# Homework Help: Coriolis Acceleration of a Mechanism

1. Nov 2, 2014

### ConnorM

1. The problem statement, all variables and given/known data
Rod AB rotates with the angular velocity and acceleration CCW as shown. Points A and D are pin connected. The collar C is pin connected to the link CD and slides over the link AB. At the instant shown the link CD is vertical and the link AB has an angular velocity of 2 rad/s, and an angular acceleration of 4 rad/s2

determine,

1) The angular velocity of link CD
2) The relative velocity of point C with reference to A
3) The velocity of point C
4) The angular acceleration of link CD
5) The relative acceleration of point C with reference to A
6) The acceleration of point C

http://imgur.com/6lTfnYV --> Here is a picture of the question so you can see what the mechanism looks like.

2. Relevant equations
v = ωr (1)
an = ω2r (2)
at = αr (3)
vb = va + vb/a (4)
ab = aa + ab/a (5)

Coriolis Accel. eqn

acorn = -2ω(vb/a t) (6)
acort = 2ω(vb/a n) (7)

3. The attempt at a solution

Since this solution involves drawing various vectors I will include a picture of my work to help make things a bit easier. All of my work is in this picture as well so you can either look at my work their or at what I have written below. http://imgur.com/PAjUJoh,4PSWKHt

VELOCITY ANALYSIS

First off I started with the velocity analysis and drew where I thought the velocity vectors of each point would be. From my vectors vc, vc', and vc/c' I found,

vc' = vc cos 30 = 1.386 m/s
vc/c' = vc sin 30 = 0.8 m/s

next I found my angular velocity of CD,

ωCD = vc' / rCD = 4.62 rad/s

ACCELERATION ANALYSIS

I started off by writing down all the relevant equations

acn = rAC ωAC2
act = rAC αAC

ac'n = rCD ωCD2
ac't = rCD αCD

ac/c'n = ?
ac/c't = ?

acorn = -2ωCD(vc/c'n)
acort = 2ωCD(vc/c't)

What I am stuck on is that I don't know where my vectors are supposed to go. I'm not sure what I've done wrong and don't really know where to go from here.

Last edited: Nov 2, 2014
2. Nov 2, 2014

### Staff: Mentor

The ω^2-related accelerations are always towards the center of rotation, the others orthogonal to it. And you know the orientations of those vectors with your 30°-sketch, this is similar to the velocities.

3. Nov 2, 2014

### ConnorM

So since I drew my vc/c' directly down my ac/c'n would be down in the same direction and I would have no tangential ac/c't?

4. Nov 2, 2014

### Staff: Mentor

I'm not sure if I understand your notation (why don't you use coordinates like x and y?). You have both horizontal and vertical acceleration at point C.

5. Nov 2, 2014

### ConnorM

It's ok I think I have found what I did wrong!