4.2 HW 2

4.2.1 Problem 1

The local body coordinates frame is as follows

Note on notations used; \(\overset{\longrightarrow }{r}_{C/O}\)is vector of point \(C\) in space. This vector originates from point \(O\) to point \(C.\)

\(O\) always represents the inertial frame of reference. \(\overset{\longrightarrow }{r}_{C/A}\) is a vector from point \(A\) to point \(C.\) In this problem there are two frames of references used. The inertial frame of reference \(XYZ\) whose origin is \(O\), and the local body frame of reference \(xyz\) whose origin is point \(A\). The unit vectors for \(XYZ\) are called \(i,j,k\) while unit vectors for local coordinates frame are \(\overset{\rightarrow }{e}_{x},\overset{\rightarrow }{e}_{y},\overset{\rightarrow }{e_{z}}\). The following is a list of complete notations used in this problem

1.

\(\overset{\longrightarrow }{r}_{A/O}\)is vector of \(A\)

2.

\(\overset{\longrightarrow }{r}_{C/O}\) is vector of \(C\)

3.

\(\overset{\longrightarrow }{r}_{C/A}\) is vector from \(A\) to \(C\)

4.

\(\overset{\centerdot }{\overset{\longrightarrow }{r}}_{C/A}\) relative velocity of position vector \(\overset{\longrightarrow }{r}_{C/A}\) as seen in local frame of reference

5.

\(\overset{\rightarrow }{\omega }_{A/O}\) is the angular velocity of coordinate system \(xyz\), whose origin is \(A\), as seen in inertial frame \(XYZ\)

6.

\(L\) is the length of the radius of the disk, which is given as \(2m\)

7.

\(L_{1}\left ( t\right ) \) is the current length from \(A\) of point \(C\). At the instance required, it is \(0.25m\)

Given the above, then, by standard vector additions \[ \overset{\rightarrow }{r}_{C/O}=\overset{\rightarrow }{r}_{C/A}+\overset{\rightarrow }{r}_{A/O}\]

Hence, taking derivatives\begin{equation} \overset{\centerdot }{\overset{\longrightarrow }{r}}_{C/O}=\overset{\centerdot }{\overset{\longrightarrow }{r}}_{C/A}+\left ( \overset{\rightarrow }{\omega }_{A/O}\times \overset{\rightarrow }{r}_{C/A}\right ) +\overset{\centerdot }{\overset{\longrightarrow }{r}}_{A/O}\tag{1} \end{equation}

Where\[ \overset{\rightarrow }{r}_{A/O}=L\cos \phi \overset{\rightarrow }{i}+L\sin \phi \overset{\rightarrow }{j}\]

And taking derivatives of the above\[ \overset{\centerdot }{\overset{\longrightarrow }{r}}_{A/O}=-L\overset{\centerdot }{\phi }\sin \phi \overset{\rightarrow }{i}+L\overset{\centerdot }{\phi }\cos \phi \overset{\rightarrow }{j}\]

The position vector of \(C\) written using local coordinates system is\[ \overset{\rightarrow }{r}_{C/A}=0\overset{\rightarrow }{e}_{x}-L_{1}\cos \theta \overset{\rightarrow }{e}_{z}+L_{1}\sin \theta \overset{\rightarrow }{e}_{y}\]

Taking derivatives\begin{equation} \overset{\centerdot }{\overset{\longrightarrow }{r}}_{C/A}=-\left ( \overset{\centerdot }{L}_{1}\cos \theta -L_{1}\overset{\centerdot }{\theta }\sin \theta \right ) \overset{\rightarrow }{e}_{z}+\left ( \overset{\centerdot }{L}_{1}\sin \theta +L_{1}\overset{\centerdot }{\theta }\cos \theta \right ) \overset{\rightarrow }{e}_{y}\tag{2} \end{equation}

And the following term is added to account for the fact that the local frame of reference itself is rotating relative to the inertial frame of reference\[ \overset{\rightarrow }{\omega }_{A/O}\times \overset{\rightarrow }{r}_{C/A}=\omega _{p}\overset{\rightarrow }{k}\times \left ( -L_{1}\cos \theta \overset{\rightarrow }{e}_{z}+L_{1}\sin \theta \overset{\rightarrow }{e}_{y}\right ) \]

