Mechanical engineering mechanics question: Motions of a triangular plate

[sophiebeth100]

Homework Statement

Shown in picture

Homework Equations

So far just used w=v/r

The Attempt at a Solution

[/B]
Could only do part a.

W=v/r
=0.3/0.2
=1.5rad/s

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 :)

 

[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 ?

 

[sophiebeth100]

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 ?

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?

 

[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 $\beta$ ?

 

[sophiebeth100]

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 $\beta$ ?

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

 

[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 ?

 

[sophiebeth100]

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 ?

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?

 

[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 ?

[edit] typo: between $\omega$ of CB and $\omega$ of BA

 

[BvU]

However, since CB=r, wouldn't the perpendicular velocity=0, since the machine only operates up and down?

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...

 

[sophiebeth100]

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 ?

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

 

[BvU]

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

 

[sophiebeth100]

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

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

 

[BvU]

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

 

[sophiebeth100]

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

The value stays the same?

 

[BvU]

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

 

[sophiebeth100]

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

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

 

[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 ?

 

[sophiebeth100]

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 ?

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
ω=π/2 rad/s

Is this right?