Newton's Second Law for Rotational Motion About a Fixed Axis?
826539
Moment of inertia:  2 opposite panes behave as 1 of 150 kg mass and 2.4 m width:\nInertia of above about center of mass = (1/12)mL² = 150(2.4²)/12 = 72 kg.m².  For 2 such panes, Itot = 144kg.m²\n\nTorque = Fr = 52N*1.2m = 62.4 Nm\n\nα = T/I = 62.4/144 = .433 rad/sec²
27041
Newtons law is called the Laws of Motions. There are three laws for it. The first law is known as the  "law of inertia" states that if the forces acting upon an object are balanced, then the acceleration of that object will be 0 m/s/s. Basically if objects are not at equilibrium they do not accelerate, they only accelerate if there is a net or unbalanced force acting upon it. The second law is mainly about the behavior of objects. That all existing forces are not balance. It states thatacceleration of an object is dependent upon two variables - the net force acting upon the object and the mass of the object. The third law is known as " for every action there is an equal and opposite reaction " this basically states that in every interaction, there is a pair of forces acting on the two interacting objects. The size of the forces on the first object equals the size of the force on the second object. The direction of the force on the first object is opposite to the direction of the force on the second object. Forces always come in pairs - equal and opposite action-reaction force pairs.
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The basic formula that we know in circular motion is:\n\nF=mv^2/r, \n\nwhere F is the centripetal force, m the mass, v is the linear velocity, and r is the radius.\n\nWe also know that v=wr\n\nwhere v is the linear speed, w (supposed to be omega) is the angular velocity, and r is the radius.\n\nSubstitute the value of v in our first formula:\n\nF=m(wr)^2/r\n F =mw^2r      Formula 1)\n\nIn this problem, we also know that w which is the angular velocity is the same for m1 as for m2 because they are located along the same line which is rotating about one end of the rod.\n\nWe are also given the value of the tension in the inner section as 3x that of the outer section. Let's draw a simple free body diagram:\n                  \nT1=3F2<----0-->T2=F2<---0\n\nThe first 0 is m1 and the second 0 is m2.\n\nThe diagram might get distorted when I send it, but let's just cross our fingers it comes out clear enough.\n\nThe net force acting on m1 is (3F2-T2). But T2=F2. \nTherefore, net force on m1 =3F2-F2=2F2. Let's call\nthis net force as F1. Thus F1=2F2. Now we also know\nthat \n\nF1=m1w^2r    \nbased on formula 1) above.  Now substitute the value of F1=2F2, and we get:\n\n2F2=m1w^2r   Equation 1)\n\nAnd based on the same formula F=mw^2r:\n\nF2=m2w^2(2r)    Equation 2)\n\nRemember that for m2 the radius is twice that of m1, which is a given condition in this problem.\n\nNow divide equation 2 by equation 1:\n\nF2=m2w^2(2r)\n------------------\n2F2=m1w^2r\n\nF2,w^2, and r will cancel out leaving:\n\n1/2=2m2/m1\n\nDivide both sides by 2:\n\n1/4=m2/m1\n\nm2/m1=1/4\n          =0.25
981952
For a sphere of constant density having mass M and radius R, the moment of inertia is\nI = (2/5) M R^2\nThe kinetic energy of a travelling and spinning sphere of uniform density is\nK = (1/2) I w^2 + (1/2) M v^2\nK = (1/2)(2/5) M R^2 w^2 + (1/2) M v^2\nK = (1/5) M R^2 w^2 + (1/2) M v^2\nWhere w is the angular speed.  And where v is the speed of the sphere's center of mass with respect to the surface on which the sphere is moving, which is equal to the tangential speed of the sphere's rotational "equator," which is in contact with the surface on which the sphere is moving, with respect to the sphere's center of mass.\nw = v / R\nK = (1/5) M R^2 w^2 + (1/2) M v^2\nK = (1/5) M v^2 + (1/2) M v^2\nK = 0.7 M v^2\nK = 0.7 M R^2 w^2\nFor your sphere, M=0.5 kg and R=0.06 m, so...\nK = (0.00126 kg m^2) w^2\nOn a slope 30 degrees from horizontal, the acceleration from gravity would be\ng = (9.8 m/s^2) cos(30) = 4.9 m/s^2\nThe work done by a ball rolling one meter along the slope is\nE = M g (1 meter) = 2.45 Joules\nK = E\nThe angular speed of the sphere after rolling 1 meter is\nw = { K / (0.7 M R^2) }^(1/2)\nw = { 2.45 Joules / (0.00126 kg m^2) }^(1/2)\nw = 44.096 s^-1\nv = wR = 2.6458 m/s\nThe average acceleration of the ball is found from the regular kinematic formula.\nVf^2 - V0^2 = 2 a dx\nWhere dx is the one meter measured along the slope, V0=0, Vf=v.  The average acceleration, therefore, is\na = v^2/(2 meters) = 3.5 m/s^2\nSo the speed of a sphere of constant density having mass 0.5 kg and radius 0.06 meters, rolling freely down a 30 degree frictionless slope, is\nv = (3.5 m/s^2) t
