Rotary to Linear Motion

## Question: Consider the diagram at the right. At the instant q=50 degrees, the slotted guide is moving upward with an acceleration of 3m/s2 and a velocity of 2m/s. Determine the angular velocity and acceleration of link AB at this instant. Note that upward motion is in the negative y direction.

One of the things we know about rotary motion is that the tangential velocity, vt, is the angular velocity, w, times the radius, r. In equation form, vt=w*r, or w=vt/r. At any instant vt may be resolved into horizontal and vertical components according to the angle q in the diagram. The vertical component of vt is given by vv=vt*sin(q). But vv is the linear velocity of the slotted guide, so plugging that value in for vv and working back up the line of reasoning should get us the angular velocity.
vt = 2m/s / sin(50) = 2 / .766 = 2.61m/s
w = 2.61 / r = 2.61 / 0.3 = 8.7rad/s.

The acceleration of the slotted guide will be equal to the vertical component of the acceleration of the point A on the AB link. Point A may undergo acceleration in both a radial and tangential direction. The magnitude of the radial acceleration is given by the expression ar = vt2/r. The vertical component of this acceleration is vt2/r*cos(q) or arv = 2.612/0.3*0.643 = 14.6m/s2. To this we must add the vertical component of any tangential acceleration to get the acceleration of the slotted guide.

The tangential acceleration of the point A of link AB is related to the angular acceleration, a, through the relationship at = a*r. The vertical component of the tangential acceleration is at times the sine of the angle q or atv = a*r*sin(q).

Adding the vertical components of the radial and tangential acceleration and equating the sum to the vertical acceleration of the slotted guide gives us an equation to solve for a when q is 50 degrees.
3=14.6+a*0.3*0.766