Centripetal Force Question?
819860
m(v)(v)/r= 7.18 mg  or m(v)(v)/11.5 = 7.18 mx9.8\nv = sqrt(7.18x9.8x11.5) = 28.446 m/s
208881
Any object in circular motin is acted on by a centripetal force.\n\nA cycle or a car or a train moving in a circular path. The frictional force provides the necessary centripetal corce.\n\nA stone tied to the end of a string and made to move in a circular path. The tension in the string provides the centripetal force.\n\nMoon rotates about the Earth. The gravitational force provides the necessary centripetal corce.
610533
answer is 350newton\nsince the system is in equilibrium, we have,\n400*1.4=W(3-1.4)\nthat is W=400*1.4/1.6=350\nand the force on the pivot would be 400-350=50newton
945234
[Edit -- two things.  #1 Oops, I made a mistake.  When I worked this, I used a mass of 4 kg instead of 0.4.  Now I'll go back through my answer and make the correction.\n\n[#2  There are two different methods here.  Look at both of them.  The <second> way is <much better> for this problem.  I think you'll see that.]\n\nSnow White has an answer, but she doesn't say how she got it.\n\nLet's start by getting the net force F = F1 + F2 = -0.6Ni + Nj with magnitude sqrt(1.36)N = 1.166 N and acceleration F/m = 2.915476 m/s^2.\n\nNow we have two ways to proceed.  This is linear motion with constant acceleration, and we know the direction ... tan theta = -1 / 0.6 = -5/3, so the angle theta in the second quadrant is 59 degrees north of west.\n\nAfter 1.6 seconds, the velocity is acceleration x time = 2.915476 x 1.6 = 4.66476 m/s in the direction indicated above.\n\nThe displacement s after 1.6 seconds is s = vt/2 or s = 1/2 at^2 = 4.66476 x 1.6 / 2 = 3.7318 m.\n\nTo get the position, we could use sines and cosines, but notice that tan theta above is -5/3, so the hypoteneuse of that triangle is sqrt(25+9) = sqrt(34).  Using that geometry, the position after 1.6 seconds is\n\nx = -3/sqrt(34) x 3.7318 = -1.92 m\n\ny = 5/sqrt(34) x 3.7318 = 3.2 m\n\nThat's your answer: (-1.92, 3.2) with velocity 4.66476 m/s and direction 59 degrees north of west.\n\n\nThere's another way to do it that may be better, especially when you're using vectors.  Do it by components, and keep the components separate.\n\nThe net force is F = F1 + F2 = -0.6Ni + Nj.\n\nThe constant acceleration a = F/m = -1.5i + 2.5j m/s^2\n\nThe velocity after 1.6 seconds is v = at = -2.4i + 4j m/s\n\nThe position after 1.6 seconds is s = vt/2 = -1.92i + 3.2j\n\nand this is the same answer we got by doing it the other way.  (You can, of course, get all the magnitudes and angles.)\n\nLooks to me that it's easier doing it by components.\n\nHope this is the same answer Snow White got.  This way, at least, you have two ways of doing it.
750653
The rotation rate is such that the centripetal acceleration equals the component of the downward acceleration of gravity that acts normal to the cylinder's wall at 68 deg. Thus, solve the following for w (rotation rate):\nw^2*r = g*sin(68).\nSince you asked what equation to use and did not ask for a worked-out answer, that's it. I hope it helps.
574012
I've alread answered two of your questiions in my last answer....\n\nIn this one, the guys above are correct - however, I'd try working it out for myself if I was you. Your tutor will soon realise something is wrong if you are giving answers to a subject you don't understand. Read the chapter, ask you teacher if you need more help. Learn Newtonian mechanics for yourself.
750239
In half the rotation, the direction of velocity will become perpendicular to original direction and will be equal in magnitude to the original as no force acts in  direction of motion(if the air-resistance is negligible).\nsuppose initial direction is along positive z-axis\n=>/&#92; V=V2-V1\n=>/&#92; V=100(-k^) -100(k^)=200(-k^)\n/&#92;V=delta V(change in velocity)\nk^=unit vector along z-axis(axis perpendicular to the x-y plane)
851078
1. Usually you have to solve these problems in terms of energy, where initial energy = final energy + friction (lost) energy. The initial and final energies are each the sum of the respective initial and final potential + kinetic energies. (Sometimes the problem doesn't explicitly give you an initial potential energy; it may just tell you that an object slides x meters downhill.) If you know the initial and/or final momentum and either the mass or the velocity, you can find the missing variable (velocity or mass) and then transform the momentum into kinetic energy. \n2. The coefficient of friction 'mu' of course affects the friction force. Sometimes you are asked to find mu so you have to work backwards. You may know the lost energy and the length over which the object encountered friction. Then you solve for friction force. Then, knowing the normal force as well (force perpendicular to the friction surface), you can find mu.\n3. Inclines with friction are a big favorite of the problem writers. Just remember that the gravitational force normal to the incline is g*m*cos(theta) and the gravitational force along the incline is g*m*sin(theta). If a ramp is tilted just enough to allow the object to slide, g*m*sin(theta) = friction force = mu*g*m*cos(theta), so mu = sin(theta)/cos(theta) = tan(theta). Or if you know mu and want to find theta, it's arctan(mu).\n4. Another fave is centripetal force on a horizontal turntable with a friction surface. We know the centripetal force is w^2*r*m (w being angular rate in rad/sec). And we know the normal force is g*m. So, if we rev up the turntable until the object slides off, or we move the object out from the center until it slides off, we know that, at the instant it lets go, w^2*r=g*mu, so we can find w or r. Easy, right?\nI hope this helps; if you have followup questions I'll be watching this question until it closes.
