Inertia on Cars: Homework help?
541091
Inertia means two things:\n\nAn object at rest will stay at rest, unless another force acts on it.\n\nAn object in motion will continue in motion, unless another force acts on it.\n\nAny force which changes the motion of an object -- the speed and/or direction -- is called acceleration.\n\n...You're riding along in your car.  Both you and the car have inertia; that is, you're both in motion, and will continue to move until something changes that.\n\nThe car runs into a wall, which certainly makes a big change in the car's motion.  You, on the other hand, and anything else inside the car that wasn't strapped down, will continue to move forward because of your own inertia.  Without a seat belt or air bag, your body will continue to move forward through the windshield, which is undignified, extremely painful, usually fatal, and a big drag to clean up.\n\nSo:\n\nA)  Seatbelts put you more "in contact" with the car; which is to say, any force which affects the inertia of the car will be more efficiently transferred to your body -- though not *so* efficiently that you go "splat" like the rest of the car; they just restrain you from continuing forward into the windshield.\n\nB)  A head rest acts as an accelerator for your head.  If the car accelerates suddenly (such as when somebody runs into the back of your car), the car's seat transfers that acceleration to your body.  The head rest acts as an extension of the seat, so that your head is also accelerated and doesn't get "left behind" because of its own inertia.\n\nC)  In practical terms, a force which is applied directly opposite to an object's direction of movement is called "deceleration."  It's still the same thing as acceleration, but we treat it as if the speed an object is going in a particular direction is being decreased in the same direction.  When the car hits the wall, as your body's inertia propels you forward, the air bag acts as resistance to that forward movement, decelerating the forward motion of your body.\n\nThe point to all three of these is to gently transfer acceleration to your body.  A car is much more rigid than your body (for instance, if you were to accelerate really hard, the rear bumper isn't going to get left behind), and is able to take much more drastic changes in its speed and direction.  Your body is much more fragile (and much more valuable) than a car, so seat belts, head rests and air bags are used to ease the changes in acceleration.\n\nAs you know, falling off a tall building isn't what kills you, it's the big fat splat at the bottom.  When you fall, your body accelerates down with a force of one G (which stands for earth's gravity), which your body can handle quite well.  But when you hit the ground, the ground acts to accelerate you in the opposite direction (up) *very* quickly, with much more acceleration than your body can stand.  If there's a big enough air bag to land on, it will slow down the up-acceleration (or down-deceleration, whichever way you look at it) to something that your body can handle.  Now, turn the whole setup sideways, and that's the reason for seatbelts, air bags and head rests.
861510
The way to solve this problem is to decompose the forces acting on the car, but not in the x and y direction. Instead, decompose the forces in the direction of the incline and the normal force.\n\nHence, we have:\n\nF(friction) = μN, moving up the incline, and μ is the coefficient of static friction\n\nThen we have gravity acting on the car, and the component that's acting down the incline is = mgsinθ, where θ is the incline of 15°.\n\nFor the car to be safe in making this turn, the forces along the incline must be equal:\n\n(a) F(friction) = μN = mgsinθ = (1450)(9.8)sin(15°) = 3677.82 N\n\nBut the forces acting perpendicular to the incline on the car must also be zero:\n\nN = mgcosθ\n\nNow substituting this into the previous equation, we get:\n\nμN = μ(mgcosθ) = mgsinθ, this implies that the coefficient of static friction is:\n\n(b) μ = sinθ/cosθ = tanθ = tan(15°) = 0.268
734736
first one is easy\n\nfirst find the normal force\n\nFf (friction) = Fn(normal force ) (.35)\n0(motionless ) = Fn .35\nFn = -.35\n\nfrom there its easy\n\nFn = mgsin(theta)\n\nsolve for theta\n\n-.35 = (200)(9.8) sin(theta)\n\nand solve it ur self\n\nsecond one, tell us wat u want us to find and we can tell u, but here are the things u can find, ill give u the formulas\n\nacceleration = 0 (constant speed)\ntheta = 15 degrees\nvelocity = constant\n\nthats pretty much it, from what u gave me
845823
the speeds are too small, so classical mechanics should be fine:\n\nthe speed of the car is 45-10 = 35mph
871488
Torque=F*D\nT=18*.35\nT=6.3\nWe know that T=a*i     [a=angular acc.\n                                       i=moment of inertia]\n         T=a*mr^2       [m=mass\n                                  r=radias]\n         6.3=a*2.65*(.35)^2\n         a=19.4 radian/sec^2\n\nProblem solned. Have a nice day.
973163
Relative to a woman in the car:\n\n(a) the translational velocity is ZERO since the woman is moving at the same translational velocity as the car relative to the ground.\n\n(b) the magnitude of the translational acceleration of the center of the wheel is also ZERO because the center of the wheel is NOT moving and NOT rotating relative to the woman.\n\n(c) v, velocity of the point at the top of the tire is = 80 km/h because that is how fast the car is moving, thus the wheel must also be rotating at the same speed.\n\n(d) a, acceleration of the point at the top of the tire is given by:\n\na = v^2/r = (22.22)^2/(0.33) = 1,496.45 m/s^2\n\n(e) v, velocity of the point at the bottom of the tire is = -80 km/h because at the bottom of the tire, the tire is moving in the opposite direction\n\n(f) a, acceleration of the point at the bottom of the tire is the same as (d) a = v^2/r = 1,496.45 m/s^2\n\nNow relative to a hitch hiker on the ground:\n\n(g) v at the center of the wheel is = 80 km/h or 22.22 m/s, since the car and the center of the wheel (which is not moving relative to the car) are moving at the same speed relative to the ground.\n\n(h) a at the center of the wheel is = v^2/r = 1,496.45 m/s^2, since the center of the wheel is rotating about the bottom point of the tire as the tire rolls on the ground.\n\n(i) v at the top of the tire is v(top) = ωd, where ω is the angular velocity of the rotating tire and d is the diameter of the wheel. The reason we use diameter of the wheel is because the top of the tire is rotating abou the bottom of the tire, not the center.\n\nω = v/r = (22.22)/(0.33) = 67.33 rad/sec\nv(top) = ωd = (67.33)(0.66) = 44.44 m/s\n\n(j) a at the top of the tire is = v(top)^2/d\n    = (44.44)^2/(0.66) = 2,992.29 m/s^2  \n\n(k) v at the bottom of the tire is equal to ZERO because the bottom of the tire is not moving relativ to the hitch hiker or the ground.\n\n(l) a at the bottom of the tire is also equal to ZERO since the velocity is zero.
