbf{i} } +(\cos t-\cos t+t\sin t) \hat{ \textbf{j} } \\ &=(t\cos t) \hat{ \textbf{i} } +(t\sin t) \hat{ \textbf{j} } \end{align} \nonumber
Next find the speed.
|v|=\sqrt{t^2\cos ^2t+t^2\sin ^2t}=\sqrt{t^2}=|t| \nonumber
When t>0, |t| simply becomes t.
We know that a_T=\dfrac{\mathrm{d} }{\mathrm{d} t}|v|, which we can use to find that \dfrac{\mathrm{d} }{\mathrm{d} t}(t)=1.
Now that we know a_T, we can use it to find a_N using Equation \ref{Normal}.
\mathbf{a}=(\cos t-t\sin t) \hat{ \textbf{i} } +(\sin t+t\cos t) \hat{ \textbf{j} } \nonumber
|\mathbf{a}|^2=t^2+1 \nonumber
a_N=\sqrt{(t^2+1)-(1)}=t \nonumber
By substituting this back into the original definition, we find that |\mathbf{a}|=(1)\mathbf{T}+(t)\mathbf{N}=\mathbf{T}+t\mathbf{N} \nonumber
Write ain the form \mathbf{a}=a_T\mathbf{T}+a_N\mathbf{N} without finding T or N.
\mathbf{r}(t)=(t+1) \hat{ \textbf{i} } +2t \hat{ \textbf{j} } +t^2 \hat{ \textbf{k} } \nonumber
\mathbf{v}=t \hat{ \textbf{i} } +2 \hat{\textbf{j}} +2t \hat{ \textbf{k} } \nonumber
|\mathbf{v}|=\sqrt{5+4t^2} \nonumber
a_T=\dfrac{1}{2}\left ( 5+4t^2 \right )^{-\dfrac{1}{2}}(8t) \nonumber
=4t(5+4t^2)^{-\dfrac{1}{2}} \nonumber
a_T(1)=\dfrac{4}{\sqrt{9}}=\dfrac{4}{3} \nonumber
\mathbf{a}=2 \hat{ \textbf{k} } \nonumber
\mathbf{a}(1)=2 \hat{ \textbf{k} } \nonumber
\mathbf{a}(t)=2 \hat{ \textbf{k} } \nonumber
Now we use Equation \ref{Normal}:
\begin{align} a_N=\sqrt{|\mathbf{a}|^2-a_T^2}&=\sqrt{2^2-\left ( \dfrac{4}{3} \right )^2} \\ & =\sqrt{\dfrac{20}{9}} \\ &=\dfrac{2\sqrt{5}}{3} \end{align} \nonumber
\mathbf{a}(1)=\dfrac{4}{3}\mathbf{T}+\dfrac{2\sqrt{5}}{3}\mathbf{N} \nonumber
References
- Weir, Maurice D., Joel Hass, and George B. Thomas. Thomas' Calculus: Early Transcendentals. Boston: Addison-Wesley, 2010. Print.