Substituting the above back to Eq.(1) results in \begin{align*} \overset{\centerdot }{\overset{\longrightarrow }{r}}_{C/O} & =-\left ( \overset{\centerdot }{L}_{1}\cos \theta -L_{1}\overset{\centerdot }{\theta }\sin \theta \right ) \overset{\rightarrow }{e}_{z}+\left ( \overset{\centerdot }{L}_{1}\sin \theta +L_{1}\overset{\centerdot }{\theta }\cos \theta \right ) \overset{\rightarrow }{e}_{y}\\ & +\left ( \omega _{p}\overset{\rightarrow }{k}\times \left ( -L_{1}\cos \theta \overset{\rightarrow }{e}_{z}+L_{1}\sin \theta \overset{\rightarrow }{e}_{y}\right ) \right ) +\left ( -L\overset{\centerdot }{\phi }\sin \phi \overset{\rightarrow }{i}+L\overset{\centerdot }{\phi }\cos \phi \overset{\rightarrow }{j}\right ) \end{align*}

At the instance given, \(\theta =0,L_{1}=0.25m,\overset{\centerdot }{L}_{1}=3m/s,\overset{\centerdot \centerdot }{L}_{1}=2m/s^{2},\overset{\centerdot }{\theta }=\omega _{m}=3rad/\sec ,\overset{\centerdot \centerdot }{\theta }=\overset{\centerdot }{\omega }_{m}=1\) rad/\(\sec ,L=2\) m\(,\) \(\overset{\rightarrow }{e}_{z}=\overset{\rightarrow }{k}\) and \(\overset{\rightarrow }{e}_{y}=\overset{\rightarrow }{j},\omega _{p}=5\) rad.sec,The above simplifies to

\begin{align*} \overset{\centerdot }{\overset{\longrightarrow }{r}}_{C/O} & =-\overset{\centerdot }{L}_{1}\overset{\rightarrow }{e}_{z}+L_{1}\overset{\centerdot }{\theta }\overset{\rightarrow }{e}_{y}+\left ( \omega _{p}\overset{\rightarrow }{k}\times \left ( -L_{1}\overset{\rightarrow }{e}_{z}\right ) \right ) +L\omega _{p}\overset{\rightarrow }{j}\\ & =-3\overset{\rightarrow }{e}_{z}+0.25\omega _{m}\overset{\rightarrow }{e}_{y}-\left ( \omega _{p}\overset{\rightarrow }{k}\times 0.25\overset{\rightarrow }{e}_{z}\right ) +2\omega _{p}\overset{\rightarrow }{j} \end{align*}

In addition, at the instance shown, \(\vec{e}_{z}=\vec{k}\) and \(\vec{e}_{y}=\vec{j}\) (but this is only at the instance given. In general it is not the case). The above simplifies to\begin{align*} \overset{\centerdot }{\overset{\longrightarrow }{r}}_{C/O} & =-3\overset{\rightarrow }{k}+0.25\omega _{m}\overset{\rightarrow }{j}-\left ( \omega _{p}\overset{\rightarrow }{k}\times 0.25\overset{\rightarrow }{k}\right ) +2\omega _{p}\overset{\rightarrow }{j}\\ & =-3\overset{\rightarrow }{k}+0.25\omega _{m}\overset{\rightarrow }{j}+2\omega _{p}\overset{\rightarrow }{j}\\ & =-3\overset{\rightarrow }{k}+0.75\overset{\rightarrow }{j}+10\overset{\rightarrow }{j}\\ & =10.75\overset{\rightarrow }{j}-3\overset{\rightarrow }{k}\text{\qquad }\left [ m/s\right ] \end{align*}

Numerically, the magnitude of the velocity vector is\[ \left \vert \overset{\centerdot }{\overset{\longrightarrow }{r}}_{C/O}\right \vert =\sqrt{10.75^{2}+9}=11.161\text{ }m/s \]

