Acceleration & velocity using instantaneous centre of zero velocity

  • Thread starter Ctom101
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  • #1
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1. A link is connected by two bearings A and B. The bearings run along slots which are at right angles to each other. The length of the link AB (length between the centres of bearings A and B) is 0.2m. At an instant of time, height H, is 0.05m, and the bearing A has a constant velocity V, of 0.1m/s upwards.
At the instant of time calculate;
The velocity of bearing B.
The angular velocity of the link AB.
The acceleration of bearing B.
The angular acceleration of the link AB





2. velocity B = distance from rotational centre x angular velocity of link AB.
angular velocity of link AB = velocity A / distance from centre of zero velocity




3. from the working out in my picture changing it into a triangle I think I have worked out the velocity of bearing B and the angular velocity of link AB, but I dont know how to find the acceleration of B or the angular acceleration of AB.
Could somebody please help me with this and let me know if what i've done so far is incorrect?


The Attempt at a Solution


a = 0.05 c = 0.2
a2 + b2 = c2
0.22 / 0.052 = 16 root 16 = 4 b = 4
velocity A / distance from centre of zero velocity
0.1 / 0.05 = 2rad/s angular velocity of link AB
velocity B = distance from rotational centre(b) x angular velocity link AB
4 x 2 = 8m/s

ωv AB = 2rad/s
v B = 8m/s
 

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Answers and Replies

  • #2
Simon Bridge
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You have to exploit the relationship between A and B ... i.e. the dimensions of the triangle.
I don't understand your working though.

If I put c=|AB|, with the position of B as x and the position of A as y, then

c^2=x^2+y^2

...differentiate through by time to get the relationship between the velocities, and again to get the accelerations.
what is the acceleration of A?
 
  • #3
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It says A has a constant velocity of 0.1m/s, so I think acceleration is 0

Looking over it again I think I worked out the length b incorrectly.

link AB = 0.2m
Height = 0.05m

Pythagoras: a^2 + b^2 = c^2

link AB = c
Height = a

c^2 - a^2 = b^2

0.2^2 - 0.05^ = 0.0375
root 0.0375 = 0.1936
b = 0.19
 
  • #4
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angular velocity of link AB = velocity A / distance from centre of zero velocity

velocity A = 0.1m/s
distance from centre of zero velocity (a) = 0.05m
0.1/0.05= 2rad/s

velocity B = distance from centre of zero velocity x angular velocity of link AB

angular velocity of AB = 2rad/s
distance from centre of zero velocity (b) = 0.19

0.19 x 2 = 0.38m/s
velocity B = 0.38m/s

I think this is now right but I don't know how to move on for;
The acceleration of bearing B.
The angular acceleration of the link AB
 
  • #5
Simon Bridge
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I still don't see how your reasoning arrives at the required quantities.

What does "center of zero velocity" mean?
You seem to be giving the velocity of A in radiens per second as if it were moving perpendicular to the corner.
Is this correct?

What's wrong with the suggestions I already gave you?
 

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