2 more centripetal force questions!?
930306
Try this site, I found some very helpful info there.\nhttp://en.wikipedia.org/wiki/Centripetal
837734
The 2nd and 3rd answers match one of the choices provided.\nThe first one is a bit off.\n\nThe component of the gravitational acceleration vector acting along the line of the string is 9.8*sin(x) m/sec^2 where x is the angle between the string and the horizontal. The centripetal acceleration is given by v^2/r.\n  \n1) a =2.7 m/sec is closest\n\n At a 40 degree angle above the horizontal the gravitational acceleration component reducing the tension in the string due to the centripetal acceleration is: 9.8xsin(40) = 6.3 m/sec^2\n\nThe centripetal acceleration = v^2/r = v^2/1.4m\n\nThe net acceleration (tension) = 3N/0.3kg = 10 m/sec^2\n\nv^2/1.4 = 10-6.3\nv^2 = 5.18\nv = 2.3 m/sec\n\n2) b. 11 N\n\nThe centripetal acceleration is: \n\nv^2/r = 8^2/2 = 32 m/sec^2\n\nThe gravitational acceleration component, at the top, which reduces the tension is 9.8 m/sec \n\nThe net acceleration is: 22.2 m/sec^2\n\nThe tension (net force) = ma = 0.5 x 22.2 = 11 N (rounded off)\n\n\n3) b. 4\n\nlet v = the wheels angular speed at the start of the 4 sec\n\nd = 80rad = v*t + 0.5 at^2 \n\nt=4sec; a = 4rad/sec^2\n\n80 = 4v+0.5*4*(4^2) = 4v+32\n\n20=v+4\n\nv=16\n\nIf the wheel starts from rest:\n\nv=a*t =16 = 4t\n\nt = 4sec
57539
Well, there are two components of acceleration here: there's centripetal acceleration, which is toward the center of the Ferris wheel, and tangential acceleration, whose direction is tangential to the wheel (in this problem, straight forward).  In order to calculate the total acceleration, you need both these components.\n\nYour centripetal acceleration is given by the velocity squared over the radius of the circle:\n\na_c = 3^2/14 = 9/14 m/s^2 = .64\n\nFrom the info in the problem, your tangential acceleration would be the .5 m/s^2.  \n\nNow draw a triangle with those two accelerations as sides.  Solve for the hypontenuse with Pythagoras to get .81 m/s^2.  Finally, the arctan of .64/.5 is 65 degrees (more or less), so your acceleration is .81 m/s^2 at a 65 degree angle.\n\nNow go back and check my work, because I just got up and I'm a bit rusty in the mornings.  AND I didn't get my newspaper this morning.  Can you believe that?
901641
Someone posted the exact same set of questions a few days back, either here or on the physics page - you ought to be able to find the methods in the responses. Have a look.\n\n\nBest of Luck - Mike
891370
Get the unbalanced force from the first 2 #s in your ? :  2.8-2.6 = .2\n\nb)  F = ma---> m = F/a = 0.2/.11 = 1.818 kg
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.
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)
262728
Simple.\nThe car tyres rotate thus the force is called centripetel force(f).It is equal to v^2/r where r is the radius of tyre.\n\nforce at 15m/s = 15*15/r\nforce at 30m/s = 30*30/r\n\nWhen we divide them we get 1:4 the answer\nit is assumed that car tyre is of same radius in both cases.
49314
Sorry, but with the time you spent typing these questions you could of looked them up in your physics book or googled them.  Sometimes getting others to do your leg work isn't always the best answer to a problem.  Good Luck in your new class!
970605
For your first problem...\n\nYou know that T = Fc = (mv^2) / r\n\nWhere T is the Tension, Fc is the Centripetal Force, m is the mass, v is the velocity, and r is the radius.\n\nYou're given m = 80 kg, r = 0.90 m, and T = 600N.\n\nSo...\n\n600 N = (80kg * v^2) / (0.90 m)\nSolve for v...\nv = +/- 2.598 m/s\n\nSo since -2.598 m/s doesn't make sense (a negative speed), the solution would be ~ 2.6 m/s.\n\nFor your second problem...\n\nIts been a long time since I've done circular motion.  This site has the equation you need to use (its too messy to type here): http://hyperphysics.phy-astr.gsu.edu/hbase/mechanics/carbank.html if you can figure out the angle.  So if you know what the angle should be, that equation will get you the coefficient of friction.\n\nIf you can't figure out the angle, at least attempting to solve it using that equation and some reasonable angle should get you some points I'd imagine.\n\nIf I can think of the correct angle I'll post an edit to this.
