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Of center of gravity and centripetal force...
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Here is a mechanics question which totally stumped me: A car together with its passengers has a combined mass of 2000kg. Its width, wheel-to-wheel is 2m and its center of gravity is 1m from the ground and midway between the wheels. It enters a roundabout which has a diameter of 10m. Calculate the speed below which it must travel to ensure that it will not flip over about the outer wheels. Is the car were to travel at half this max speed, what is the combined frictional forces on the wheels? Here are a couple of things that I'm quite confused about if you don't want to bother with the calculations: a) what causes the car to flip in the first place? b) in this case, if centripetal force is the resultant force of circular motion, what causes it in the first place? c) does the difference of normal force on both sides of the car cause the car to overturn? Please please help!!! |
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Re: Of center of gravity and centripetal force...
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by kudoushinichi
on 2007-12-11 05:22:13
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Erm... here's my attempt at the calculations questions. Max speed before the car flips = 2 sqrt g m/s Combined frictional force if car travels half this speed = 500g N But I'm not too sure... Unless it's correct, I can offer some sort of explanation. If it's not, then forget it.
Shinjitsu wa itsumo hitotsu!
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Re: Of center of gravity and centripetal force...
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Um i'm not sure that this is at my level of physics but here is what I think A) the motion of the car as it is turning changes the point about which the combined mass of the people and the car act which puts the car off balance as it turns above its maximum speed B)Would the resultant force not be the drive of the car in order for it to move in the first place? So it and friction (between wheels and the ground) would become the centripetal force? C)I'm not to sure about this at all Thats my go Sorry if that doesnt help But now I'm determined to figure this out I'll get back to you later if I think of something else Cana |
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Re: Of center of gravity and centripetal force...
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by kudoushinichi
on 2007-12-11 07:08:10
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Cana, I think centripetal motion is not your level yet... My tentative answers to the conceptual questions would be as follow; a) As the car is rotating about the outer wheels, there must be some torque that causes this. For this to happen, the center of mass of the car must go over the outer wheels. Now, what puzzles me is what force causes this torque? The best I've got is - some sort of centrifugal force. But I hear that centrifugal force is baloney... So I dunno anymore. b) Centripetal force is always the resultant force. In this case, the centripetal force is provided by the frictional force exerted by the road, on the tires. c) Well, I'm not sure if we should talk about normal forces in the first place... That's about what I can help...
Shinjitsu wa itsumo hitotsu!
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Re: Of center of gravity and centripetal force...
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The best way to answer the questions is to solve the problem yourself. Solve it by drawing a picture and identifying every applicable force and labeling them with variables. You then use all the equations you can regarding this until you have as many equations as you have variables. You then do tons of substituting and get one final equation in which you then reinsert the actual values. This is a little longer to solve it but it will garentee a correct solution so long as you got the equations right and your math is right. I'll do this myself when I get home from work tonight (around Midnight EST) if I am not too sleepy. If I can't do it then, then I'll do it the next morning. |
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Re: Of center of gravity and centripetal force...
