Velocity & Accel of Pt C: Rod/Stick's Homework

[Nugso]

Homework Statement

http://i.imgur.com/toGAIGP.png?1
BC is vertical at the given position

Calculate the velocity and acceleration of the point C.

Homework Equations

V = ω*r

The Attempt at a Solution

I guess, AB, BC and CD rods have the same angular speed ω = 12 rad/s. And now I'm sort of up the creek and have no idea what to do. I, however, think that I can write the following equations:

V_a/(30√3) = V_b/30V_c = V_C/B + V_C/D + V_B + V_D

 

[tiny-tim]

Hi Nugso!

For a start, what are the lengths of AB BC and CD ?

 

[Nugso]

Hi tiny-tim and merry Christmas!

AB = 20$\sqrt{}3$ cm

BC = 30 cm

CD = 16$\sqrt{}3$ cm

 

[tiny-tim]

CD = 16$\sqrt{}3$ cm

hmm … am i missing something?

the diagram doesn't seem to say how high up D is

 

[Nugso]

hmm … am i missing something?

the diagram doesn't seem to say how high up D is

Ummm, well, how about CD² = (30-10$\sqrt{}3$)² + 24²

I think the equation is correct only when A and D have the same height.

 

[tiny-tim]

I think the equation is correct only when A and D have the same height.

yes, but i don't think they do have the same height

anyway, we can make a start …

we know completely the velocity of B

and we can find at least one of the components of the velocity of C relative to B

 

[Nugso]

yes, but i don't think they do have the same height

anyway, we can make a start …

we know completely the velocity of B

and we can find at least one of the components of the velocity of C relative to B

Can we calculate the length of CD without assuming they have the same length or AB and CD are parallel to each other?

How do we know the velocity? From the equation V = ω*r which is then 12*20$\sqrt{}3$? Or do you mean the angular velocity of B which is 12 rad/s?

 

[tiny-tim]

How do we know the velocity? From the equation V = ω*r

yes, V = ω*r gives you the speed, and the direction is … ?

 

[Nugso]

The direction is upwards I think but when want to do it with i, j, k I can't find it.

w= -12k

r = (30i - 10√ 3j)

Hence when we multiply, we get 2 directions one is - j and the other one is +i , right?

 

[tiny-tim]

you mean v_B = 12(-30j + 10√3i) ?

actually, no

but wouldn't it be simpler just to say that it's obviously perpendicular to AB? ​

 

[Nugso]

Well when I think of it logically I guess you are right, but why can't I write it with I j k versions? Also shouldn't we put k after w since its direction is in k?

 

[tiny-tim]

Well when I think of it logically I guess you are right, but why can't I write it with I j k versions?

you can, but you're much more likely to make a mistake (with all those minuses) if you do!

Also shouldn't we put k after w since its direction is in k?

using your method, yes

 

[Nugso]

you can, but you're much more likely to make a mistake (with all those minuses) if you do!


using your method, yes

I don't seem to get the right direction with my method

 

[tiny-tim]

you got a minus wrong!

 

[Nugso]

you got a minus wrong!

All right, I'd not bother with minuses from now on especially when there's much easier way! So we know the velocity of B and now how do we calculate the C's velocity? V_c = 12*30? so V_C/B = 12*30 - 12*20√3

 

[tiny-tim]

So we know the velocity of B and now how do we calculate the C's velocity?

try calculating the velocity of C relative to B …

what can you say about it?

 

[Nugso]

V_c/b= 12*20√3 - 12*30?