Rotational Dynamics and Static Equilibrium. What is the acceleration of the ccart?
412737
force = m*a + I*alpha\n\nthe first part is for the cart and the second is for the pulley\n\nThen relate the angular acceleration to the linear acceleration as follows:\nalpha=angular acceleration= w^2*r   in rad/s\nso a(accleration of cart)= w^2*r^2   m/s\n\nso force=m*a + I*(a/r)\n\nfrom which you can find a (acceleration)
804891
For a moment of inertia I, an angular acceleration A and a torque T we have T = IA.\n\nThe moment of inertia is equal to the mass of an object multiplied by the square of a representative length called the radius of gyration, which is in fact the root-mean-square distance of its elements from the axis of rotation. If the axis of rotation passes through the vertical part of the Y section then the elements of the section will be symmetrical around the axis of rotation and the root-mean-square distances can be measured and the radius of gyration could be calculated.
807395
http://eta.physics.uoguelph.ca/tutorials/torque/Q.torque.inertia.html\n\ngives 2I=mr^2  so I = 163*0.306*0.306/2\n\nhttp://buphy.bu.edu/~duffy/py105/notes/AngularMo.html\n\ngives 2 * kinetic energy = Iw^2\n\nso KE at 1500rpm = I*4/2*pi^2*(1500/60)^2 joules - everything is frictionless so this is the work done in accelerating the flywheel to 1500 rpm\n\nyou can work out the KE at 1181 rpm and find the difference that was dissipated as heat.\n\nbest of luck - Mike
984692
The moment of inertia of a solid cylinder is 1/2 mr ^2..\nTorque is the product of moment of inertia and angular acceleration.  It can also be expressed as the product of the force applied to the edge of the disk and the radius.  Set the two products equal to each other and solve for m.\n\nT = I α = r F\n\n1/2 m r ^2 α = r F\n\nm = (r F) / (1/2 r ^2 α)  = 2 F / r α\nm = 2 (40) / 0.16 (90)\n\nm = 5.56 kg
753766
First we determine the total moment of inertia of the system (with everyone on the rim).\n\nPeople: I = m r^2 = (65*150) (96^2) = 8.986 * 10^7\nSpace Station: I = 5.08 * 10^8 (it says 5.08 108 in your question, so I'm gonna assume that this is what you meant)\nSum = 5.679 * 10^8 kg*m^2\n\nIf the apparent acceleration = 1g\nThen, using a = v^2 / r , we can get the linear velocity of the edge\n\n9.8 = v^2 / 96\nv = 30.672 m/s\n\nTo get the angular velocity, use w = v / r \n(w is angular vel.)\n\nw = 30.672 / 96 = 0.3195 radians/sec\n\nSince angular momentum is conserved, we can equate the following:\nI * w (before meeting) = I * w (during meeting)\n\nObtain I (during meeting)\n\nPeople: I = m * r^2 = (65*50) * (96^2) = 2.995 * 10^7\nSpace Station: I = remains the same\nSum = 5.380 * 10^8 kg*m^2\n\nConservation of angular momentum:\nBefore = During\n(5.679*10^8) * (0.3195) = (5.380*10^8 ) * w\nw = 0.337 radians/sec\n\nFind the linear velocity:\nv = w * r = 0.337 * 96 = 32.377 m/s\n\nAcceleration to maintain circular motion:\na = v^2 / r\na = 32.377^2 / 96 = 10.92 m/s^2\n\n(that's 1.11g, if you're wondering)\n\nTada!
256104
Read Newton's 3rd law of motion.\nAlso, use F=ma to figure out your acceleration.
969112
First you need the mass moment of inertia of the ring.  Since it is thin, I think it is just the mass times the radius squared.\nNext, you need the torque applied to the ring.  That would just be the force times the radius of the ring.\nTorque = Inertia x angular acceleration\n(solve for angular acceleration)\n\nAssuming constant angular acceleration for the entire 5 seconds:\nangular velocity = angular acceleration x time
210676
Use free body diagram to balance the forces and u can derive the equation of motion \nIf u want i can give u full derivation.\n"::
723260
Question 1:\n\nTorque is a force applied at a distance,\nT = F * r\nWhere T is the torque, F is the force, and r is the radius the force is applied away from the center or rotation.\n\nT = 40 Newtons * .15 meters = 6 Newton meters\n\nA torque causes an angular acceration,\nT = alpha * I\nwhere T is the torque,  alpha is the angular acceleration, and I is the rotational inertia ("moment of inertia").\n\nAn angular acceleration relates to a linear acceleration,\nalpha = a / R\nwhere alpha is the angular acceleration, a is the linear acceleration, and R is the radius of rotation (the wheel's radius in this case).\n\nThe linear acceleration can be found by finding the change in linear velocity with respect to time,\na = delta v / delta t\na = 2.25 m/s / 3 s = .75 m/s^2\n\nalpha = .75 m/s^2 / .5 meters = 1.5 radians per second^2\n\nT = alpha * I \n--> I = T / alpha\nI = 6 Nm / 1.5 rad/s^2 = 4 kg m^2\n \nDo your own homework next time (like on question 2).\nhttp://sln.fi.edu/qa97/spotlight3/spotlight3.html
943735
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
737218
Given a rotating hub or wheel all the centrifugal forces are along the radii from every point on the hub to the center. Because of circular symmetry, all the these forces will cancel; one pointing up is cancelled by one pointing down. Therefore the net external force is zero; there is no way the linear motion is influenced.
35590
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.
140474
Use the formula Ac = v^2/r = r w^2, where w = angular speed.\n\nSince you already have the angular speed, you can use the second form of the equation above.\n\nAc= r w^2 =(1.0 m) (2.0 rad/s)^2
782945
I won't do your homework for you, but you should look here:\nhttp://hyperphysics.phy-astr.gsu.edu/hbase/carcr.html#cc1\n\nGood luck. :)
986259
(a) Displacement is given by the equation:\n\nx = 6 cos[3π(t+(1/9))], so at t = 2 sec\n\nx = 6 cos[3π(2+(1/9))] = 3 m\n\n(b) Velocity is v = dx/dt = -18πsin[3π(t+(1/9))], so at t = 2 sec\n\nv = -18πsin[3π(2+(1/9))] = -48.97 m/s (in the negative x direction)\n\n(c) Acceleration is a = dv/dt = -54π^2cos[3π(t+(1/9))]\n\nso at t = 2 sec, a =  -266.48 m/s^2\n\n(d) Phase of the motion is θ = 3π(t+(1/9)), for t = 2\n\nθ = 19.89 radians\n\n(e) The frequency = ω/2π = 3π/2π = 1.5 revolutions per sec\n\n(f) Period = 2π/ω = 2/3 sec
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;
796588
Centripetal acceleration A = w^2*r, where v is velocity, w is the rotation rate in rad/s and r is the radius. Since velocity v = w*r, we can substitute v/r for w in the equation for A: A = v^2/r. \nSo in this case, A = 2000^2/0.153 = 2.6144 * 10^7 m/s^2 (the answer to (a). For (b), F = M*A = .001 *  A = 26144 N.