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The translational energy is equal to 0.5 * mass * (velocity)^2, or\n\n0.5 * 12 * 12^2 = 864 J\n\nThe rotational kinetic energy is equal to 0.5 * (moment of inertia) * (angular velocity)^2.\n\nThe moment of inertia for a cylinder is 0.5 * (mass)*(radius)^2\n\nThe angular velocity is the (velocity)/(radius).\n\nso substituting in gives:\n\n0.5 (0.5 * mass * radius^2) * (velocity/radius)^2\n\nthe radius^2 factor will cancel leaving\n\n0.25 * mass * velocity^2 or\n\n0.25 * 12 * 12^2 = 432 J\n\nTotal energy is the sum = 432 + 864 = 1296 J.
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Try this: http://en.wikipedia.org/wiki/Newton's_method
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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
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Although merry-go-rounds (normally) do not experience any linear motion, they do experience ANGULAR, or rotational motion. Angular equations are the same set-up as linear equations, except for their variables. Here's a table to help:\nd ==> theta (angle in radians)\nv ==> omega (velocity in radians/second)\na ==> alpha (radians/second^2)\n= = = = = = = = = = = = = = = = = = = = = = = =\nThe relationship between linear and angular motion is:\nd = theta*r\nv = omega*r\na = alpha*r\nWhere r is the distance from the center of rotation. (This relationship is not the same for other variables, such as force or momentum)
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0.4*(100kg*g+F*sin(x)) = F*cos(x). ignoring any rotational component
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If you are assuming "uniformly accelerated rotation" then the following apply;  (please excuse the crude variables)\n\nw = dθ / dt\na = dw / dt = d^2 θ / dt^2\n\nwhere "w" stands for omega <-- angular velocity\nand "a" stands for alpha <--  angular acceleration.\n\nIf you have had any calculus you can take the first derivitive of your θ expression to get the angular velocity, and then the second derivitive to get the angular acceleration.\n\nsince we have a a simple polynomial the following says\n\nd/dx [x^n] = nx^(n-1)  , however in our case we will be deriving t\n\nTo yeild       w = 8t + 6t^2   , for angular velocity\nand              a = 8 + 12t      , for angular acceleration\n\n\n(a) if you plug in 0 for t then you get θ = 2radians\n(b) its angular velocity is zero when t = 0\n(c) w = 8(4) + 6(4^2) = 128 radians/second\n(d) a = 8 + 12(2) = 32radians/second^2\n(e)  I don't believe its angular acceleratin is constant.  Because if you do some testing with plugging in times for acceleratin and velocity they don't quite add up...   \n\nBut that could be because I am totally wrong.  Anyway that is my stab at it.  Hope it helps
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There are a number of angular equvalents in physics.  Here are some:\n\nDistance, x (meter) is angle, b (radian)\nForce, F (Newton) is torque, T (Newton-meter)\nVelocity, v(meter/second) is angular velocity, w (radian/second)\nAcceleration, a (meter/second^2) is angular acceleration, z (radian/second^2)\n\nSo far these are all straight forward.  \n\nTo make Newton's law (F=m*a) work in angular terms, we need to define an angular mass.  It is defined by considering all the differential masses in an object and how they under go angular acceleration when subjected to a torque (an angular force).  Without too much trouble, you can show that this angular mass is:\n\nS r^2 dm (S means integration)\n\nThis mass property is called moment of inertia and is represented by I.  It just takes into account the fact that mass located farther from the axis is harder to accelerate.\n\nWith that defined, all the other properties fall into place.\n\nMass m (kg)  becomes moment of inertia I (kg-m^2) \nNewton's Law F=ma becomes T=Iz\nKinetic energy E = (m*v^2)/2 becomes E = (I*w^2)/2\nMomentum P = m*v, is angular momentum H = I * w
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I concur with previous answerer. The acceleration is 1.5 m/s^2. Rearrange Newton's Second Law, to solve for the acceleration. Then, substitute what you know.
