Might be the sketch most useful first:
The prism is sliding horizontally from left to right under condition : S(t)=4*t*t+5*t+1[cm]
and the bar OA is rotating, point A is sliding over the prism's slope and point O is the immobilized center of bar's rotation, OA=10cm.
What would be the angular velocity and acceleration of the bar at the moment t=1s,
if at that moment the bar is declined at β=60deg from the vertical?
Idea is the point A is once transferred by the prism's horizontal motion to the left ( Ve and Ae ),
and twice transferred over the prism's slope in relative motion ( Vr and Ar ).
It's absolute motion ( Va and Aa ) comes also from the circular path with radius OA.
The Attempt at a Solution
Ve(t)=dS(t)/dt, so Ve(1s)=13cm/s
point A is following a circle so Va is 60deg above Ox ( τ direction ),
otherwise Vr is parallel to the prism's slope ( η direction ).
This way vector solving of Va=Ve+Vr results in Va=13cm/s, Vr=13cm/s.
Va is perpendicular to OA so ω=Va/OA=1.3s-1
( this is my first problem, examples booklet answer is 1.0s-1 )
Ae(t)=dVe(t)/dt, so Ae(1s)=8cm/s.s ,
Ar is parallel to the prism's slope ( η direction ),
Aa has accelerating part αxOA ( τ direction ) and centripetal part ωxωxOA ( η direction ).
Solving Aa=Ae+Ar in vectors results in α=0.176s-2
( this is my second problem, examples booklet answer is 0.19s-1 )
I could attach my Mathcad sheet if needed.
(My results for Aa=17cm/s.s , Ar=11.5cm/s.s )