Physics question involving angular 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
723976
Question is: why on earth would you want to make that calculation?  m/s^2 is the standard form for acceleration.
791900
Ok.  From the hint start by figuring out the ratio of wheel speeds\nCircumference of A =2*pi*10\nCircumference of C = 2*pi*25\nso C/A=2.5  ie A must go around 2.5 times to make C go around 1 time.\nThere are 2*pi radians to 1 revolution so 100 rev/min =200*pi rad/min.  A must go 2.5 times as fast as this or 2.5*200*pi=500 rad/min\nAccelleration is the derivative of velocity which means that we must integrate the angular acceleration to obtain the speed at any given point of time.\nIntegrating 1.6rad/s^2 with respect to time and setting velocity equal to zero at time equals zero we obtain angular velocity= 1.6*time elapsed (rad/sec).  Note that this speed is in "per seconds" and we need it to be in "per minutes" so we must multiply by 60 for the units conversion.\nThis gives us the angular speed in rad/min for any chosen passage of time.  We shall set this equal to the goal speed of 500*pi rad/min as mentioned above. So\n1.6*time elapsed*60 =500*pi\nSolving for time elapsed we obtain a time of 1.63 minutes.
701707
The Average Linear Speed is {total distance / total time taken} = 65/ 5.7 = 11.4 m/s. \n\nDivide this value by the radius of the wheel to get the ‘The Average Angular Speed’. \n\nThe Average Angular Speed is = 11.4 / 0.02 = 570 radian per second.\n\nSince the motion is uniform,\n\n The Average Angular Speed = {Initial Angular Speed + Final Angular Speed} / 2   \n\nBut the Initial Angular Speed is zero.\n\nTherefore, the Final Angular Speed   is twice the Average Angular Speed\n= 2 x 570 \n=1140 radian/ second. (The answer for the second question)\n\nSince the Initial Angular Speed is zero,\n\nThe Angular acceleration is, “Final Angular Speed/ total time”.\n= 1140/5.7 = 200 [radian / second] / second. The answer for the first question)\n\n\nThe number of revolutions is found by “total distance / (pi x diameter of the wheel)” = 65 / pi 0.04 = 517.25 revolutions.
985616
All are true. For the last one,\nsuppose acceleration = a.\nThen the velocity is at + c\nand the position is at^2/2 + ct + d,\nwhere c and d are constants.
382082
use d=V1*t + [(a*t*t)/2]\nd=distance=490m\nV1=initial velocity = 0m/s\nt=time = ?\na=acceleration due to gravity = 9.81m/s*s\n\nusing equation above, t will come up to be:\nt=(square root)(2*d/a)\nt=(square root)(2*490m / 9.81m/s*s)\nt=9.99 sec or 10 sec.
701875
We're given r = 10 m, t = 5.0 s, and that \n\nθ = αt^2, where α = 0.3 rad/s^2\n\n(a) Rotational velocity, ω = θ/t = αt^2/t = αt = (0.3)(5.0) = 1.5 rad/sec\n\n(b) Tranlational velocity, v = ωr = (1.5)(10) = 15 m/s\n\n(c) Tangential acceleration, a = ω^2r = (1.5)^2(10) = 22.5 m/s^2\n\n(d) Radial acceleration, α = ω/t = 0.3 rad/s^2
871488
Torque=F*D\nT=18*.35\nT=6.3\nWe know that T=a*i     [a=angular acc.\n                                       i=moment of inertia]\n         T=a*mr^2       [m=mass\n                                  r=radias]\n         6.3=a*2.65*(.35)^2\n         a=19.4 radian/sec^2\n\nProblem solned. Have a nice day.
