rigid body rotation: homework help please.?
826674
use conservation of angular momentum in these kind of problems\nNOT energy conservation\nif you dont know how to use the conservation equation, better read up on examples
800064
For these problems, it would be helpful to draw a free-body diagram to represent all the forces on this problem.  Here you have 4:\n\n1) Weight in the vertical direction going down, let's denote it W\n2) Force, moving in the x-direction, denote it F_1\n3) frictional force, moving opposite of Force, denote it F_2\n4) Normal force, perpendicular place of contact (opposite direction of W), denote it N\n\nNow use Newton's 2nd law to write equations in each direction:\n\ny-direction: \nN - W =0 (because the block does not move in the y-direction).  Therefore:\n\nN = W, where W = m*g\n\nx-direction:\nF_1 - F_2 = 0 [because the block is just about to move when F_1 is applied].  Therefore:\n\nF_1 = F_2, where F_2 = u*N\n\nMaking the necessary substitutions for F_1 and N, you get:\n\n30 = u*(4.4)(9.8).  Solving for u, you get --------->\n\nu = 0.7\n\n--------\n\nHope this helps
769161
Its a pain to do this with X and Y components, but there is an easier way....\n\nRange Formula - used when start and end heights are the same\nQ = Theta\nd = ((Vo^2)(sin 2Q))/g\n\n301.5 = ((Vo^2)(sin 2(25)))/9.81\n\nVo = 62.13 m/s\n\nNow that u have initial velocity, find the y component\n\nVo Sin Q\n\n62.13 sin 25 = 26.26 m/s\n\nThen No Time Formula\n\nVf^2 = Vo^2 + 2gd\n\n0 = (26.26^2) + 2(-9.81)d\n\nd = 35.15 m\n\nHope this helps
716372
1 and 2 are easy. I'll help with #3.\n\n3. You need I, the rotational inertia. From a table in my physics book, I = MR^2 for the reel. You need to convert the force to torque, T = F*R. \n\na. Then you can get the angular acceleration, alpha, from \nT = I*alpha\n\nb. In t = 0.25 s, omega will reach alpha*t. The angular displacement, theta, will be (1/2)*omega*t. From that and the circumference you can get the line drawn from the reel.
835037
Let the two case be identified as having  moment-of-inertia I1 and I2. \nTorque T on to the each of the two bodies produces angular accelerations (alfa1 and alfa2) given  by the relation T= I x alpha \n\nalfa1 = T/ I1 and alfa2 = T/I2\n\nStarting form rest(wi =0), let the time taken to reach final angular velocity wf be t1\nusing:  wf =- wi + alfa x t  \n \nwf/alfa1 = t1 and wf/alfa2 = t2 \nleading to \n\nt2  = alfa1x t1/alfa2 \n\n     = I2 xt1/I1\n     = (I2/I1) x t1\n     =. ......Ans \n(Hoping you will put the values and get the answer you want)
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!
748368
first get the time from the horizontal motion  t = .51 / 1.2 s (uniform motion, no acceleration here)\n\nthen use h = 1/2gt^2 for the vertical motion (with initial velocity zero)
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
918781
Where is your freebody diagram? There are other forces acting on the block than just gravity...\n\nGood Luck
841543
Are you taking into account the fact that the moment of inertia changes once the child jumps on?  In that case you would get that the intial moment of inertia is 1/2MR^2, then once the child jumps on (treating the child as a particle) the moment of inertia becomes 1/2MR^2 + McR^2 where Mc is the mass of the child.  \n\nFrom the information you've provided, that seems to be where you went wrong.  You could also probably convert the linear velocity that you are given to angular velocity and then just say \n\nI(initial) w(initial) = I(final) w(final)\n\nHope that helped.
772638
try the formulas:\n\nforce = mass * aceleration\n\nwork = force * distance\n\n\n\n\nWell i dont speak english very well, but i hope i help u.\n\nKisses
705060
If I understand the question, you are giving a velocity to the cube and don't want it to touch the hemisphere at any time in its flight. I think you want the minimum, not the maximum hor. velocity to do this, right? Because any arbitrarily high velocity will do it. I'll use a planar analysis where the hemisphere is a semicircle.\nAssuming the semicircle is centered at x=y=0, the trajectory will be a parabola y=r-kx^2 and at y=0, |x| must be >r. Another constraint arises when we consider the region of the semicircle near the top. Here the semicircle approximates a parabola, since 1-cos(small angle) approaches small angle^2/2. Thus deltay = r(1-cos(small angle)) = r-y approaches r*small angle^2/2 = r(x/r)^2/2, or x approaches sqrt(2/r*deltay*r^2) = sqrt(2r*deltay).\nDownward acceleration is g and horizontal velocity is v. When y=0, 0.5gt^2=r, so t=sqrt(2r/g), and x=vt must be >r. Thus the 1st condition (for clearance at the bottom) is v > r/sqrt(2r/g), or\nv > sqrt(rg/2).\nThe 2nd condition requires that when r-y=deltay is small, then and x=vt must be >sqrt(2r*deltay) where t=sqrt(2deltay/g), or  \nv > sqrt(2r*deltay) / sqrt(2deltay/g) or \nv > sqrt(rg).\nThis latter is the more stringent of the two requirements and thus is the minimum clearance velocity.
