Maximizing Velocity and Acceleration in a Slider Crank Mechanism

AI Thread Summary
The discussion focuses on calculating the velocity, angular velocity, and acceleration in a slider crank mechanism at a 45° angle. The initial calculations yield a piston velocity of 1.11 m/s, an angular velocity of link AB at 5.55 rad/s, and a radial acceleration of 49.3 m/s² for point AO. The user seeks assistance with determining the angle at which the velocity of point B is zero and when the angular velocity of link AB is maximized, suggesting that these conditions occur when the crank and link are aligned or at right angles. The conversation emphasizes a desire for guidance rather than direct answers, indicating a commitment to learning the underlying concepts.
hopkid
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Homework Statement


a) For the mechanism shown in figure 1 determine for the angle (45°):
i) the velocity (VBO) of the piston relative to the fixed point (O).
ii) the angular velocity of AB about point A.
iii) the acceleration of point B relative to A.
*Note: Link AB is horizontal when angle = 45°.
b) Determine the value of the angle (measured from vertical) when:
i) the velocity of point b = 0.
ii) the angular velocity of link AB a maximum.
c) what is the maximum angular velocity of link AB?

Homework Equations


v=lw (a)r = v^2/l

The Attempt at a Solution


l of AO = 50mm = 0.05m
l of AB = 200mm = 0.2m
w of AO = 300 x 2pi/60 = 31.42 rads^-1

ai) V of AO = l of AO x w of AO.
V of A0 = 0.05 x 31.42
V of AO = 1.57ms^-1

*Use this value to draw a velocity diagram (attached).

To confirm using trig,

Sin(45°) = V of BO/1.57
V of BO = 1.57 x Sin(45°)
V of BO = 1.11ms^-1

aii) v=lw therefore w=v/l

w of AB = v of AB/l of AB
w of AB = 1.11/0.2
w of AB = 5.55 rads^-1

aiii) (a of BA)r = v of BA^2/l of BA
(a of BA)r = 1.11^2/0.2
(a of BA)r = 6.16ms^-2

*Tangential velocity is calculated in the acceleration diagram (attached).

Radial acceleration of AO calculated below to use in the diagram.
(a of AO)r = v of AO^2/l of AO
(a of AO)r = 1.57^2/0.05
(a of AO)r = 49.3ms^-2

So, as per the diagram, (a of BA)t = 34.5ms^-2.

This is the part where I am starting to struggle, I have had conformation back from my tutor that the first 3 parts to the question look good but I am struggling with parts b & c.

If someone could help push me in the right direction then it would be very much appreciated. I am after guidance, not answers, I want to learn.
 

Attachments

  • Offset Crank Slider.JPG
    Offset Crank Slider.JPG
    12.7 KB · Views: 2,738
  • Velocity Diagram.JPG
    Velocity Diagram.JPG
    20.7 KB · Views: 1,752
  • Acceleration Diagram.JPG
    Acceleration Diagram.JPG
    32.5 KB · Views: 1,537
No I have no new information, I've tried to be as thorough as possible with my post because in the past people have asked for more information. I'm really struggling with understanding this, and I'm not too sure why I am not getting any responses?!
 
Hi,

Did you have any luck finding out about Q1b and Q1c? I am inclined to think that velocity B=0 occurs when the crank arm and link create a straight line. And max acceleration occurs when they are at right angles. Then its just a case of drawing out the triangle to find theta? Is that anywhere along the right line?

-Mikey
 
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