755334
Analyze the forces at the top of the loop:\n\nThe force of gravity pulling the ball downward,\nthe upward centrifigal force keeping the ball on the track,\nand the force of the track pushing the ball downward.\n\nThe answer to the question is when the centrifigal force is exactly equal to the weight, so the force of the track equals zero.\nmg=mw^2R\nw is the angular velocity\nw=sqrt(g/R)\n\nOnce you know the angular velocity, compute the equivalent linear velocity by mutlipying by R\n\nthen use mgh=1/2mv^2\nto compute h\n\nWhere is point Q?\n\nj
904251
Let's say F1 and F2 are the tensions (forces) in the inner and outer sections of the rod respectively, w is the angular velocity, and r the distance to the center (half the length of the rod). Equating the radial force to (mass x centripetal acceleration) for each mass gives:\n\nr*w^2*m1 = F1 - F2 = 2F2\n(2r)*w^2*m2 = F2\n\nDividing the second equation by the first all variables cancel except for the ratio of the masses:\n\nm2/m1 = 1/4
672475
Since the proton is moving in a circular pattern, the force given is the centripetal force.  I know that it it worded weirdly, but I believe that is just to throw you off.  The equation for the centripetal force due to circular motion is F = m * (v^2) / r  \n\nwhere F is the force, m is the mass of the object, v is the velocity, and r is the radius of the circular path.\n\nSo we set up the following:\n\n8.00 * 10^-13 N = (1.67 * 10 ^-27 kg) * (v^2) / (0.8 m) \n\nI used 0.8m instead of 80cm because we need to be in standard SI units.  This will make sense later. Rearrange and solve for v.\n\nV = sqrt[ (8.00 * 10^-13 N) * (0.8 m) / (1.67 * 10^-27 kg) ]\n\nSince N = (kg*m)/(s^2)\n\nwe can re-write as:\n\nV = sqrt[ (8.00 * 10^-13 (kg*m)/(s^2)) * (0.8 m) / (1.67 * 10^-27 kg) ]\n\nthe kg cancel out:\n\nV = sqrt[ (8.00 * 10^-13 (m)/(s^2)) * (0.8 m) / (1.67 * 10^-27) ]\n\nSimplifying you get:\n\nV = sqrt(3.83 * 10^-14 m^2/s^2)\n\nSince they tell you that the sqareroot of 3.83 is 1.96, then we get our final answer to be:\n\nV = 1.96*10^7 m/s\n\nLet me know if you need furhter clarification on this problem.
954885
First lets assign the direction of f(1) as being at 0 degrees, then f(2) is at 57 degrees.\n\nResolve both forces in x, and y\n\nF(x) = f(1) + f(2) cos (57)\n\nF(y) = f(2) sin (57)             f(1) is at 0 degrees so has no y component.\n\nThe resultant force F = sqrt ( F(x)^2 + F(x)^2)\n\nThe angle of the resultant force \n\nArctan( F(y)/ F(x))\n\nYou do the number crunching
908729
(a) ω = ωo + αt\n0 = 150 + α * 132\nα = -150/132\n≈ -1.136 rev/s²\n\n(b) θ = ωot + ½αt²\n= 150 *132 - ½*1.136*132²\n≈ 9903 revolutions (nearest rev)\n\n(c) See (b)\n\n(d) 75 rpm = 75*2π rad/min\nα = rω²\n= 50 *(75*2π)²\n≈ 11103300 cm/min²\n= 111033 m/min²\n= 111033/3600 m/s²\n≈ 30.84 m/s²\n\n(e) ???? I do not understand what it is you are seeking.
874727
You need to sum the components. Let Fr, Fl, Fu, and Fd be to right, left, upward, and downward directed forces, respectively. Then\n\nFx = Fr - Fl\n\nand\n\nFy = Fu - Fd.\n\nThe right and upward directions are taken as the positive direction so that the minus sign indicates that the forces (Fl and Fd) are directed to the left and downward.\n\nThe magnitude of the resulting force is then computed as\n\nF = sqrt(Fx^2 + Fy^2)\n\nand the direction can be found from\n\ncos(q) = Fx/F\n\nUse some reasoning to be sure that the direction you compute is consistent with the orientation of the force components.
901635
1\na.  distance, slower\nb. weaker gravitational pull farther away\n2\na. 750.353 km\nb. 4.67 cm\nc. Not enough info\n\nPS please award best answer