
#1
Nov1312, 01:45 PM

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1. The problem statement, all variables and given/known data
(see attachment) 2. Relevant equations 3. The attempt at a solution I am clueless here. I can conserve angular momentum but how will i find the "impulsive reaction"? 



#2
Nov1312, 03:57 PM

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Let the impulsive reaction at A be I, a vector.
How will the rod move just after the impact? Put in an unknown for that. You can now write down three equations: conservation of linear momentum (one vertical, one horizontal) and conservation of angular momentum. 



#3
Nov1412, 01:59 AM

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mv=mv_{x}' (v_{x}' is the horizontal velocity of rod after impact) v_{x}'=v (is this correct?) I am not sure how conserving the linear momentum in vertical direction would help? Initially it is zero and after the impact, even if we assume that the rod has got some velocity in vertical direction, the velocity of rod in vertical direction turns out to be zero. 0=mv_{y}' (v_{y}' is the velocity of rod in vertical direction) I am still not sure how to proceed further. 



#4
Nov1412, 02:16 AM

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Impulsive reaction 



#7
Nov1412, 11:14 AM

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The angular momentum is conserved as there is no external torque acting on the system rodparticle. But the linear momentum is not conserved, as there is a force at the hinge during the impact.
The change of linear momentum is m(rod) vcm(rod)mv=ΔI where ΔI is the impulse of the force at the hinge. You can call it "impulsive reaction". So find the angular velocity of the rod after the impact,, you get the linear velocity of the CM from it. ehild 



#9
Nov1412, 11:27 AM

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Glad to hear. I never heard about "impulsive reaction" in this sense. Google gave such things as http://www.vsatbay.com/1/post/2012/0...reactions.html, for example
ehild 



#10
Nov1412, 11:32 AM

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But, is haruspex wrong with his/her suggestion of conserving linear momentum? 



#11
Nov1412, 03:12 PM

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The change of linear momentum is equal to the impulse of the external force. Here the force comes from the hinge, but that is opposite to the force the rod exerts on the hinge. ΔP=F(external)Δt can also be written as Pi=Pf+F(system)Δt, and that looks as conservation of impulse or momentum, just like when they speak about conservation of energy when friction does work. I would not say it conservation of energy or conservation of momentum, but other people do. Haruspex wanted you to include the impulse from the hinge when you wrote the change of linear momentum, he only called the law ΔP=F(external)Δt "conservation of momentum".
ehild 



#12
Nov1412, 03:50 PM

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Conservation of linear momentum (only need to consider horizontal since there is no vertical motion involved) gives you an equation involving the reactive impulse, I, and the speed of the rod directly after impact. The second equation is from conservation of angular momentum, but what equation you get will depend on where you take moments. If you take moments about the hinge it will not involve I; if you take them about the c of g of the rod it will. Either way, you end up with the same answer. 



#13
Nov1412, 04:18 PM

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ehild 



#14
Nov1412, 05:27 PM

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#15
Nov1412, 10:26 PM

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Thanks you both for the replies, i understand it better after this discussion.




#16
Nov1412, 10:28 PM

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ehild 



#17
Nov1512, 07:53 AM

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Δp = _{t1} ^{t2}∫F_{ex}dt Where ∫F_{ex}dt is just impulse.... If impulse =0 , we get Δp=0 , which is law of conservation of linear momentum... 


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