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/s^{2}
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.
## Answer:

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

v_{t} = 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 a_{r} =
v_{t}^{2}/r. The vertical component of
this acceleration is
v_{t}^{2}/r*cos(q)
or a_{rv} =
2.61^{2}/0.3*0.643 = 14.6m/s^{2}. 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 a_{t} = a*r. The vertical component of the
tangential acceleration is a_{t} times the sine of the
angle q or
a_{tv} = 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

a=(3-14.6)/(.3*.766) =
-50.5rad/s^{2}

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