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# P8 - 3 4 30 sin 30 cos 2 2 L a y x L a A G G A AG AG A...

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Discussion 8: Name: TAM212, Spring, 2003 The vertical bar is fixed. The other bar AB (of uniform mass m and length L ) is initially held at the angle shown. It is attached to a collar that can slide vertically without friction on the vertical rod. End B of bar AB is attached to a vertical cable BC. If the bar is released from rest, what is its angular acceleration immediately after release? I c = mL 2 /12. Hint: set up three “kinetic” equations and two “kinematic” equations. For the latter, note point A of the bar can only accelerate vertically, and point B of the bar (at the moment shown) can only accelerate horizontally. = = x m F F A x & & (1) = = y m mg T F y & & (2) ( )( ) ( ) ( ) ( )( ) ( ) ( ) k i j i j j i M α = × + + × = c zz A C I F L T L 30 sin 30 cos 2 / 30 sin 30 cos 2 / ( ) α = 12 / 4 / 4 / 3 2 mL L F TL A ( ) α = 3 / 3 mL F T A (3) ( )( ) ( ) ( ) i j j i i j i k j r r k a a + α + = + + × α + = ω × α + = 3 4 / 0 30 sin
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Unformatted text preview: 3 4 / 30 sin 30 cos 2 / 2 L a y x L a A G G A AG AG A G & & & & v v v v x-direction 4 / L x G α = ⇒ & & (4a) y-direction 4 / 3 L a y A G α + = ⇒ & & (4b) ( ) [ ] [ ] ( ) ( ) [ ] ( ) i j j i j i j i k j i r r k a a − α − + = + ⇒ + + − × α + ρ + = ω − × α + = 3 4 / 30 sin 30 cos 2 / 2 2 L s y x L s s B G G B B BG BG B G & & & & & & & & & v v v v x-direction 4 / L s x B G α − = ⇒ & & & & (5a) y-direction 4 / 3 L y G α − = ⇒ & & (5b) Plug (4a) into (1) 4 / L m F A α = ⇒ ; Plug (5b) into (2) 4 / 3 L m mg T α − = ⇒ Plugging these into (3) ( ) ( ) ( ) L g L g mL L m L m mg 4 / 3 3 3 / 4 3 3 / 4 / 3 4 / 3 = α ⇒ α = ⇒ α = α − α − ⇒ j i k mg G T F A...
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