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by SuicidopoliS
on 2007-12-11 12:07:41 (edited 2007-12-11 12:28:30)
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I don't like the centripetal force... I believe it's easier to look at things using angular velocities en stuff, but hey, that's just me. One of the reasons i don't like the centripetal force (CF), is that it kindda mixes cause and consequence. It is said that the CF is the resulting force in a circular motion. This, for one, depends on which reference frame you choose, so first of all, let's assume we take an inertial reference frame in which the "circle" is at rest, and only the car moves. So in this case, for the car to turn, it needs to accelerate towards the center of the circle it is driving on, assuming it's velocity in the frontal direction is constant. This acceleration, multiplied by the mass of the car, then gives you the centripetal force. Some people say the centripetal force is caused by friction, but i think this is a wrong way of looking at things. The friction doesn't cause the CF to be, the fact that you accelerate towards the center of the circle does. But you can only accelerate towards this center if there is friction, because if there wasn't any friction, you wouldn't turn, even if you turned your steering wheel... you'd just continue to move forward. so, once again, what causes the centripetal force is the acceleration towards the center of the circle, but in order to accelerate towards this center, you need friction. The amount of friction will put an upper limit to the amount of acceleration: if you accelerate more than this upper limit, you will partially move in a circular way, but you will also be thrown out of the bend, making you spiral outwards. This also implies that your car can only flip over if you have sufficient friction. If not, you'll glide outwards once again, but not flip. Let's switch views for a moment, making things easier. Suppose we now consider a referential frame in which the car is at rest, and thus it is the "circle" that's moving (in the opposite way as our car was first moving in the frame in which the circle was at rest). Now, we don't have a centripetal force anymore, but we do have a centrifugal force. The centrifugal force pushes our car outwards, but friction keeps it firmly on track. The thing is, the CF acts on the center of mass of our system, whilst frictions acts upon our tires, forming a force couple. If this couple starts working a bit too well, our car will flip over... If you're having trouble visualizing this, think of standing upwards in/on a train, while this train is taking a cornering. You, standing tall, will feel getting pushed outwards, so what do you do? Yes, indeed, you apply more force to one of your legs, so as to, thanks to friction, compensate the force that pushes you outwards. This way, you still stand on the same spot, and do not fall down like an idiot or start jumping around like a madman. However, should this train take this cornering faster and faster, and would you still, stubborn as you are, be standing upwards an applying more and more force on that leg of yours, there will be a point where you simply flip over that side of you body... And as i was just walking away, ready to go make my evening supper, i realized something should be pointed out about the train example though... What happens in fact is that, as your standing upwards with your two legs more or less spread out, as pointed out earlier, due to the circular motion of the train (or car) there will be a force couple acting upon you: centrifugal force acting on the mass center of your body, and friction telling you to move the other way. This makes you tilt a bit, so you compensate for this tilting by applying more force to one of your legs than the other. This is: you push yourself away from the floor on one side, so as to still keep standing upwards. Your car does the same thing: as you exert a, let's say counterclockwise circular motion, there will be more force on the right tires of your car than on the left side due to this tilting, or if you prefer, the normal on the right side will be bigger than on the left side. So in a way, your question c) is true, but not entirely: the car does indeed flip over because the normal on the right side is (much) bigger than on the left side, but this is not the cause of the flipping over. The cause is just the circular motion... In fact, your car is always tilting, but up to a certain point, it manages to push itself back straight up again... Once that point cross, it will flip over. I hope i have expressed myself more or less clearly, if not, i apologize for the confusion i will undoubtedly have provoked! .oO° Life's THE CURE, the rest are details! °Oo. |
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Re: Of center of gravity and centripetal force...
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According to my logic, the car will flip over when the centriptal force (Fc) is greater than the gravitational force holding cancelling out the moment(a.k.a. torque). Because the question makes the car look square life is made easy. The car will start to overturn when the tan(a)>45 degrees. tan(a) = Fc/Fg = [(mv^2)/r]/(mg) = v^2/(rg) v = ((5)(9.8)tan(45))^0.5 = 7 m/s Please point out any places that are confusing or holes in the logic. The second part I'm not as confident with :S I believe the frictional force will be equal to the acceleating force, which is the centripedal force (Fc). Ff = Fc = mv^2/r = (2000)(3.5)^2/(5) = 4900 N. |
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Re: Of center of gravity and centripetal force...
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They want the maximum velovity (linear i belive) that the car can travel aroun the curve. In other word Fg = Fc or the Weight force of the car that is acting upon the track has to equal the Centripedal force as it travels around the bend. So Fg=ma or Fg=mg, Fc=(m(v^2))/r --> Fc=Fg so 2000*9.81=(2000*(v^2))/5 there fore v=(((2000*9.81)*5)/2000)^0.5 v=7.003570518 v=7.00m/s (3sf) My Friction is only a guess We can model the cars motion by the use of a vetor diagram. Because Fc and Fg are at right angles to each other I persume that we can say that the addition of the vectors is the combined friction. So Fc=(2000*(3.5^2))/5 Fc=4905N Fg=2000*9.81 Fg=19620N Thus Ff?=((19620^2)+(4905^2))^0.5 Ff=20234.8N (which makes some scence as the total friction force is greater than the gravitational force acting upon the car) I do hope Gendou can correct me |
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Re: Of center of gravity and centripetal force...
