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## Homework Statement

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?

## Homework Equations

Va=ωxOA

Aa=αxOA+ωxVa

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.

Va=Ve+Vr

Aa=Ae+Ar

## 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 )

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