To find the acceleration, derivative of Eq. (1) is now taken\begin{align} \overset{\centerdot }{\overset{\longrightarrow }{r}}_{C/O} & =\overset{\centerdot }{\overset{\longrightarrow }{r}}_{C/A}+\left ( \overset{\rightarrow }{\omega }_{A/O}\times \vec{r}_{C/A}\right ) +\overset{\centerdot }{\overset{\longrightarrow }{r}}_{A/O}\nonumber \\ a & =\overset{\centerdot \centerdot }{\overset{\rightarrow }{r}}_{C/A}+\left ( \overset{\rightarrow }{\omega }_{A/O}\times \overset{\centerdot }{\overset{\longrightarrow }{r}}_{C/A}\right ) +\left ( \overset{\centerdot }{\overset{\rightarrow }{\omega }}_{A/O}\times \overset{\rightarrow }{r}_{C/A}+\omega _{A/O}\times \left ( \overset{\centerdot }{\overset{\longrightarrow }{r}}_{C/A}+\omega _{A/O}\times \overset{\rightarrow }{r}_{C/A}\right ) \right ) +\ddot{\vec{r}}_{A/O}\nonumber \\ & =\overset{\centerdot \centerdot }{\overset{\rightarrow }{r}}_{C/A}+\left ( \omega _{A/O}\times \overset{\centerdot }{\overset{\longrightarrow }{r}}_{C/A}\right ) +\overset{\centerdot }{\overset{\rightarrow }{\omega }}_{A/O}\times \overset{\rightarrow }{r}_{C/A}+\omega _{A/O}\times \overset{\centerdot }{\overset{\longrightarrow }{r}}_{C/A}+\omega _{A/O}\times \left ( \omega _{A/O}\times \overset{\rightarrow }{r}_{C/A}\right ) +\ddot{\vec{r}}_{A/O}\nonumber \\ & =\overset{\centerdot \centerdot }{\overset{\rightarrow }{r}}_{C/A}+2\left ( \omega _{A/O}\times \overset{\centerdot }{\overset{\longrightarrow }{r}}_{C/A}\right ) +\left ( \overset{\centerdot }{\overset{\rightarrow }{\omega }}_{A/O}\times \overset{\rightarrow }{r}_{C/A}\right ) +\omega _{A/O}\times \left ( \omega _{A/O}\times \overset{\rightarrow }{r}_{C/A}\right ) +\overset{\centerdot \centerdot }{\overset{\rightarrow }{r}}_{A/O} \tag{3} \end{align}

\(\overset{\centerdot \centerdot }{\overset{\rightarrow }{r}}_{C/A}\) is found by differentiating Eq. (2) in the local frame giving\begin{align*} \overset{\centerdot }{\overset{\longrightarrow }{r}}_{C/A} & =-\left ( \overset{\centerdot }{L}_{1}\cos \theta -L_{1}\overset{\centerdot }{\theta }\sin \theta \right ) \overset{\rightarrow }{e}_{z}+\left ( \overset{\centerdot }{L}_{1}\sin \theta +L_{1}\overset{\centerdot }{\theta }\cos \theta \right ) \overset{\rightarrow }{e}_{y}\\ \overset{\centerdot \centerdot }{\overset{\rightarrow }{r}}_{C/A} & =-\left ( \overset{\centerdot \centerdot }{L}_{1}\cos \theta -\overset{\centerdot }{L}_{1}\overset{\centerdot }{\theta }\sin \theta \right ) \overset{\rightarrow }{e}_{z}+\left ( \overset{\centerdot }{L}_{1}\overset{\centerdot }{\theta }\sin \theta +L_{1}\overset{\centerdot \centerdot }{\theta }\sin \theta +L_{1}\overset{\centerdot }{\theta }^{2}\cos \theta \right ) \overset{\rightarrow }{e}_{z}\\ & +\left ( \overset{\centerdot \centerdot }{L}_{1}\sin \theta +\overset{\centerdot }{L}_{1}\overset{\centerdot }{\theta }\cos \theta +\overset{\centerdot }{L}_{1}\overset{\centerdot }{\theta }\cos \theta +L_{1}\overset{\centerdot \centerdot }{\theta }\cos \theta -L_{1}\overset{\centerdot }{\theta }^{2}\sin \theta \right ) \overset{\rightarrow }{e}_{y} \end{align*}

At the instance given the above becomes\begin{align*} \overset{\centerdot \centerdot }{\overset{\rightarrow }{r}}_{C/A} & =-2\overset{\rightarrow }{k}+0.25\omega _{m}^{2}\overset{\rightarrow }{k}+\left ( 3\omega _{m}+3\omega _{m}+0.25\overset{\centerdot }{\omega }_{m}\right ) \overset{\rightarrow }{j}\\ & =-2\overset{\rightarrow }{k}+0.25\left ( 9\right ) \overset{\rightarrow }{k}+\left ( 9+9+0.25\right ) \overset{\rightarrow }{j}\\ & =18.25\overset{\rightarrow }{j}+0.25\overset{\rightarrow }{k} \end{align*}

And \begin{align*} \overset{\centerdot }{\overset{\longrightarrow }{r}}_{C/A} & =-\left ( \overset{\centerdot }{L}_{1}\cos \theta -L_{1}\overset{\centerdot }{\theta }\sin \theta \right ) \overset{\rightarrow }{e}_{z}+\left ( \overset{\centerdot }{L}_{1}\sin \theta +L_{1}\overset{\centerdot }{\theta }\cos \theta \right ) \overset{\rightarrow }{e}_{y}\\ & =0.75\overset{\rightarrow }{j}-3\overset{\rightarrow }{k} \end{align*}

