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The question asks me to find the angular velocity.

figure

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.

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2 Answers 2

up vote 1 down vote accepted

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

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

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When you say ar=a*er, where er=cos(th)i + sign(th)j ... does that mean I have to multiply the i-component in a with the i component of er ? and similarly to j ? –  Fendi Apr 12 '12 at 14:13
1  
Yes. When you do scalar product of two vectors, it eventually comes to multiplying its components, e.g. $\vec{a} \cdot \vec{b} = a_x b_x + a_y b_y$. But that is already mathematics. –  Pygmalion Apr 12 '12 at 15:12
    
One more question please, why is er=+cos +sin while et= +cos -sin ... did you take i to the right and j upwards ? or another convention ? –  Fendi Apr 13 '12 at 13:05
    
The choice of $\vec{e}_\text{r}$ and $\vec{e}_\text{t}$ comes from the polar (cylindrical) coordinate system ($r, \theta$). See, for example, picture No. 15 on en.wikipedia.org/wiki/Polar_coordinate_system. It just happens that unit vector $\hat{r} = \vec{e}_\text{r}$ always point in the direction of radial acceleration and unit vector $\hat{\theta} = \vec{e}_\text{t}$ always in the direction of tangential acceleration, regardless of $\theta$ and $r$. To get radial and tangential part of acceleration, you just have to multiply it with these unit vectors. –  Pygmalion Apr 13 '12 at 15:56
    
Thanks for the info. I have been trying to solve that now but with no avail, the answer keeps coming out wrong. Can you further explain how I get to choose the correct angle for cos and sin and relative position in terms of i and j ? I know ar always points along the radius r of 300mm, and at is tangent (90 degrees) to that r. However how does this affect the solving ? I think I'm getting confused with the whole coordinates system. –  Fendi Apr 13 '12 at 16:02

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

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

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This is a good answer, and more physical one, the only problem is that it is mathematically much more complex to solve (vector product and double vector product). –  Pygmalion Apr 14 '12 at 7:31
    
I agree. So I edited the answer with the planar projection of the cross products to make constructing the answer easier. –  ja72 Apr 15 '12 at 19:08

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