How Do You Solve the Offset Slider-Crank Mechanism at θ = 45°?

[oxon88]

Homework Statement

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°

Homework Equations

The Attempt at a Solution

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

 

[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

 

[oxon88]

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

Angular Velocity, ω = 10∏ rads/sec

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

θ = 45
V_AO = 1.571 m/s

Sin(45) = V_BO / 1.571

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


Is this correct so far?

 

[oxon88]

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

W_AB = V_BO/r = 1.11 / 0.2 = 5.55 rad/s^-1

 

[oxon88]

can anyone confirm if the above is correct please?

 

[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 I am still stuck

 

[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.

 

[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?

 

[billy_joule]

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

 

[agreaces]

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

 

[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
r = crank radius
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!

 

[adamhere]

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

 

[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!

 

[adamhere]

Hey OldEngr63, I'm sure your way of solving works but I'm looking for help solving using vector diagrams.

Thanks anyways!