Solution: 

Solution:
 
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Given:
 
■ Length of AD = 125 mm = 0.125 m
 
■ Length of AB = 62.5 mm = 0.0625 m
 
■ Length of BC = CD = 75 mm = 0.075 m
 
■ Speed of link AB, NAB = 10 rpm
 
■ Angular velocity at link AB ,ωAB = (2πN) / 60 = (2 x π x 10) / 60 =1.0471 rad/s
 
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Objective:
 
Find the Angular velocities of
 
(1) Link BC
(2) Link CD
 
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Step 1 of 4 │Determine the number of Instantaneous centers
 
 
The given four bar mechanism have four links (i.e n=4), Therefore number of Instantaneous centers is given by
 

 

 
Therefore Six instantaneous centers are in the given four bar mechanics. They are
 
I12, I14are the fixed instantaneous centres and I13, I23, I24, I34are varying instantaneous centres (Changes with configuration of links)
 
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Step 2 of 4 │Locate the instantaneous centres on the figure
 
 
Draw the Fourbar mechanism (Refer Figure.1) as per the dimension and locate the instantaneous centres. Measure the distance of the instantaneous from the joints.
 
Figure.1

 
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Step 3 of 4 │Determine the angular velocity of the link BC
 
 
We know that
 

 
 
Therefore velocity at point B in link AB is given by
 

 
Substituting known values
 

 

 
 
Since the point B is also a point on link BC, therefore velocity of point B on link BC is
 

 

 
 
From the Figure.1,
 
The distance I13.B =103.83 mm = 0.1038 m
 
 
Therefore
 

 

 
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Step 4 of 4 │Determine the angular velocity of the link BC
 
 
The instantaneous centre I13 is common for joints B and C, Therefore we can write
 

 

 
From the Figure.1,
 
The distance I13.B =103.83 mm = 0.1038 m
The distance I13.C =75 mm = 0.075 m
 
Substituting values
 

 

 
 
We know that
 

 

 
Therefore
 

 

 
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Answer:
 

 

 
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