what is the force constant of a spiral spring?
113199
When a body is deformed,certain forces are developed in the body by which it tries to restore its undeformed shape or size.These forces constitute the elastic restoring force.\nElastic restoring force is always equal in magnitude and oposite in direction to the applied deforming force.
757648
Friction force:\nF = m g c\n\nm = mass of car\ng = gravitational accel\nc = coeff. of friction\n\nCentripetal force:\nF = m v^2 / r\n\nm = mass of car\nv = velocity\nr = radius of curve\n\nSet the two forces equal to each other:\nm g c = m v^2 / r\n\nCancel m, plug in given values:\n9.8 m/s^2 * c = (12 m/s)^2 / 52 m\nc = 0.28257\n\nSo, c has to be at least 0.28257 to keep the car from slipping.
865501
This is letting spring energy = gravitational potential energy.\n\nkx²/2 = mgh→ k = 2mgh/x² = 2*1*9.8*.6/.02² = 29400 N/m
784311
If M0 is the mass at the origin and My is the mass on the y-axis, and Mx is the mass on the x-axis then\n\nFy = G*M0*My/Ry^2\n\nwhere Ry is the distance between M0 and My. Likewise,\n\nFx = G*M0*My/Rx^2\n\nwhere Rx is the distance between M0 and Mx. The magnitude of the resultant force (F) is then\n\nF = sqrt(Fx^2 + Fy^2)\n\nand the angle (q) can be computed from\n\ntan(q) = Fy/Fx.
903133
From what I learned in physics class, the force you are speaking of is measured in Newtons.
949222
This question requires you to equate the rotational and translational force equations to find the answer. Start by drawing your free body diagrams:\n\n1. The mass is acted upon by the gravitational force, mg, and a tension T in the string.\n\n2. The cylinder is acted upon the the gravitational force (through its center of mass), the normal force upwards on the fixture (whatever is holding the cylinder in place), and the tension force T at its outer edge. Note two things: neither the graviational or normal forces acting through the center of mass will not affect the cylinder's motion.\n\nAdditionally, the tension force that acts at the cylinder's outer edge is a torque. That is:\n\ntau = T * r\n\nThe tension force acting on the cylinder is simply the value of the block's weight:\n\nT = m*g\n\nThus, the torque acting on the cylinder is:\n\ntau = m*g*r\n\nFinally, from kinematics, the rotational motion of a body about its ceneter of mass due to a torque is:\n\ntau = I * alpha.\n\nwhere I is the moment of ineria. For a cylinder, this is simply:\n\nI = (1/2) * M * r²\n\nThus, the force on the block can be related to the rotational motion of the cylinder by equating the the expressions derived for the torque:\n\nm*g*r = (1/2) * M * r² * alpha\n\nalpha = m*g*r / ((1/2) * M * r² )\n\nalpha = mg/ ((1/2) * M * r )\n\nalpha = 2mg/(Mr)\n\nalpha = 2(.670 kg)(9.81 m/s²)/(1.95 kg * .133 m)\n\nalpha = 50.69 rad/s\n\n"How far does m travel downward between 0.470 s and 0.670 s after the motion begins?"\n\nAssuming the cylinder starts from rest, then the angular displacement of the cylinder is described by:\n\nR(t) = (1/2) * alpha * t²\n\nSo between t1 and t2, the cylinder's angular displacement is:\n\nR(t2) - R(t1) = 1/2*alpha*(t2² - t1²)\n\nSubstituting the times provides:\n\nR = (1/2)(50.69)[(.67)² - (.47)²]\n\nR = 5.778 rad\n\nThe arc length traced by the cylinder during this period is:\n\nS = R * r\n\nS = (5.778 rad) * (.133 m)\n\nS = .769 m\n\nThus, the mass drops .769 m during this period.\n\n"The cylinder is changed to one with the same mass and radius, but a different moment of inertia. Starting from rest, the mass now moves a distance 0.440 m in a time of 0.570 s. Find Icm of the new cylinder."\n\nTo solve this, we now work backwards to find I:\n\nR = R(t*) - R(0) = R(t*) - 0 = R(t*) \n\nR(t) = (1/2)*alpha * (t*)²\n\nS = R * r = (1/2)*alpha * r * (t*)²\n\nWe can now plug in values to get our first equation:\n\n0.440 m = (1/2)*alpha * (.133) * (.57)²\n\nalpha = (1/.133) * 2(.44)/(.57)²\n\nalpha = 20.365 rad/s²\n\nRecall that:\n\ntau = I * alpha.\n\nwhere tau = m*g*r. Using these expressions, we can solve for I as:\n\nm*g*r = I * alpha\n\nI = m*g*r / alpha\n\nI = (.67)(9.8)(.133) / (20.365)\n\nI = .04288 kg-m²\n\n\nThat's it!