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by kudoushinichi
on 2007-12-11 20:46:21
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Nee, Suicidopolis... I learned that acceleration is caused by a force, that is, Force is the cause and Acceleration is the effect. Simply put, you can't accelerate without a force. So, you saying that centripetal force is caused by the acceleration towards the center is well... wrong! Centripetal force is just like any other forces, only that it changes direction of motion instead of the object's magnitude of velocity. But then you said that the centripetal force is cause by the acceleration towards the center which is caused by friction. -_-" This is a redundancy, imo. The frictional force IS what causes the centripetal force~ How tight a turn a car can take is ultimately depends on two things, one is how low the center of gravity of the car, and the other is the frictional force between the road and the tires. Let's say if the center of gravity is so low that it's impossible for the car to flip about the outer wheels, then the frictional force between the road and the tires limit how small the radius of the circular motion be. Yo, ANTALIFE... Don't you have the answers to the question? I am now more and more certain with my answer... only waiting for someone to confirm my calculations. :P
Shinjitsu wa itsumo hitotsu!
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Re: Of center of gravity and centripetal force...
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My anwsers kudoushinichi came out to be v=7m/s I see my mistake the Ff should equal the acceleration thus Ff=Fc. So Ff=(2000*(3.5^2))/5 Ff=4905N |
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Re: Of center of gravity and centripetal force...
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by kudoushinichi
on 2007-12-12 06:21:51 (edited 2007-12-12 06:33:03)
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Oh, you used r = 5m. I used r = 4m because I thought the diagram is something like below; ._ _ _ O . O |.........| | | |.........| | Outer wheel |.........| | |.........| Center of gravity of car |.........| |.........Inner wheel | center of roundabout Diagram when the car is viewed from the butt of the car. Since roundabout has diameter of 10m, so the distance from the center of the roundabout to the outer wheels should be 5m. But since for simplification, we model the car as a particle, so the centripetal force can be thought to act at the center of mass (center of gravity) of the car, which is only r = 4m (since the CM is halfway between the inner and outer wheels). And that is how I got my answers in terms of g.
Shinjitsu wa itsumo hitotsu!
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Re: Of center of gravity and centripetal force...
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Thanks for the help, especially with the tan(a)>45 degrees & tan(a) = Fc/Fg idea. But I was thinking it should be more like this: Since centripetal force is caused by frictional force, mv^2/r = uN, u being the frictional coefficient. Since the inner wheel is 3m away from the center, and the outer wheel is 5m away from the center, and that w is the angular velocity, the equation can be rewritten as: m(w^2*r^2)/r = uN rmw^2 = uN For the inner wheel, 3mw^2 = uNi -- 1 For the outer wheel, 5mw^2 = uNo -- 2 Dividing both equation -- 2/1 No/Ni = 5/3 No = 5/3 Ni From this equation, I can see that the normal force on the outer wheel is more than the inner wheel. This inbalance in force causes a torque about its center of gravity. Nnet = No - Ni torque = rNnet Can I assume it this way? I was unsure about certain parts in my argument above: 1) Is the m of the outer and inner wheel the same? (If it is, then m can cancel eaach other out) 2) If so, how much torque must be applied b4 the car starts flipping? |
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Re: Of center of gravity and centripetal force...
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by kudoushinichi
on 2007-12-12 07:38:34
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Well, from those calculations, the frictional force on the inner wheels is indeed greater than on the outer wheels. But these forces are directed towards the center of the roundabout, direction that well... can't produce a torque about the outer wheels...
Shinjitsu wa itsumo hitotsu!
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Re: Of center of gravity and centripetal force...
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| LOL!! Normal force is perpendicular to the ground, meaning it's directed upwards! So it DOES apply... The centripetal force is horizontal, yes. The frictional force is also horizontal, yes. But the normal force affecting the frictional force is vertical. So torque can be produced... |
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Re: Of center of gravity and centripetal force...