And\begin{align*} \vec{r}_{C/A} & =0\overset{\rightarrow }{e}_{x}-L_{1}\cos \theta \overset{\rightarrow }{e}_{z}+L_{1}\sin \theta \overset{\rightarrow }{e}_{y}\\ & =-0.25\overset{\rightarrow }{k} \end{align*}

And\begin{align*} \overset{\centerdot }{\overset{\longrightarrow }{r}}_{A/O} & =-L\overset{\centerdot }{\phi }\sin \phi \overset{\rightarrow }{i}+L\overset{\centerdot }{\phi }\cos \phi \overset{\rightarrow }{j}\\ \overset{\centerdot \centerdot }{\overset{\rightarrow }{r}}_{A/O} & =-\left ( L\overset{\centerdot \centerdot }{\phi }\sin \phi +L\overset{\centerdot }{\phi }^{2}\cos \phi \right ) \overset{\rightarrow }{i}+\left ( L\overset{\centerdot \centerdot }{\phi }\cos \phi -L\overset{\centerdot }{\phi }^{2}\sin \phi \right ) \overset{\rightarrow }{j} \end{align*}

Hence at the instance given\begin{align*} \overset{\centerdot \centerdot }{\overset{\rightarrow }{r}}_{A/O} & =-L\overset{\centerdot }{\phi }^{2}\overset{\rightarrow }{i}+L\overset{\centerdot \centerdot }{\phi }\overset{\rightarrow }{j}\\ & =-2\omega _{p}^{2}\overset{\rightarrow }{i}+2\overset{\centerdot }{\omega }_{p}\overset{\rightarrow }{j}\\ & =-50\overset{\rightarrow }{i}+4\overset{\rightarrow }{j} \end{align*}

And\begin{align*} \overset{\rightarrow }{\omega }_{A/O} & =\omega _{p}\overset{\rightarrow }{k}\\ & =5\overset{\rightarrow }{k}\text{ rad/sec} \end{align*}

And\begin{align*} \overset{\centerdot }{\overset{\rightarrow }{\omega }}_{A/O} & =\overset{\centerdot }{\omega }_{p}\overset{\rightarrow }{k}\\ & =2\vec{k}\text{ rad/sec}^{2} \end{align*}

And\begin{align*} \overset{\rightarrow }{r}_{C/A} & =0\overset{\rightarrow }{e}_{x}-L_{1}\cos \theta \overset{\rightarrow }{e}_{z}+L_{1}\sin \theta \overset{\rightarrow }{e}_{y}\\ & =-0.25\overset{\rightarrow }{k} \end{align*}

Therefore, Eq. (3) becomes\begin{align*} a & =\overset{\centerdot \centerdot }{\overset{\rightarrow }{r}}_{C/A}+2\left ( \overset{\rightarrow }{\omega }_{A/O}\times \overset{\centerdot }{\overset{\longrightarrow }{r}}_{C/A}\right ) +\left ( \overset{\centerdot }{\overset{\rightarrow }{\omega }}_{A/O}\times \overset{\rightarrow }{r}_{C/A}\right ) +\overset{\rightarrow }{\omega }_{A/O}\times \left ( \overset{\rightarrow }{\omega }_{A/O}\times \overset{\rightarrow }{r}_{C/A}\right ) +\overset{\centerdot \centerdot }{\overset{\rightarrow }{r}}_{A/O}\\ & \\ & =\left ( 18.25\overset{\rightarrow }{j}+0.25\overset{\rightarrow }{k}\right ) +2\left ( 5\overset{\rightarrow }{k}\times \left ( 0.75\overset{\rightarrow }{j}-3\overset{\rightarrow }{k}\right ) \right ) +\left ( 2\overset{\rightarrow }{k}\times \left ( -0.25\overset{\rightarrow }{k}\right ) \right ) \\ & +5\overset{\rightarrow }{k}\times \left ( 5\overset{\rightarrow }{k}\times \left ( -0.25\overset{\rightarrow }{k}\right ) \right ) +\left ( -50\overset{\rightarrow }{i}+4\overset{\rightarrow }{j}\right ) \\ & \\ & =18.25\overset{\rightarrow }{j}+0.25\overset{\rightarrow }{k}+2\left ( 5\overset{\rightarrow }{k}\times 0.75\overset{\rightarrow }{j}-5\overset{\rightarrow }{k}\times 3\overset{\rightarrow }{k}\right ) -50\overset{\rightarrow }{i}+4\overset{\rightarrow }{j}\\ & =18.25\overset{\rightarrow }{j}+0.25\overset{\rightarrow }{k}+2\left ( 5\overset{\rightarrow }{k}\times 0.75\overset{\rightarrow }{j}\right ) -50\overset{\rightarrow }{i}+4\overset{\rightarrow }{j}\\ & =18.25\overset{\rightarrow }{j}+0.25\overset{\rightarrow }{k}+2\left ( -3.75\overset{\rightarrow }{i}\right ) -50\overset{\rightarrow }{i}+4\overset{\rightarrow }{j}\\ & =-107.5\overset{\rightarrow }{i}+22.25\overset{\rightarrow }{j}+0.25\overset{\rightarrow }{k} \end{align*}

