Linear Velocity from Compound Rotating
Links

## Question:

Consider the diagram at the right. The
shaper mechanism is designed to give a slow cutting stroke and a
quick return to a blade attached to slider at C. Determine the
velocity of the slider block at the instant
q is 60 degrees.
## Answer:

The tangential velocity of the point B is the angular velocity
times the 0.3m radius, or v_{t} = 4*0.3
= 1.2m/s. The velocity of the point C is the horizontal
component of the velocity of point B plus the horizontal
component of any tangential velocity about the point B. The
horizontal component of the tangential velocity at B is 1.2m/s*sin(q) =
1.2*0.866 = 1.04m/s. The link BC is rotating such that the
vertical component of the tangential velocity of B relative to C
is the vertical component of the velocity of point B. The
horizontal component of the tangential velocity of B relative to
C will be equal to the vertical component since the BC link is at
45 degrees. Each component of the tangential velocity of B
relative to C is then 1.2*cos(60) =
0.6m/s. The horizontal component of this velocity
contributes to the velocity of C so the velocity of C is 1.64m/s.
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