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)
612059
Consider two objects having masses 1kg and 5 kg are in a horizontal plane.\n\nWe apply a force of 10 newton to the 1 kg mass and 50 newton to the 5kg mass. \n\nThat is, we apply less force for less mass and greater force for greater mass.\n\nNow we find the acceleration of the two bodies.\n\nThe1 kg mass will have an acceleration of 10/1 = 10 m/ s^2. That is its speed will be going on increasing by 10meter / second in every second. \n\nSimilarly the 5 kg mass will have an acceleration of 50/5 = 10 m/s^2. That is its speed will be going on increasing by 10meter / second in every second. \n\nBoth have the same acceleration and both have the same speed at any time after start.\n\nNow returning to gravitational pull, let us place the 1kg mass in one pan of the balance and the 5kg mass in another pan. \n\nThe pan having 5kg mass will come down and the pan having 1kg mass will go up.\n\nThis is because the earth is pulling the 5 kg mass with a force of 50 newton and 1kg mass with a force of 10 newton.\n\nIf  we drop the two objects from the same place, as we have seen already the 1 kg mass will fall down with an acceleration of 10 m/ s^2 and the 5kg mass will also fall down with 10 m/ s^2. But the forces acting on them are different.\n\nThe Earth’s pull is such that it gives less force for less mass and greater force for greater mass.\n\nTo be more specific it gives a force of 10 newton for every kilogram of mass and hence the acceleration is for any mass is the same as approximately 10m/s^2.
912569
I think F is the answer. Taking acceleration due to gravity is constant at 9.81ms^-2, but this assumes that air resistance is neglected, and g is a constant.\n\nIf air resistance is NOT neglected, and acceleration due to gravity is not taken to be constant, H is the correct answer. The rest of the options are wrong.
771534
The problem is about conservation of angular momentum.  The tricky part is (b). To solve this use this definition of angular momentum and draw a picture to find out the value of r sin θ\n\nThe angular momentum of a particle of mass m with respect to a chosen origin is given by vector product L = r x p= mvr sin θ
37325
first thing you need to do is calculate the moment of inertia of the wheel.  this can be done by treating the poles and the main wheel separately.  the moment of inertia of a circular disc (the wheel) is (1/2)MR^2, and of the rods (also a circular disc!) is (1/2)mr^2.  add them together to get the total moment of inertia.  \n\nyou'll need to use conservation of energy.  specifically, when the wheel drops a height h, the potential energy lost is -mgh.  this energy gets split into two parts, the rotational energy of the wheel and the translational energy (since it's moving downwards).  you can then write an equation setting the potential energy loss equal to the sum of the rotational and kinetic energies:\n\n-E_p = E_rot + E_trans\n-mgh = (1/2)Iw^2 + (1/2)(m + M)v^2\n\nhere I used (1/2)Iw^2 for the rotational kinetic energy and (1/2)mv^2 for the translational kinetic energy, where I substituted (m + M) for the total mass of your object.\n\nthe key point is now to realize that because the wheel unwinds along a string that is wrapped around the pole, the angular velocity w = v/r where r is the pole's radius and v is the translational velocity of the pole.  so you can substitute that into your expression.  it's not clear from your question whether you want the angular acceleration or the translational acceleration.  so you can either substitute the rw in for v, or v/r in for w.  \n\nin either case you then have a formula for the velocity as a function of the height of the wheel, which you can then solve for the velocity, and then take the derivative to get the acceleration.\n\ngood luck!
846820
The formulae for linear acceleration and angular acceleration are very similar:\n\nF=ma\nT = I a  (the T and a here are usually written Tau & alpha in greek)\n\nYour 'torque' of 60N can't be right though, a torque value would be in N-m. Assuming its 60N-m then\n\na = T/I = 60 /12 = 5 rad/s^2
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
441442
you have your units wrong, do you mean 5m/s or 5m/s^2 and you mean 20m from the ground?\nIf acc=5 when height=20 speed= 14.14 in upwards direction (using 2ax=v^2-u^2)\nIf initial speed is therefore -14.14m/s then at t=1 speed=-4.34m/s so it is still going upwards\nt=2 speed= 5.45m/s it is now going down. Both last results worked out using v=u + at\n(Sorry I assumed you meant 5m/s^2 straight up, you need the angle at which it accelerates aswell)
784184
θ =0.3t² rad\nω = dθ/dt = 0.6t rad/s\nα = d²θ/dt² = 0.6 rad/s²\n\nWhen t = 5\na. ω = 0.6t\n= 0.6*5 = 3 rad/s\n\nb. v = rω\n= 10*3 = 30m/s\n\nc. α = 0.6 rad/s²\n\nd a = v²/r (= rω² = 10*3² = 90 m/s²)\n= 30²/10\n= 90 m/s²