Hence\begin{align*} \left \vert a\right \vert & =\sqrt{107.5^{2}+22.25^{2}+0.25^{2}}\\ & =109.78\qquad m/s^{2} \end{align*}

4.2.2 problem 2

Note on notations used; \(\overset{\longrightarrow }{r}_{P/O}\)is vector of point \(P\) in space that originates from point \(O\), which is the origin of the inertial frame of reference. \(O\) always represents the inertial frame of reference. Hence \(\overset{\longrightarrow }{r}_{P/A}\) is a vector from point \(A\) to point \(P.\) In this problem there are two frames of references used. The inertial frame of reference \(XYZ\) whose origin is called \(O\), and the local body frame of reference \(xyz\) attached to point \(A\) which in this problem happens to be the same as \(O\) point shown above. Hence the origin of \(xyz\) is \(A.\) The unit vectors for \(XYZ\) are always called \(\overset{\rightarrow }{i},\overset{\rightarrow }{j},\overset{\rightarrow }{k}\) while unit vectors for local coordinates frame are \(\overset{\rightarrow }{e}_{x},\overset{\rightarrow }{e}_{y},\overset{\rightarrow }{e_{z}}\). The following is a list of complete notations used in this problem

1.

\(\overset{\longrightarrow }{r}_{P/A}\) is vector from \(A\) to \(P\)

2.

\(\overset{\rightarrow }{\omega }_{A/O}\) is the angular velocity of vector coordinate system \(xyz\), whose origin is \(A\), as seen in inertial frame \(XYZ\)

3.

\(y\left ( x\right ) \) is the \(y\) coordinates of point \(P\) as seen in local coordinates system

4.

\(x\) is the \(x\) coordinates of point \(P\) as seen in local coordinates system

Let \(p\) be the point, and as seen in the local frame \(xyz\) it will appear as follows

Using vector addition,\[ \overset{\longrightarrow }{r}_{P/O}=\overset{\longrightarrow }{r}_{P/A}+\overset{\longrightarrow }{r}_{A/O}\]

Where, the position of \(p\) expressed in local frame is \begin{align} \overset{\rightarrow }{r}_{P/A} & =0\overset{\rightarrow }{e}_{z}+x\left ( t\right ) \overset{\rightarrow }{e}_{x}+y\left ( t\right ) \overset{\rightarrow }{e}_{y}\nonumber \\ & =x\left ( t\right ) \overset{\rightarrow }{e}_{x}+\frac{1}{2}\left ( 3+x^{2}\right ) \overset{\rightarrow }{e}_{y} \end{align}

But \begin{align*} ds & =\sqrt{dx^{2}+dy^{2}}\\ \frac{ds}{dt} & =\sqrt{\left ( \frac{dx}{dt}\right ) ^{2}+\left ( \frac{dy}{dt}\right ) ^{2}}\\ & =\sqrt{\left ( \frac{dx}{dt}\right ) ^{2}+\left ( \frac{dy}{dx}\frac{dx}{dt}\right ) ^{2}}\\ & =\left ( \frac{dx}{dt}\right ) \sqrt{1+\left ( \frac{dy}{dx}\right ) ^{2}} \end{align*}

But \(\frac{dy}{dx}=\frac{d}{dx}\left ( \frac{1}{2}\left ( 3+x^{2}\right ) \right ) =x\), hence the above becomes\[ \frac{ds}{dt}=\overset{\centerdot }{x}\sqrt{1+x^{2}}\]

But \(\frac{ds}{dt}\) is constant and given by \(2\) ft/sec, therefore\begin{align*} 2 & =\overset{\centerdot }{x}\sqrt{1+x^{2}}\\ \overset{\centerdot }{x} & =\frac{2}{\sqrt{1+x^{2}}} \end{align*}

And since \begin{align*} \frac{dy}{dt} & =\frac{dy}{dx}\frac{dx}{dt}\\ & =x\overset{\centerdot }{x} \end{align*}

Hence\[ \overset{\centerdot }{y}=\frac{2x}{\sqrt{1+x^{2}}}\]

