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.

The tangential velocity of the point B is the angular velocity times the 0.3m radius, or vt = 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.

This information is brought to you by M. Casco Associates, a company dedicated to helping humankind reach the stars through understanding how the universe works. My name is James D. Jones. If I can be of more help, please let me know.

JDJ