703117
here u go!\n a):\nuse the principal of F=ma to find the frictional force.  \nIn order to do this, you need to determine acceleration, so use the kinematic formula: v^2=vo^2+2a(x-xo).  \nwith this equation, you should find a to be equal to -1.5m/s^2.  \nPlug this answer into the F=ma equation, with m equal to .090kg.  Since Friction is the only force in the horizontal direction, F is equal to the frictional force.\nSolving for F=(.090)(-1.5)\nF=-.135N, so its magnitude is .135N\n\nb.):\nOne physics principal states that a frictional force is equal to mew (the coefficient of kinetic friction) times the normal force.\nSince you now know the frictional froce, just solve that equation.\nThe normal force in this problem is equal to the opposite of the gravitational force, which is (9.8)(.090)=.882\nSo, when you have the equation \n(.882)(coefficient of friction)=.135\njust divide .135 by .882 and you will get the coffecient of friction to be .153!!!!!!! yay!!!
850378
Centripetal acceleration a=r*w^2, r=0.35 m, w=2500*2pi/60 rad/s is angular speed, v=w*r is linear speed;
772772
The centrifugal force on the yo-yo is m*v^2/r; it is directed at the angle ø with the vertical. The vertical component of the force is m*v^2/r * cos(ø).  This must overcome the gravitational force on the yo-yo which is m*g.  Therefore\n\nm*v^2/r * cos(ø) = m*g\n\ncos(ø) = v^2/(r*g)\n\nThe stretching force on the string is the centrifugal force m*v^2/r.  The string will match that force by k*∆s, where k is the spring constant and ∆s is the amount it stretches.\n\nk*∆s = m*v^2/r\n\n∆s = m*v^2/(r*k)\n\nThe radius was given as 6.5cm while it is spinning (string is stretched).  The string length unstretched is then 6.5cm - ∆s.\n\nKeep track of the units in solving: convert everything to kg, m, sec for computation and then convert back if necessary when done.
643953
depends on the direction of the force
121693
cmon...dont u know basic physics...just calculate the resultant of these 2 forces\nanswer= [(132)^2 + (181)^2]^1/2\nif this makes sense!!
1122895
First you find the centripetal acceleration by:\n\na = v^2/r\n\nThen you take that answer and divde by the acceleration due to gravity.  \n\nFor example, let's say the radius of the circle is 10 feet, and the speed is 20 ft/s.  \n\na = 20^2/10 = 40.\n\nThen using 32 ft/s^2 for g you have:\n\n40/32 = 1.25 g's of force.
437681
You have to calculate the centripetal force.\nFc = Mrw^2\n   = (M V^2)/r\n   = [100*(10m/s)^2]/30\n   = 10000/30 m/s^2\n   = 1000/3 N = 333.33 N\n\nM = Mass of the car\nV= velocity\nr = radius\n\nFor banking curve \ntan theta = v^2/g*r\ntherefore\ntheta = arctan [10 m/s]^2/[9.8 m/s^2  * 30m]\n      = arctan[10*10]/[9.8*30]\n      = arctan[100/294]\n      =    degrees (please chack your tables and get the arctan)
512623
I will state this here, in yet another question: THERE IS NO SUCH THING AS CENTRFUGAL FORCE!! Look it up in a basic physics textbook like Giancoli or Serway. Or even do a google search. You have to understand what reference frame you're talking about. The fact that your body tends to want to move to the outside of the curve when your car goes around in a circle has nothing to do with anything other than the fact that your body wants to keep going along in a straight line (conservation of momentum), but the car gets in front of it at every turn (semi pun intended). What is actually happening is that the car is an accelerating reference frame.\n\nAs for the moon/earth question... imagine a man 6' tall throws a tennis ball horizonatally with respect to the earth's surface. The ball will go forward a certain distance (because of it's velocity tangential to the earth's surface) as it falls the ~6' to the ground (because gravity pulls it down). The faster he throws the ball (horizontally), the further it will go before it hits the ground. If he throws it fast enough then the amount that the ball travels forward and the amount that it drops towards the earth's surface can match the curvature of the earth, and the ball will orbit the planet (neglecting air resistance, as we physicists always do) and come around and bonk the thrower in the back of the head. The moon is just a much larger tennis ball, thrown by a much larger man.\n\nYour eqn. is correct, but it only describes the force acting upon the moon. It is not the equation of motion, which detirmines what actually happens to the moon. The equation of motion is:\n\nF = m * a  (F = force, m = mass of moon, a = acceleration of moon around earth)\n\nIn the case of the circular motion of the moon around the earth, "a" is defined by:\n\na = (v^2)/r (where v is the magnitude of the velocity of the moon and r is the center-to-center distance from the earth to the moon)\n\nPlugging one into the other you get:\n\nG * M * m/r^2 = m * (v^2)/r (where M is the earth's mass)\n--> G * M/r = v^2\n\nSo, as long as the tangential velocity, v = sqrt(G * M/r), the moon falls just enough towards the earth as it travels just enough tangentially second by second, that it continues to orbit around the earth. (Note: there is NO CENTRIFUGAL FORCE anywhere in any of these eqns.!!)
155890
*is a center-directed force\n\n\nnot "*acts as an unbalanced force on the satellite" because, it is balanced by centrifugal force. \n\nnot "*changes the direction of the satellite" because direction is changed by only tangential forces and not the force acting towards center.