822312
wish I could solve em for you but ill try and help still.\n\n1) The acceleration should be the derivative of the equation for the car's velocity.  The original distance/time equation would be the anti-derivative of the velocity.  \n\n2 The equation to find out how long it will take to get there is h+ax-16x^2 in feet.  (a is the original speed and x is number of seconds and h can be the height of the building.)\n\nHope this helps
739092
PE at release point = KE + PE at the top of the loop.\n\nmgh = 1/2 m v^2 + mg 2R   the 2R is he height of the loop.\n\nSolve for KE\n\nmgh-mg2R = KE = mg(h-2R)\n\nNow for the second one I believe you have to use Net F = ma\n\nAssume the normal force is 0, because the car is barly staying on the track. so Fgrav = m v^2/R\n\nso mg = mv^2/R\nso g = v^2/R\n\nNow going back to the energy equations, mgh = 1/2 mv^2 + mg2R\n\nm's cancel so gh= 1/2 v^2 + g2R\n\nsolve for h and you have h =    (.5v^2 + g2R)/ g  make that into 2 fractions and you have \n.5v^2/g  +  g2R/g  and v^2/g from the equation above = R\n\nso .5R + 2R = h  so 2.5R = h\n\nHope thats right!
984567
a=(delta V) /t = (26/24)=1.08 m/s2\n\nF= mxa,   \n\nwe know weight of car is 11420 N, we need mass\n\nm=11420 / g= 11420 / 9.81 = 1164 kg\n\nso F= 1164 x 1.08 = 1257.25 N to stop the car\n\nfriction = F/Fn =  1257.25/11420 = .11
906572
50 Km is the answer\n\nIf the car moved 50 km north and then 50 south its displacement would be 0\n\nif it go in a direction like sw or ne or nw then you have to use a triangle to solve its displacement
396138
a=F/m, lim x->inf x/x = 1. The object will accelerate.
882327
easy ;-)\n\nthe friction force, call it F, is equal to the normal force (the force perpendicular to a surface - in this case, the car's weight), times the friction coefficient, call it k. And the weight would be equal to the mass, m, times g.  So you'd have\n\nF = k * m * g\n\nnow the acceleration is the force divided by the mass, so:\n\na = F / m = k*g = (0.642)*(9.81)  (in metric)\n\nthe formula you suggest to use is the correct one ;-)\n\nnow i'm used to metric so i'll convert your mph to m/s, 40mph is 17.8 m/s.\n\nso you have\n\nd = (17.8)^2 / 2*(0.642)*(9.81)\nd = 25 metres\n\nif you want feet, and don't want metric in between, then you'd have to use 32 ft/s2 and not 9.81 m/s2 for g.  And you'd have to convert your 40mph in ft/sec (about 58.3 ft/s).\n\nhope this helps
257087
If something weighs about a ton you will need much inertia to stop the object. If you have a marble and you want to stop it you need less inertia. So inertia is force needed.
889055
Power:\nP=dE/dt = dW/dt\n\nForce:\nf=ma\n\nEnergy:\ne=mc^2
136454
I will give you a direction to solve the problem...\n\nFirst I will do the problem from a rectilinear formulation(using force, acceleration, velocity) then transfer it to a curvilinear formulation(using Torque, angular accelaration, angular velocity).\n\ngiven F=ma (or F=m(v/t) or more specifically F=m(dv/dt) or even more specifically F=d(mv)/dt).\nWhere F=force, m=mass, v=velocity, and t=time.\n\ngives you Ft=mv (conservation of energy=>conservation of momentum or the impulse-momentum equation).\n\nThis states that it will take a certain amount of force for a certian period of time to produce a certain change in velocity of an object assuming that the mass remains constant.\n\nTransmitting to curvilinear motion force becomes torque, v becomes rv(radius time velocity or w), mass becomes moment of inertia (or I) and t stays the same.
857622
1) I think your 2nd c.f. should say c. acceleration; that's the units you're using------ the c force of a mass m with acceleration a is f = ma, so youhave 1.5 kg*18 m/s² = 27 N\n\n2) First find the velocity: v = d/t = 2πr/t = 2π*.75m/.8sec = 5.89 m/s\nThen, acceleration: a = v²/r = 5.89²/.75 = 46.256 m/s²\nThen, the tension: T = ma = .1kg*26.256m/s² = 2.626 N\n\n3)  v = 2πr/t = 2π*90/12.3sec = 45.974m/s\na = v²/r = 45.97²/90 = 23.485 m/s²\nF = ma = 900*23.485 = 21,136.5 N\nThe lateral friction of the tires on the track
791525
questions are very easy. You must try them by yourself. I will only give hints-\n1> Instantaneous accl  = dv/dt\nAvg accl = (v2-v1)/(t2-t1)\n\n2> draw the vectors, take the components separately along x and y axes.\nremember accl component along x axis increases velocity along x axis, and accl component along y axis increases velocity along y axis.\n3> use the basic kinematics equations like v^2=u^2+2as etc