780831
That is known as the "solid height" of a spring; when all the wires are stacked on top of eachother.  You can calculate it by just multiplying the number of turns of the spring by the diameter of the wire.
543978
Frictional force, f = uR\n\nwhere u - frictional constant\n(it is actually "miu", but I can't typr it out.)\n\nand R - the normal/reaction force (which is the reaction caused by the action of the object's weight.)
801237
A mass attached to a spring follows a simusoidal acceleration vs. position function. At equilibrium, the acceleration of the spring is 0 and velocity is maximum, at the two extreme postions, the acceleration is maximum and the velocity is 0. So at Xo/2 and -Xo/2, the acceleration will be half the maximum acceleration.
774877
The equation doesn't differ for different springs, only the value of spring stiffness k does. A different value of k means the deflection x resulting from a given force Fm will differ also since Fm = k*x.
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.
834149
The period T is 0.9s\nT = 1/f = 2pi/w so w = 2pi/T\n\nAs the movement is circular, the total of the forces is in this plan and to the centre and the value is mw²R.\n\nSo P + S = mw²R where P is the weight and S the tension of the string and R the radius and a the angle between the string and the vertical\nIn projection, we have :\nmw²R = S sin a\nmg = S cos a\n\nDividing the 2 equations :\ntg a = w²R/g\nand R = L sin a\n\nso w² = g/(Lcos a)\nso cos a = g/Lw² = g T²/(L 4 pi²) \n\nso cos a= 9.8 0.9²/ (0.3 4 3.14²) = 0.6702\n so a = 47.91°  or a =0.83 rad\n\nBut this angle is the angle between the string and the vertical so angle between the string and the horizontal is 90-a = 42.09°
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
641340
it would double. imagine you stretch a spring with constant k a distance x using force f=kx. now while you are holding the end at displacement x, also hold the middle of the spring, which has been displaced x/2. now cut the spring in the middle while holding it, so the second half falls off and the first half is still stretched. since the first half is not stretched any farther than it was before, the force is the same. the displacement, however is x/2. so, the new spring constant is 2k.\n\nspring constant is really a misnomer; its not constant for the material, but only for that spring of that length. it should really be given as a constant independent of the length of the spring, constant per unit length, but alas it is not.
545558
Force = mass * acceleration\nForce = 1300 kg * 2.6 m/s2\nForce = 3380 Newtons
378184
When the cylinder rolls, there is a force F=15N at the top and a static frictional force f preventing slipping.\neverything in usual notation,\nacceleration A=R*a (a=angular acceleration)\nNewton's second law implies'\nMA=F-f         (resultant force)\nalso for torques,\nIa=FR+fR         (for the axis passing through the centre)\nFRom the two equatins,\n2F=mA+Ia/R\ni.e.  2F=maR+Ia/r\n   a=2F/(mR+I/R)\nI=0.5MR^2=0.05\nplugging hte values,\na=16 \ntherefore, A=aR\n            =16*0.1\n           A=1.6
917169
L = μ(0) μ(r) N^2 A/l\n\nwhere\n \nμ(0) is the permeability of free space (4π × 10-7 henrys per metre) \nμ(r) is the relative permeability of the core (dimensionless) \nN is the number of turns. \nA is the cross sectional area of the coil in square metres. \nl is the length of the coil in metres\n\nHere we assume that μ(r) = 1, since the solenoid doesn't enclose anything like a metal core. Then\n\nL = [μ(0) N^2 (πR^2)] / l, where R is the radius of the solenoid\n\n= [(4π × 10-^7)(2920)^2(π(0.005)^2)] / (0.1)\n\n= 8.44 × 10-^3 henries