Finding the acceleration in a gear rotating and constrained by a link.

[theBEAST]

Homework Statement

I am having difficulties with finding the acceleration at A. In the picture above are the questions that I need help with.

 

[Simon Bridge]

$a_0=r\alpha$

 

[mishek]

Don't you have both translation and rotation here?

This third member in your third row (ω²*r_O/IC) should be centripetal acceleration? By definition, it is directed towards the center of the path? When your gear moves from one position to another, doesn't his center makes only translation?

 

[theBEAST]

Don't you have both translation and rotation here?

This third member in your third row (ω²*r_O/IC) should be centripetal acceleration? By definition, it is directed towards the center of the path? When your gear moves from one position to another, doesn't his center makes only translation?

Alright now I understand why (ω²*r_O/IC) is zero. But what about a_I?

 

[Simon Bridge]

This is why I don't like to see just a set of equations and working out ... the reasoning is also needed if the working out etc is to make sense.

The trick to understanding the equations is to think through what is the author trying to calculate? Once you know the reasoning the rest follows.

So, off your question, what about a_I? What does it tell you? What would be it's physical meaning? Can you put it into words?

 

[theBEAST]

This is why I don't like to see just a set of equations and working out ... the reasoning is also needed if the working out etc is to make sense.

The trick to understanding the equations is to think through what is the author trying to calculate? Once you know the reasoning the rest follows.

So, off your question, what about a_I? What does it tell you? What would be it's physical meaning? Can you put it into words?

a_I is the acceleration tangential to the gear + the acceleration of the whole gear horizontally? Would those cancel each other out? I am not sure :S

 

[Simon Bridge]

acceleration of what?
you need to start out from the beginning.

I like to redefine things so I can think about them properly.
I don't like subscripts ... so I try not to use them if I can help it.
I also don't like putting the numbers into the equations so soon.

Let the points be as labelled.
Add point D, directly below O and A, same height as B.

So,
the radius of the gear is R=|DO|
the radius to the peg is r=|OA|
the distance along the arm to the peg is L=|BA|

angular speed of the gear is $\omega$
angular acceleration of the gear is $\alpha$

(all these things have values that are given to us)

need to find the angular speed $\omega_a$ and acceleration $\alpha_a$ of the arm about point B.

It may be useful to the calculation to define the angle ∠ABD the arm makes with the horizontal to be $\theta$ so that $\sin(\theta)=(R+r)/L$.

we can see that peg A is moving horizontally to the left at the time shown. So let's make that direction positive, and, while we're at it, measure positive angles anti-clockwise as well.
The peg's motion comes from the whole gear moving to the left and also from the gear's rotation.

The whole gear is going left at the speed of point O, so that would be $v_O=+R\omega$ but the peg is also moving wrt point O, with speed $v_{AO}=+r\omega$ [1]

Therefore, you can work out the velocity of A wrt to the ground, $v_{A}$.
Only the tangential part matters to the arm though, because $v_t=v_A\sin(\theta)=L\omega_a$

I hope from this example you can see the value of writing out what everything means using words. You should be able to work the problem from there. Then - having understood the problem, you will be in a better place to figure out what all those chicken-scratchings on that page are about.

(Be careful to check my working - don't take it for granted that I have done stuff right.)

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[1] we hates the subscriptes we does! Here, a double subscript $p_{XY}$ is property p of point X measured wrt point Y. If there is no Y then it's wrt to the ground.)