Now taking derivatives of Eq. (1), and noting that \(\overset{\centerdot }{\overset{\longrightarrow }{r}}_{A/O}=0\) since the origin of the local frame coincides with the origin of the inertial frame, hence \(\overset{\longrightarrow }{r}_{A/O}=\vec{0}\)\begin{align} \overset{\centerdot }{\overset{\longrightarrow }{r}}_{P/O} & =\overset{\centerdot }{\overset{\longrightarrow }{r}}_{P/A}+\left ( \overset{\rightarrow }{\omega }_{A/O}\times \overset{\rightarrow }{r}_{P/A}\right ) +\overset{\centerdot }{\overset{\longrightarrow }{r}}_{A/O}\tag{2}\\ & =\left ( \overset{\centerdot }{x}\overset{\rightarrow }{e}_{x}+\overset{\centerdot }{y}\overset{\rightarrow }{e}_{y}\right ) +\left ( \omega \left ( t\right ) \overset{\rightarrow }{j}\times \left ( x\left ( t\right ) \overset{\rightarrow }{e}_{x}+y\left ( t\right ) \overset{\rightarrow }{e}_{y}\right ) \right ) +\vec{0}\nonumber \end{align}

At the instance shown, \(\overset{\rightarrow }{e}_{x}\) is aligned with \(\overset{\rightarrow }{i}\) and \(\overset{\rightarrow }{e}_{y}\)is aligned with \(\overset{\rightarrow }{j}\),hence the above becomes\begin{align*} \overset{\centerdot }{\overset{\longrightarrow }{r}}_{P/O} & =\left ( \overset{\centerdot }{x}\overset{\rightarrow }{i}+\overset{\centerdot }{y}\overset{\rightarrow }{j}\right ) +\left ( \omega \overset{\rightarrow }{j}\times \text{ }x\left ( t\right ) \overset{\rightarrow }{i}\right ) +\left ( \omega \overset{\rightarrow }{j}\times y\left ( t\right ) \overset{\rightarrow }{j}\right ) \\ & =\left ( \overset{\centerdot }{x}\overset{\rightarrow }{i}+\overset{\centerdot }{y}\overset{\rightarrow }{j}\right ) -\omega x\left ( t\right ) \overset{\rightarrow }{k} \end{align*}

Substituting for \(\overset{\centerdot }{x},\overset{\centerdot }{y}\) in the above\[ \overset{\centerdot }{\overset{\longrightarrow }{r}}_{P/O}=\left ( \frac{2}{\sqrt{1+x^{2}}}\overset{\rightarrow }{i}+\frac{2x}{\sqrt{1+x^{2}}}\overset{\rightarrow }{j}\right ) -3t^{2}x\left ( t\right ) \overset{\rightarrow }{k}\]

At this instance, \(t=2\) sec\(,x=1\)ft, hence\begin{align*} \overset{\centerdot }{\overset{\longrightarrow }{r}}_{P/O} & =\frac{2}{\sqrt{2}}\overset{\rightarrow }{i}+\frac{2}{\sqrt{2}}\overset{\rightarrow }{j}-3\left ( 4\right ) \overset{\rightarrow }{k}\qquad \left [ ft/\sec \right ] \\ & =1.414\,\overset{\rightarrow }{i}+1.414\overset{\rightarrow }{j}-12\overset{\rightarrow }{k} \end{align*}

and \begin{align*} \left \vert \overset{\centerdot }{\overset{\longrightarrow }{r}}_{P/O}\right \vert & =\sqrt{\left ( \frac{2}{\sqrt{2}}\right ) ^{2}+\left ( \frac{2}{\sqrt{2}}\right ) ^{2}+12^{2}}\\ & =12.166\qquad ft/\sec \end{align*}

4.2.2.1 Extra (finding the acceleration)

This is not required, but for practice. Now the total acceleration is found. From Eq. (2) above it was found that\[ \overset{\centerdot }{\overset{\longrightarrow }{r}}_{P/O}=\overset{\centerdot }{\overset{\longrightarrow }{r}}_{P/A}+\left ( \overset{\rightarrow }{\omega }_{A/O}\times \overset{\rightarrow }{r}_{P/A}\right ) +\overset{\centerdot }{\overset{\longrightarrow }{r}}_{A/O}\]

