# Acceleration & velocity using instantaneous centre of zero velocity

**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

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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