Biomechanics 2023/4

Quiz 1

Have a go at these questions and don't worry if you get anything wrong, the main purpose is to shape teaching this term

Q1

Which if the following is the correct formulation of Newton's third law?

  1. $\displaystyle F=ma$
  2. $\displaystyle F=m\frac{dv}{dt}$
  3. $\displaystyle F=m v \frac{dv}{dx}$
  4. $\displaystyle F=m\ddot{x}$
  5. $\displaystyle F= \frac{dp}{dt}$

Where $F$ is force, $m$ is mass, $a$ is acceleration, $x$ is position, $p$ is momentum, $v$ is velocity

Q2

Given movement of a point mass under constant acceleration (e.g. gravity) how does the start velocity relate to the final velocity? (some may know this as a suvat equation, and it is possible to get this answer from Newton's third law)

Q3

Given the following four vectors

\begin{align*} \vec{f}_1& =\begin{bmatrix}2 &1.6\end{bmatrix}^T\\ \vec{f}_2& =\begin{bmatrix}-3 &1\end{bmatrix}^T\\ \vec{f}_3& =\begin{bmatrix}-0.8 &-1.2\end{bmatrix}^T\\ \vec{f}_4& =\begin{bmatrix}1.8 &-1.4\end{bmatrix}^T \end{align*}

What do you notice and what can you conclude?

Q4

a) The vector norm

Given a vector $\vec{k}=\begin{bmatrix}k_1 &k_2 & k_3 \end{bmatrix}^T$ the length is defined as $|\vec{k} | \text{ (norm of k) } = \sqrt{k_1^2+k_2^2+k_3^2}$

If $\vec{k}=\begin{bmatrix}2 &3 & 7 \end{bmatrix}^T$ calculate $|\vec{k}|$ and hence calculate the unit vector $\hat{\vec{k}}$ which has the same direction as $\vec{k}$ but a length of 1.

b) The dot product

If a vector $\vec{k}$ with $r$ elements is expressed as a matrix with a single column and $r$ rows, then the dot product of $\vec{k}$ with a vector $f$ also with $r$ rows is defined as

\[ \vec{r}\cdot\vec{f}=\vec{r}^T\vec{f} \]

See if you can calculate the dot product of the unit vector $\hat{\vec{k}}$ from question (a) and the vector $\vec{f}=\begin{bmatrix}14 & -7 & 7 \end{bmatrix}^T$

The dot product can also be considered as the projection of $\vec{f}$ onto $\vec{k}$ or the projection of $\vec{k}$ onto $\vec{f}$. This is because another way to define it is $\vec{r}\cdot\vec{f}=|\vec{r}||\vec{f}|\cos(\theta)$ where $\cos{\theta}$ is the angle between these two vectors. A challenge is to show that this is the case.

c)

if $\vec{r}=\begin{bmatrix} 2 & -1 & \frac12 \end{bmatrix}^T$

Calculate the cross product of $\vec{r}$ and $\vec{f}$

If you don't know how to calculate a cross product then

create a matrix R based on the values of $\vec{r}$ so that

\[ R= \begin{bmatrix} 0 & -r_{3} & r_{2}\\ r_{3} & 0 & -r_{1}\\ -r_{2} & r_{1} & 0 \end{bmatrix} \]

Now use the identity

\[ \vec{r}\times\vec{f}=R\vec{f} \]

if $\vec{f}$ is considered as a column vector

A useful property is that if $\vec{r}$ is a position and $\vec{f}$ is a force the the cross product is considered as a torque.

Can you think about some examples where the idea of torque might be helpful? (think about things that rotate)

Q 5

Google `Newton-Euler equations of motion'

How do these equations compare to Newton's third law in equation 1?

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Q6

Basis vectors are used to create coordinate frames. A common set of basis vectors are $\vec{i}$ (not the square root of -1) $\vec{j}$ (also not the square root of -1) and $\vec{k}$ (also not ...)

Draw the following coordinate frames onto the grid below assuming in all cases $\vec{k}=\begin{bmatrix}0 &0 & 1 \end{bmatrix}^T$

\begin{align*} \vec{i}&=\begin{bmatrix}1 &0 & 0 \end{bmatrix}^T\\ \vec{j}&=\begin{bmatrix}0 &1 & 0 \end{bmatrix}^T \end{align*} \begin{align*} \vec{i}&=\begin{bmatrix}2 &1 & 0 \end{bmatrix}^T\\ \vec{j}&=\begin{bmatrix}-1 &2 & 0 \end{bmatrix}^T \end{align*} \begin{align*} \vec{i}&=\begin{bmatrix}1 &-2 & 0 \end{bmatrix}^T\\ \vec{j}&=\begin{bmatrix}-2 &-1 & 0 \end{bmatrix}^T \end{align*} \begin{align*} \vec{i}&=\begin{bmatrix}\sqrt{3} &-1 & 0 \end{bmatrix}^T\\ \vec{j}&=\begin{bmatrix}1 &\sqrt{3} & 0 \end{bmatrix}^T \end{align*}

What can you observe about these coordinate frames?