Taking derivative of the above\begin{align} \overset{\centerdot \centerdot }{\overset{\longrightarrow }{r}}_{P/O} & =\overset{\centerdot \centerdot }{\overset{\longrightarrow }{r}}_{P/A}+\left ( \overset{\rightarrow }{\omega }_{A/O}\times \overset{\centerdot }{\overset{\longrightarrow }{r}}_{P/A}\right ) +\left ( \overset{\centerdot }{\overset{\rightarrow }{\omega }}_{A/O}\times \overset{\rightarrow }{r}_{P/A}+\overset{\rightarrow }{\omega }_{A/O}\times \left ( \overset{\centerdot }{\overset{\longrightarrow }{r}}_{P/A}+\left ( \overset{\rightarrow }{\omega }_{A/O}\times \overset{\rightarrow }{r}_{P/A}\right ) \right ) \right ) +\overset{\centerdot \centerdot }{\overset{\longrightarrow }{r}}_{A/O}\nonumber \\ & =\overset{\centerdot \centerdot }{\overset{\longrightarrow }{r}}_{P/A}+\left ( \overset{\rightarrow }{\omega }_{A/O}\times \overset{\centerdot }{\overset{\longrightarrow }{r}}_{P/A}\right ) +\left ( \overset{\centerdot }{\overset{\rightarrow }{\omega }}_{A/O}\times \overset{\rightarrow }{r}_{P/A}+\overset{\rightarrow }{\omega }_{A/O}\times \overset{\centerdot }{\overset{\longrightarrow }{r}}_{P/A}+\overset{\rightarrow }{\omega }_{A/O}\times \left ( \overset{\rightarrow }{\omega }_{A/O}\times \overset{\rightarrow }{r}_{P/A}\right ) \right ) +\overset{\centerdot \centerdot }{\overset{\longrightarrow }{r}}_{A/O}\nonumber \\ & =\overset{\centerdot \centerdot }{\overset{\longrightarrow }{r}}_{P/A}+2\left ( \overset{\rightarrow }{\omega }_{A/O}\times \overset{\centerdot }{\overset{\longrightarrow }{r}}_{P/A}\right ) +\overset{\centerdot }{\overset{\rightarrow }{\omega }}_{A/O}\times \overset{\rightarrow }{r}_{P/A}+\overset{\rightarrow }{\omega }_{A/O}\times \left ( \overset{\rightarrow }{\omega }_{A/O}\times \overset{\rightarrow }{r}_{P/A}\right ) +\overset{\centerdot \centerdot }{\overset{\longrightarrow }{r}}_{A/O}\tag{4} \end{align}

But \[ \overset{\rightarrow }{r}_{P/A}=x\left ( t\right ) \overset{\rightarrow }{e}_{x}+\frac{1}{2}\left ( 3+x^{2}\right ) \overset{\rightarrow }{e}_{y}\]

Hence \[ \overset{\centerdot }{\overset{\longrightarrow }{r}}_{P/A}=\overset{\centerdot }{x}\overset{\rightarrow }{e}_{x}+\overset{\centerdot }{y}\overset{\rightarrow }{e}_{y}\]

And\[ \overset{\centerdot \centerdot }{\overset{\longrightarrow }{r}}_{P/A}=\overset{\centerdot \centerdot }{x}\overset{\rightarrow }{e}_{x}+\overset{\centerdot \centerdot }{y}\overset{\rightarrow }{e}_{y}\]

And \[ \left ( \overset{\rightarrow }{\omega }_{A/O}\times \overset{\centerdot }{\overset{\longrightarrow }{r}}_{P/A}\right ) =\omega \left ( t\right ) \overset{\rightarrow }{j}\times \left ( \overset{\centerdot }{x}\overset{\rightarrow }{e}_{x}+\overset{\centerdot }{y}\overset{\rightarrow }{e}_{y}\right ) \]

And\begin{align*} \overset{\centerdot }{\overset{\rightarrow }{\omega }}_{A/O}\times \overset{\rightarrow }{r}_{P/A} & =\frac{d}{dt}3t^{2}\overset{\rightarrow }{j}\times \left ( x\left ( t\right ) \overset{\rightarrow }{e}_{x}+\frac{1}{2}\left ( 3+x^{2}\right ) \overset{\rightarrow }{e}_{y}\right ) \\ & =6t\overset{\rightarrow }{j}\times \left ( x\left ( t\right ) \overset{\rightarrow }{e}_{x}+\frac{1}{2}\left ( 3+x^{2}\right ) \overset{\rightarrow }{e}_{y}\right ) \end{align*}

And\[ \overset{\rightarrow }{\omega }_{A/O}\times \overset{\rightarrow }{r}_{P/A}=\omega \left ( t\right ) \overset{\rightarrow }{j}\times \left ( x\left ( t\right ) \overset{\rightarrow }{e}_{x}+y\left ( t\right ) \overset{\rightarrow }{e}_{y}\right ) \]