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The point P moves in such a way that its Cartesian coordinates with respect to an origin O at time t is given by x= e^(-t) , y= 2te^(-t). the distance OP is denoted by r and the angle between OP and the x-axis by θ.\n(a)Using an analytical method, find the value of t when r^2 is stationary. Determine the nature of the stationary point.\n\nr² = Δx² + Δy²\n= (e^(-t) - 0)² + (2te^(-t) - 0)²\n= e^(-2t) + 4t²e^(-2t)\n= e^(-2t)(1 + 4t²)\nd(r²)/dt = -2e^(-2t)(1 + 4t²) +8te^(-2t)\n= 2e^(-2t)(1 - 4t + 4t²)\n= 0 for stationary points\nSo 1 - 4t + 4t² = 0 as e^(-2t) > 0 for all t\nie (1 - 2t)² = 0, ie t = ½, ½\nWhen t <½, d(r²)/dt > 0 and when t >½, d(r²)/dt > 0\nThus when t = ½ there is a stationary inflexion\n\n(b) Find, in terms of t, the rate of change of θ with respect to\n\ntanθ = Δy/Δx\n= 2te^(-t)/e^(-t)\n=2t\n\nsec²θ * dθ/dt = 2\nie\n\ndθ/dt = 2/sec²θ\n= 2/(1 + tan²θ)\n= 2/(1 + 4t²)\n\nCan you show the non analytical method for part (a) too!\n\nx = e^(-t) → -t = lnx ie t = -lnx\n\ny = 2te^(-t) \n= -2xlnx\n\nx: .. 0 ...... 0.25 ... 0.5 .... 0.75 . 1.0 .. 1.25 .. 1.5\ny: undef . -0.15 . -0.36 . -0.22 ... 0 .... 0.28 . 0.61\n... (→0)\n\nLim x → 0 (x lnx) = 0 and OP increases with x \n(limiting min value of OP = 0)\n\nMin value of xlnx occurs when lnx + 1 = 0 ie x = -1/e and at this point OP² will show its inflexion as it cannot be a minimum and nor can it be a maximum so all it can do is slow its rate of growth.
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Read Newton's 3rd law of motion.\nAlso, use F=ma to figure out your acceleration.
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a) T = dL / dt = 16\nL = 16 t \n\nI don't think you can do any better than that since they don't ask for the angular momentum at a particular moment in time, and it certainly is not constant.\n\nb) L = Iw\nw = L / I = 16 t / I = 16(33 * 10^-3) / (1.2 * 10^-3) = 440 rad/s
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You can find the math related to this topic in "rotational motion or Circular motion" in any physics text book.\n\nYou need to search a good high standars book  for it..\n\nIf it contains above mentioned chapters, you will find some questions related to it..\n\nHope this helps you..\nGood Luck!
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All objects react as if there were only a single force acting upon them. Let's say an object on your desk had a force of ten newtons pushing it to the east. Let's call that force positive 10 newtons. According to Newton's second law, it would accelerate (pick up speed) moving in an eastward direction. It had a single force acting upon it and since there was no opposing force we say that it was an unbalanced force. Now say the same object had another force of 10 newtons acting on it in a westward direction. Lets call that force negative 10 newtons. Now we have two forces and since they are equal but have opposite signs the algebraic sum is zero ([+10N] + [-10N] = 0). We now have balanced forces - equal but opposite. Remember I said that all objects react as if there were only one force acting upon them? Even though there are two forces acting in opposite directions on the object on your desk, that object reacts as if there were only one force acting upon it - a force of zero newtons. BTW, that sum force is called the resultant. So according to Newton's first law since a zero force is applied to the object it will remain at rest. In short, unbalanced forces result in movement, balanced forces don't.