Kinematics Problem

Question

The question asks me to find the angular velocity.

Now I do not want you to solve my homework, I want explanation please.

It states that the acceleration of point P is $\vec{a}= -3.02 \vec{i} -1.624 \vec{j}$ when $\theta=60$ It also states that the diameter of the flywheel is 600mm.

My question is :

I know that the acceleration is split into 2 components, tangential and normal.

I know that $a_n=r \alpha$ and $a_t=-r(\omega)^2$.

• Is everything I mentioned until now correct ?
• How can I know which value does an and at take from a given above ?
• How do I decide the i and j terms respectively ?

Again, please do not answer the question and find the angular velocity, but please explain the correct approach and whether my deductions are correct.

Answer 1

You have the acceleration vector already specified. You have to separate it into tangentual and radial components, and only after you obtain $a_\textrm{t}$ and $a_\textrm{r}$ you use expression $a_\textrm{t} = r \alpha$ and $a_\textrm{r} = r \omega^2$. Therefore, you can obtain $\alpha$ and $\omega$ of the flywheel.

You can separate $\vec{a} = a_x \vec{i} + a_y \vec{j}$ by multiplying it (by virtue of scalar product) with unit vectors for the point position

$$a_\textrm{r} = \vec{a} \cdot \vec{e_\textrm{r}}, a_\textrm{t} = \vec{a} \cdot \vec{e_\textrm{t}},$$ with

$$\vec{e_\textrm{r}} = \cos(\theta) \vec{i} + \sin(\theta) \vec{j}, \vec{e_\textrm{t}} = - \sin(\theta) \vec{i} + \cos(\theta) \vec{j}.$$ .

Answer 2

For your reference working backwards, if the acceleration of the origin is zero then the acceleration at P is

$$ \vec{a}_P = \vec{\alpha}\times\vec{r}_P + \vec{\omega}\times(\vec{\omega}\times\vec{r}_P) $$

where $\vec{r}_P = r\cos(\theta) \hat{i} + r \sin(\theta) \hat{j} $, $\vec{\omega}=\omega \hat{k}$ and $\vec{\alpha}=\alpha \hat{k}$. Then you equate the left hand side components (known) with the right hand side for the unknown $\omega$ and $\alpha$.

Note $\times$ is the vector cross product. Projected into the xy-plane these are

$$ \begin{pmatrix} 0 \\ 0 \\ \alpha \end{pmatrix} \times \begin{pmatrix} x \\ y \\ 0 \end{pmatrix} = \begin{pmatrix} -y \,\alpha \\ x \,\alpha \\ 0 \end{pmatrix} $$

and

$$ \begin{pmatrix} 0 \\ 0 \\ \omega \end{pmatrix} \times \left( \begin{pmatrix} 0 \\ 0 \\ \omega \end{pmatrix} \times \begin{pmatrix} x \\ y \\ 0 \end{pmatrix} \right) = \begin{pmatrix} -x \,\omega^2 \\ -y \,\omega^2 \\ 0 \end{pmatrix} $$

making the above vector equation into a planar one

$$ \begin{pmatrix} a_x \\ a_y \end{pmatrix} = \begin{pmatrix} -y \, \alpha \\ x\,\alpha \end{pmatrix} + \begin{pmatrix} -x \, \omega^2 \\ -y \,\omega^2 \end{pmatrix} $$