2.6: Tangential And Normal Components Of Acceleration

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:

\[\begin{align} \mathbf{a}&=\dfrac{\mathrm{d\mathbf{v}} }{\mathrm{d} t}\\ &=\dfrac{\mathrm{d} }{\mathrm{d} t}(\mathbf{T}\dfrac{\mathrm{d\mathit{s}} }{\mathrm{d} t}) \\ &=\dfrac{\mathrm{d} ^2\mathit{s}}{\mathrm{d} t^2}\mathbf{T}+\dfrac{\mathrm{d} s}{\mathrm{d} t}\dfrac{\mathrm{d} \mathbf{T}}{\mathrm{d} t} \\ & =\dfrac{\mathrm{d} ^2s}{\mathrm{d} t^2}\mathbf{T}+\dfrac{\mathrm{d} s}{\mathrm{d} t}\left ( \dfrac{\mathrm{d} \mathbf{T}}{\mathrm{d} s}\dfrac{\mathrm{d} s}{\mathrm{d} t} \right ) \\ & =\dfrac{\mathrm{d} ^2s}{\mathrm{d} t^2}\mathbf{T}+\dfrac{\mathrm{d} s}{\mathrm{d} t}\left ( \kappa \mathbf{N}\dfrac{\mathrm{d} s}{\mathrm{d} t} \right ) \\ & =\dfrac{\mathrm{d} ^2s}{\mathrm{d} t^2}\mathbf{T}+\kappa \left ( \dfrac{\mathrm{d} s}{\mathrm{d} t} \right )^2\mathbf{N} \end{align} \nonumber \]

Note

\[\dfrac{\mathrm{d} \mathbf{T}}{\mathrm{d} s}=\kappa \mathbf{N} \nonumber \]

This, in turn, gives us the definition for acceleration by components.

Definition: acceleration vector

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.

Example \(\PageIndex{1}\)

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 \[|\mathbf{a}|=(1)\mathbf{T}+(t)\mathbf{N}=\mathbf{T}+t\mathbf{N} \nonumber \]

Example \(\PageIndex{2}\)

Write \(a\)in 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 \(\ref{Normal}\):

\[\begin{align} a_N=\sqrt{|\mathbf{a}|^2-a_T^2}&=\sqrt{2^2-\left ( \dfrac{4}{3} \right )^2} \\ & =\sqrt{\dfrac{20}{9}} \\ &=\dfrac{2\sqrt{5}}{3} \end{align} \nonumber \]

\[\mathbf{a}(1)=\dfrac{4}{3}\mathbf{T}+\dfrac{2\sqrt{5}}{3}\mathbf{N} \nonumber \]