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Simple Mechanics, relative motion example

  1. Nov 21, 2011 #1
    1. The problem statement, all variables and given/known data
    Might be the sketch most useful first:
    ex.3.3.jpg
    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?

    2. Relevant 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

    3. 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 )
     
    Last edited: Nov 21, 2011
  2. jcsd
  3. Nov 21, 2011 #2
    Later edit, sorry for the typinng bug.
     
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