Mechanical engineering mechanics question: Motions of a triangular plate

  • #1

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 :)
 

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  • #2
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Hi beth, :welcome:
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 :smile:

You sure about your a answer ? Does it mean ##\omega## is in fact constant ?
 
  • #3
Hi beth, :welcome:
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 :smile:

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?
 
  • #4
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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
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
 
  • #6
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Oh boy, back to the drawing board :nb).

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
Oh boy, back to the drawing board :nb).

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?
 
  • #8
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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 ? :smile:

[edit] typo: between ##\omega## of CB and ##\omega## of BA
 
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  • #9
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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....
 
  • #10
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 ? :smile:
Do you mean of CB and BA? But hmm are they the same?
 
  • #11
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Very good. Next step: any relation with ##\beta## ?
 
  • #12
Very good. Next step: any relation with ##\beta## ?
I really can't find a relationship between those two things
 
  • #13
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What can you say about ##\beta## if the triangle does not rotate (only translates) ?
 
  • #14
What can you say about ##\beta## if the triangle does not rotate (only translates) ?
The value stays the same?
 
  • #15
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Yes. So what is ##{d\over dt} \beta## when it rotates ?
 
  • #16
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?
 
  • #17
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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
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?
 

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