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by kudoushinichi
on 2007-12-12 09:52:15 (edited 2007-12-12 09:55:38)
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The normal force is cancelled out by the weight, deshou? So there is a zero net force vertically... Furthermore, there's no such thing as the vertical component of the frictional force, since the frictional force is purely horizontal. Perpendicular forces are independent of one another. I went googling and this is what I found in some RC site; -------------------------------------------------------------------------------- The CG Shifts as the Car Moves Now, just because your car may have equal weight on all four tires when it is standing still, doesn’t mean that it stays that way. The weight will shift as the car moves. As you accelerate, the CG will shift to the rear of a car just like when you feel like you get pushed into the seat of a real car when it accelerates. Because of this shift, the rear tires will carry more of the weight of the car on them than the front when the car is accelerating. When you brake or slow down, the opposite occurs and the front tires have more weight on them than the back. As you turn, the chassis will roll and the CG will shift to the outside of the car (if the car is turning left, the weight shifts over to the right). The tires that are on the outside of the car will thus have greater weight on them than those on the inside. Because your car can, and does, turn while accelerating or braking, the CG will usually move towards one tire. For instance, if you are slowing down to enter a right turn, the CG moves forward to the front tires. As you turn right, the CG will shift to the left. Since you are slowing down and turning right, the weight on the left front tire is more than any of the other tires. The right front and left rear have about the same amount of weight. This may seem not to make sense, but remember: the car is turning right and putting more weight on the left side tires, thus making the left rear heavier, but the car is also slowing down at the same time putting more weight on the front tires. This almost equal weight between opposite tires (either left front and right rear or right front and left rear) is often referred to as corner weight. In our example, the tire with the least amount of weight on it would be the right rear tire. In general, the more weight that is on a tire the more traction ability it has. -------------------------------------------------------------------------------- http://www.hpiracing.com/tuning/chap_5.htm -------------------------------------------------------------------------------- So, what causes the torque is settled. It is because the center of gravity of the car moves over the outer tires! Why does the center of gravity moves anyway? I guess it has to do with the inertia of the car... it's tendency to move in a straight line. But how do you describe this mathematically?
Shinjitsu wa itsumo hitotsu!
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Re: Of center of gravity and centripetal force...
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Do you know what frictional force is? Ff = uN, in which u is the frictional coeffiecient, and N being the normal force, a force PERPENDICULAR to the ground!!(meaning it's acting upwards!) N is not entirely cancelled out by gravitational force, due to the fact that its a reaction of forces, not a force by itself. If you wanna know why, just look through my equation again.. I managed to show that No/Ni = 5/3, tell me why thi is wrong. I have to admit that your argument about the COG shifting is strong, but unless you can come up with a mathematical description of that argument, my solution is as plausible as yours. PS. I just thought of something. Maybe kudoshinichi idea and my idea can be reconciled. Think about it. He said that N is cancelled out by W(weight), and I managed to show that the N of the outer wheel is bigger than that of the N of the inner wheel. Maybe the COG shifts towards the outer wheel, thereby cancelling out this inbalance of N. Not a bad idea... What do you think? kudoshinichi, no hard feelings. Doing thins for the sake of science =) |
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Re: Of center of gravity and centripetal force...
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by kudoushinichi
on 2007-12-13 03:07:00 (edited 2007-12-13 06:55:53)
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I asked the the same question in a different forum... http://physicsforums.com/showthread.php?t=204223 and after some calculations, speed limit = sqrt(5g) m/s and combined frictional force = 2000g N
Shinjitsu wa itsumo hitotsu!
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Re: Of center of gravity and centripetal force...
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kudoushinichi In the other forum when calculating the Frictional Force acting upon the car. You forgot to square the velocity. So the final anwser should look more like this 2000*(g/4)N I belive |
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Re: Of center of gravity and centripetal force...
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by kudoushinichi
on 2007-12-13 15:39:39 (edited 2007-12-13 15:46:20)
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Ah, right... that was the first few times I was using LaTeX, so pardon the mistake... :P So, yeah, the answer should be 500g N. EDIT: Wait a second... This figure agrees with the value of friction I gave much much earlier in this thread! So, I think I was rather correct at that time. The critical velocity car before it flips is sqrt(5g) on the outer side of the car!. But if we model the car of the particle, the critical speed at the center of gravity is indeed 2 sqrt g! So, in essence, I was correct the first time! By sheer dumb luck, perhaps...
Shinjitsu wa itsumo hitotsu!
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Re: Of center of gravity and centripetal force...
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| Can you please explain why does (f1+f2)*h = (R2-R1)*b/2? I understand the right side of the equation- that shows the moment (perpendicular distance*force). But why the (f1+f2)*h part? Is it actually derivation using geometry or some physics principle that i don't know about? |
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