# Mechanical engineering mechanics question: Motions of a triangular plate

1. May 4, 2016

### sophiebeth100

1. The problem statement, all variables and given/known data
Shown in picture

2. Relevant equations
So far just used w=v/r

3. The attempt at a solution

Could only do part a.

W=v/r
=0.3/0.2

For part b I would assume the velocity is zero? But I don't think that's correct. Any help with part b and c would be great, and is part a correct? Thank you :)

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2. May 4, 2016

### BvU

Hi beth,
funny to see 'Shown in picture' for the problem statement, right next to the picture. It doesn't say which half of the picture constitutes the problem

You sure about your a answer ? Does it mean $\omega$ is in fact constant ?

3. May 4, 2016

### sophiebeth100

Could you be more specific in regards to my answer for a? Did I use the right formula? Is there a different formula I should be using?

4. May 4, 2016

### BvU

I asked first ! Where are your v and r and what is your axis of rotation ? Don't you become suspicious if the answer doesn't seem to depend on $\beta$ ?

5. May 4, 2016

### sophiebeth100

I used the v for the hydraulic cylinder, 0.3, and the radius 0.2, as the edge of the triangle is.

I think I have found my mistake - I had assumed B was the centre of the circular motion but I now see that B can move and this is wrong. Since there is no fixed point on the triangle, I don't know where to go from here

6. May 4, 2016

### BvU

Oh boy, back to the drawing board .

There was something else I wanted to bring up: your equation comes from $\ \ \vec v = \vec \omega \times \vec r\ \$, a vector product. You familiar with that ? In other words, if you want to write $\omega = v/r$ you can't just take any $v$: you need the tangential velocity (the component perpendicular to $\vec r$).

With that, do you now see an opening to get going with this exercise ?

7. May 4, 2016

### sophiebeth100

I see. So when beta=30, the side CB is vertical. However, since CB=r, wouldn't the perpendicular velocity=0, since the machine only operates up and down?

8. May 4, 2016

### BvU

Yes, they made a deceiving drawing. On purpose ?
Even if CB is vertical, the thing is rotating when A moves upwards: B moves to the right. So $\omega \ne 0$.

Can you see a relation between $\omega$ of CD and $\omega$ of BA ?

 typo: between $\omega$ of CB and $\omega$ of BA

Last edited: May 4, 2016
9. May 4, 2016

### BvU

Not because the machine only operates up and down. But because B only moves horizontally. So if you answer b) first you do have an answer for a) -- but you have to take into account that C also moves !
But perhaps the composer of the exercise wasn't that devious after all and gave you part a) first for a reason....

10. May 4, 2016

### sophiebeth100

Do you mean of CB and BA? But hmm are they the same?

11. May 4, 2016

### BvU

Very good. Next step: any relation with $\beta$ ?

12. May 4, 2016

### sophiebeth100

I really can't find a relationship between those two things

13. May 4, 2016

### BvU

What can you say about $\beta$ if the triangle does not rotate (only translates) ?

14. May 4, 2016

### sophiebeth100

The value stays the same?

15. May 4, 2016

### BvU

Yes. So what is ${d\over dt} \beta$ when it rotates ?

16. May 4, 2016

### sophiebeth100

Is that the change in β related to time? Where does time come into it?

17. May 4, 2016

### BvU

You want $\omega$ at the moment that $\beta = {\pi/6}$. A is moving, remember ?
Do you know the definition of $\omega$ ? It is the time derivative of an angle. What angle in your exercise ?

18. May 4, 2016

### sophiebeth100

Ah I see, so ω=dθ/dt

So to calculate the time, use time = distance / speed.
Not 100% sure on this, but is the distance 0.1m? Since it is half the height of one side of the triangle?
If that's right, time = 0.1/0.3 = 0.33s

ω=dθ/dt
ω=(π/6)/0.33