# Offset slider-crank problem

1. Aug 8, 2014

### oxon88

1. The problem statement, all variables and given/known data
For the mechanism shown in FIGURE 1 determine for the angle θ = 45°:
(i) the velocity of the piston relative to the fixed point O (VBO)
(ii) the angular velocity of AB about point A (i.e. ωAB)
(iii) the acceleration of point B relative to A (aBA).

Note: Link AB is horizontal when θ = 45°

2. Relevant equations

3. The attempt at a solution

this is a self study question which im struggling with, can anyone offer some guidance please?

2. Aug 8, 2014

### billy_joule

Where are you stuck?

If you haven't found the velocity and acceleration of point A you should revise uniform circular motion.

Once you have that the rest should follow pretty easily.
you only need v=wr and some trig. to answer i and ii

3. Aug 10, 2014

### oxon88

(i) the velocity of the piston relative to the fixed point O (VBO)

Angular Velocity, ω = 10∏ rads/sec

VAO = ω x r = 0.05m x 10∏ rads/sec = 1.571 m/s

θ = 45
VAO = 1.571 m/s

Sin(45) = VBO / 1.571

VBO = Sin(45) x 1.571 = 1.11 m/s

Is this correct so far?

4. Aug 10, 2014

### oxon88

(ii) the angular velocity of AB about point A (i.e. ωAB)

WAB = VBO/r = 1.11 / 0.2 = 5.55 rad/s-1

5. Aug 14, 2014

### oxon88

can anyone confirm if the above is correct please?

6. Sep 9, 2014

### bonsai

I'm stuck on the same questions! Have you drawn it out with scaling? Eg. 1cm = 1m/s & 1cm = 10mm?

someone suggested this to me but im still stuck

7. Sep 9, 2014

### oxon88

i was totally stuck with this. Not looked at it recently. i did draw it and got the same result as i recall. drop me a pm if you like.

8. Sep 11, 2014

### 26m

Yes I'm also stuck on this, I get the same value for the velocity of the piston and the link AB, it's the acceleration I'm struggling with? Anybody managed to solve this part?

9. Sep 11, 2014

### billy_joule

Are you familiar with circular motion?
do you know the equation for centripetal acceleration?

10. Sep 23, 2014

### agreaces

can you just use w^2 X r or (V^2)/r for the acceleration

11. Oct 2, 2014

### OldEngr63

Forget about the connecting rod being level at this instant, and instead think of the situation with very small theta, so that the crank pin is above the wrist pin, giving the connecting rod an inclination phi.
Now write these two loop closure equations:
r*sin(theta) + L*cos(phi) - x = 0
r*cos(theta) - L*sin(phi) - e =0
where
theta = crank angle from vertical as given
L = connecting rod length
phi = angle of connecting rod with the horizontal
x = horizontal distance from crank pivot to wrist pin (on the piston)
e = offset from crank pivot up to piston center line.

For any assigned value of theta, you can solve the second equation for phi.
With those two values (theta, phi), you can evaluate x from the first equation.

Differentiate these equations once to find velocities, and differentiate again to find accelerations. It is really very, very simple, provided you are not afraid to differentiate!

12. Jan 25, 2015

For part (iii) I am unsure how to solve using vector diagrams.. I think the crank being offset has thrown me

I worked out aOA = 49.349m/s^2 and aBA(radial) = 6.161m/s^2

13. Jan 25, 2015

### OldEngr63

Did you try the process I described above? Somehow I doubt it because it never fails if you do it correctly. Since I don't know any better way to do the problem, I'll have to leave you to your own devices. Good Luck!!

14. Jan 27, 2015