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Acceleration & velocity using instantaneous centre of zero velocity

  1. Apr 18, 2014 #1
    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?


    3. 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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    Last edited: Apr 18, 2014
  2. jcsd
  3. Apr 18, 2014 #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?
     
  4. Apr 18, 2014 #3
    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
     
  5. Apr 18, 2014 #4
    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
     
  6. Apr 18, 2014 #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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