And\[ \overset{\rightarrow }{\omega }_{A/O}\times \left ( \overset{\rightarrow }{\omega }_{A/O}\times \overset{\rightarrow }{r}_{P/A}\right ) =\omega \left ( t\right ) \overset{\rightarrow }{j}\times \left ( \omega \left ( t\right ) \overset{\rightarrow }{j}\times \left ( x\left ( t\right ) \overset{\rightarrow }{e}_{x}+y\left ( t\right ) \overset{\rightarrow }{e}_{y}\right ) \right ) \]

And since \(A\) is attached to \(O\), hence\[ \overset{\centerdot \centerdot }{\overset{\longrightarrow }{r}}_{A/O}=0 \]

Now the above is evaluated at the instance given where, \(\overset{\rightarrow }{e}_{x}\) is aligned with \(\overset{\rightarrow }{i}\) and \(\overset{\rightarrow }{e}_{y}\)is aligned with \(\overset{\rightarrow }{j}\), hence Eq. (4) becomes\[ \overset{\centerdot \centerdot }{\overset{\longrightarrow }{r}}_{P/O}=\overset{\centerdot \centerdot }{x}\overset{\rightarrow }{i}+\overset{\centerdot \centerdot }{y}\overset{\rightarrow }{j}+2\left ( \omega \overset{\rightarrow }{j}\times \left ( \overset{\centerdot }{x}\overset{\rightarrow }{i}+\overset{\centerdot }{y}\overset{\rightarrow }{j}\right ) \right ) +\left ( 6t\overset{\rightarrow }{j}\times \left ( x\overset{\rightarrow }{i}+\frac{1}{2}\left ( 3+x^{2}\right ) \overset{\rightarrow }{j}\right ) \right ) +\omega \overset{\rightarrow }{j}\times \left ( \omega \overset{\rightarrow }{j}\times \left ( x\overset{\rightarrow }{i}+y\overset{\rightarrow }{j}\right ) \right ) \]

At the instance shown \(x=1\) ft\(,t=2\) sec and hence \(\omega =3t^{2}=12\) rad/sec. Since speed of particle is constant, then \(\overset{\centerdot \centerdot }{x}=0\) and \(\overset{\centerdot \centerdot }{y}=0\), then the above simplifies to\begin{align*} \overset{\centerdot \centerdot }{\overset{\longrightarrow }{r}}_{P/O} & =2\left ( 12\overset{\rightarrow }{j}\times \left ( \frac{2}{\sqrt{2}}\overset{\rightarrow }{i}+\frac{2}{\sqrt{2}}\overset{\rightarrow }{j}\right ) \right ) +\left ( 12\overset{\rightarrow }{j}\times \left ( \overset{\rightarrow }{i}+2\overset{\rightarrow }{j}\right ) \right ) +12\overset{\rightarrow }{j}\times \left ( 12\overset{\rightarrow }{j}\times \left ( \overset{\rightarrow }{i}+2\overset{\rightarrow }{j}\right ) \right ) \\ & \\ & =2\left ( \left ( 12\overset{\rightarrow }{j}\times \frac{2}{\sqrt{2}}\overset{\rightarrow }{i}\right ) +\left ( 12\overset{\rightarrow }{j}\times \frac{2}{\sqrt{2}}\overset{\rightarrow }{j}\right ) \right ) +\left ( 12\overset{\rightarrow }{j}\times \overset{\rightarrow }{i}\right ) \\ & +\left ( 12\overset{\rightarrow }{j}\times 2\overset{\rightarrow }{j}\right ) +12\overset{\rightarrow }{j}\times \left ( \left ( 12\overset{\rightarrow }{j}\times \overset{\rightarrow }{i}\right ) +\left ( 12\overset{\rightarrow }{j\times }2\overset{\rightarrow }{j}\right ) \right ) \\ & \\ & =2\left ( -\frac{24}{\sqrt{2}}\overset{\rightarrow }{k}\right ) -12\overset{\rightarrow }{k}+12\overset{\rightarrow }{j}\times \left ( -12\overset{\rightarrow }{k}\right ) \\ & =-\frac{48}{\sqrt{2}}\overset{\rightarrow }{k}-12\overset{\rightarrow }{k}-144\overset{\rightarrow }{i}\\ & =-144\overset{\rightarrow }{i}-45.941\overset{\rightarrow }{k} \end{align*}

Hence \[ \left \vert \overset{\centerdot \centerdot }{\overset{\longrightarrow }{r}}_{P/O}\right \vert =\sqrt{144^{2}+45.941^{2}}=\allowbreak 151.\,\allowbreak 15m/\sec ^{2}\]

4.2.3 key solution

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