Suggestions:This problem is naturally broken up into two parts: from immediately before impact ("1") to immediately after impact with the curb ("2"), and from immediately after impact ("2") to the position with G directly over the contact point of the tire with curb ("3").
From 1 to 2:
The impact force of tire with curb, FC, is an impulsive force. Compared to this force we can ignore the non-impulsive forces of NL, NR and mg. Angular momentum about the contact point of the tire with the curb (call this point O) is therefore conserved. Before impact, the auto moves as a particle -- use the expression of angular momentum for a particle. After impact, the auto moves as a rigid body -- use the expression of angular momentum for a rigid body.
From 2 to 3:
Energy is conserved. During this period of time the auto rotates as a rigid body about O.
Let us know if you have questions on this problem.
12 comments:
I found an angular impulse momentum eqx and a work energy eqx. are these two related through omega?
In stage 2-3, immediately after the impact there would be no velocity, and angular velocity right? So T1 = 0. Would you only have V1, V2, and T2? Am I on the right track? Also were given a mass moment of inertia about G, is this Ig?
bryon, I thought it was the other way around. At stage two, the point of tipping, g would have a velocity and an omega but when g is above o, there wouldnt be a velocity or an omega. so T2 = 0. Not sure if this is correct but this was my approach.
I was thinking the same thing with T2 = 0, but maybe I'm missing something else with the potential energy, because the answer I'm getting is WAY too big. If I put the datum at G when the car is on the ground, then V2 would have a height of the sqrt(.88^2 + .76^2)-.76, right?
So your V1=0 then? Your height seems right. How big is your answer? I'm getting an answer of just a few m/s.
I was able to first use work-energy eqn's to solve for omega_1 (between states 2 and 3). I did just as Sam said, with my datum at initial height so that V1=0 and T2=0 (because there is no omega at state 3). Then I used conservation of angular momentum and the omega value from above to solve for initial speed. I got a reasonable answer by this method. Hope this helps and let me know if I missed something.
Oops. My bad. I was taking the 900 kg*m^2 to be k, not I_o. I should have been checking my units!
When using conservation of angular momentum, I used an r_G/O value of (-.88i - .76j). I'm not sure if this is right. Is V_G in the positive i-direction?
J.T.,
The j-component should be positive because you have to move left and up to get from O to G.
anyone knows why we r treating the car as a particle before impact and as a rigid body immediately after?
My understanding is that before the impact it is moving all together with no angular momentum, so your allowed to think of it as a particle. After the impact it is rotating, so it must be treated as a rigid body. I think you could treat it as a rigid body before and after impact, but its easier to treat it as a particle before impact.
thanks john
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