Pendulum & base motion

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P3 - Motion with support (dec)accelerationYCar accelerationaoXgRear-view mirrorOθLgm, Ig2I g = m⋅r kmass moment of inertia about cgr k = radius_of_gyrationL g = distance_from_cg_to_pivot_O→a o→⎯b og→= a o ⋅ k ao : acceleration of pivotIf ao
(c) Find maximum angular displacementfor maximum angular displacement β=θmax, set ω=0 in eq. [8] to obtain: cos β( ) z ⋅ sin( β)cos βsquare both sides of equation to obtainUsing the identity( ) 2 1 − sin( β) 2cos β( ( ) − 1) z ⋅ sin( β)− = 0− = 1 [9] solve this equation with a calculator or:( ) 2 2 ⋅ zcos β( )( )( ) 2− ⋅ cos β ⋅ sin β + z 2 ⋅ sin β = 1( ) sin β= above, reduces Eq. above to z 2 − 1 ⋅ − 2 ⋅ z⋅cos β ⋅ sin β = 0( ) 2( )( )or( ) ⎡⎣ z 2 − 1sin β( ) sin β( )( ) ⎤⎦⋅ ⋅ − 2 ⋅ z⋅cos β = 0 [10]This equation has two roots, sin(β)=0, i.e. β=0 which is a trivial solution; and z 2 − 1( ) sin β( )( )⋅ − 2 ⋅ z⋅cos β = 0from whichFor the problem data given,2 ⋅ z( ) =⎛ ⎞⎜⎝ z 2 ⎟− 1 ⎠tan βcar initial speed vel := 100 ⋅ kmhwherez =a ogkmh :=1000 ⋅ m3600 ⋅ sdecelerationa o := −20⋅kmhsa o = −5.556ms -2β := atan⎛⎜⎝a oz := z = −0.567gMaximum angle θmax2 ⋅ z ⎞ 180z 2 ⎟⎠⋅π β = 59.064 degrees− 1NOT for exam: Graph the max-angle for a number of acceleration ratios (respect to g)θ max ( z) := atan⎛⎜⎝2 ⋅ zz 2 − 1⎞⎟⎠⋅180πmaxmum angular displacement10075502502550751001 0.75 0.5 0.25 0 0.25 0.5 0.75 1z: acceleration ratio (a/g)when z=1, g 9.807 m s -2=g= 35.304 kph ( acceleration rate)sθ max (.999) = −89.94390 degree
Page 1: Problem 3 (30 points)A car has been

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