Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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.

share|improve this question

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

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

share|improve this answer
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
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

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


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

making the above vector equation into a planar one

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

share|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.