917347
It sounds like you know what to do.  From your description, it looks like a basic statically determinate problem.\n\nYou have a known applied force P, and two unknown reaction forces, lets say A and B for the top and bottom reactions respectively.  You also have two equations of static equlibrium, specifically the sum of the forces in the "x" direction, and sum of the moments about any point you wish.\n\n(Ok, really you also have the force of gravity and the normal force, but those are the only forces in the "y" direction so that equation won't help you, especially since the mass or weight isn't given).\n\nAssuming to the right is positive, assume A is to the left and B is to the right (this is intuitive, but if directions are incorrectly assumed the answer will just have a negative sign, thats all).\n\n                    _ ------------> P\n                    |   |\n                    |   |\n                    |   |     10.5 m\n                    |   |\n                    |   |\n                    |   |\n                    |   |<----------A\n                    |   |\n                    |   |       1.5 m\n                    |_|--------------> B\n                    \n                 (FBD!)\n\nSum of forces in "x" gives: P - A + B = 0\nSum of moments about B: A(1.5) - P(12) = 0\n\nI took moments about an unknown reaction, B, to eliminate it.  Also, I assumed counter-clockwise to be positive.\n\nSubbing in the value of P = 5.40 N into the second equation:\n\nA = (5.40*12)/1.5 = 43.2 N\n\nThen, put that into the first equation:\n\nB = A - P = 43.2 - 5.40 = 37.8 N\n\nThat should do it.  Both come out positive because their assumed directions were correct, however on the Free-Body-Diagram they are in opposite directions necessary for equilibrium.\n\nCheers,
77860
hey, didnt i answer something very similar to this for you? you need to do F=ma so 43000000N/2100000kg then you get an acceleration, but then you have to subtract gravity from this because thats the acceleration produced by the rocket but there is still gravity pulling with 9.8 m/s^2 in the opposite direction. so just do (43000000/2100000) - 9.8, and that's your answer :)
827798
You haven't told us what point it's pivoting about.\nThis is a Physical Pendulum problem:\n\nT = 2π√(I/mgL)......for small angular oscillations....\n\nI = mR²  where R is the mean radius of the hoop\nL = the distance from the pivot to the center of mass of the hoop\n\nGOTTA have L............
750653
The rotation rate is such that the centripetal acceleration equals the component of the downward acceleration of gravity that acts normal to the cylinder's wall at 68 deg. Thus, solve the following for w (rotation rate):\nw^2*r = g*sin(68).\nSince you asked what equation to use and did not ask for a worked-out answer, that's it. I hope it helps.
945234
[Edit -- two things.  #1 Oops, I made a mistake.  When I worked this, I used a mass of 4 kg instead of 0.4.  Now I'll go back through my answer and make the correction.\n\n[#2  There are two different methods here.  Look at both of them.  The <second> way is <much better> for this problem.  I think you'll see that.]\n\nSnow White has an answer, but she doesn't say how she got it.\n\nLet's start by getting the net force F = F1 + F2 = -0.6Ni + Nj with magnitude sqrt(1.36)N = 1.166 N and acceleration F/m = 2.915476 m/s^2.\n\nNow we have two ways to proceed.  This is linear motion with constant acceleration, and we know the direction ... tan theta = -1 / 0.6 = -5/3, so the angle theta in the second quadrant is 59 degrees north of west.\n\nAfter 1.6 seconds, the velocity is acceleration x time = 2.915476 x 1.6 = 4.66476 m/s in the direction indicated above.\n\nThe displacement s after 1.6 seconds is s = vt/2 or s = 1/2 at^2 = 4.66476 x 1.6 / 2 = 3.7318 m.\n\nTo get the position, we could use sines and cosines, but notice that tan theta above is -5/3, so the hypoteneuse of that triangle is sqrt(25+9) = sqrt(34).  Using that geometry, the position after 1.6 seconds is\n\nx = -3/sqrt(34) x 3.7318 = -1.92 m\n\ny = 5/sqrt(34) x 3.7318 = 3.2 m\n\nThat's your answer: (-1.92, 3.2) with velocity 4.66476 m/s and direction 59 degrees north of west.\n\n\nThere's another way to do it that may be better, especially when you're using vectors.  Do it by components, and keep the components separate.\n\nThe net force is F = F1 + F2 = -0.6Ni + Nj.\n\nThe constant acceleration a = F/m = -1.5i + 2.5j m/s^2\n\nThe velocity after 1.6 seconds is v = at = -2.4i + 4j m/s\n\nThe position after 1.6 seconds is s = vt/2 = -1.92i + 3.2j\n\nand this is the same answer we got by doing it the other way.  (You can, of course, get all the magnitudes and angles.)\n\nLooks to me that it's easier doing it by components.\n\nHope this is the same answer Snow White got.  This way, at least, you have two ways of doing it.