Can anyone explain torsional oscillation?
85890
Get a thin flexible rod. Let it hang down. twist the bottom of the rod, and release it. The rod will twist back and forth. This is called a torsional oscillation.
1165249
This is the part of the wikipedia that I understood.  I can explain it to you if you want to send me a message.  I think it makes alot of sense.\nNow let’s see what happens to our accelerating particles when a torque is applied to the spinning wheel. Assume the axis of rotation created by the torque is through the center of the wheel at 90 degrees to the primary rotation of the wheel. Let’s look at a particle that is on this axis of rotation. Since the particle is on the axis of rotation there is no direct motion applied to the particle at the instant of the applied torque. But let’s look at what will need to happen at the next moment in time. The particle is now going to be forced to curve again. This time in the direction of the curve so as to accommodate the tilt of the wheel. Now we have a particle that is already moving and it wants to keep moving in a straight line. So the particle will exert a force on the wheel. If you look at a particle on the other side of the wheel you will see that the force of the second particle is in the opposite direction of the first particle. That pair of unmatched forced is what causes the precession torque that is 90 degrees to the applied torque
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
12398
The rotational energy you impart to the object when spinning overcomes the force of gravity that would topple a stationary object.  \n\nA total layman's explanation:\n\nA perfect stationary top perfectly balanced will not fall over.  In real life though,it's impossible to perfectly balance a top.  It's shape practically guarantees that its center of gravity will not be directly over the balance point for any practical length of time.  Imperfections in the material will cause a slightly uneven weight distribution, air currents will introduce a small force, and mechanical vibrations will be transmitted through the medium the top rests on.  The upshot of all this is that as soon as the center of mass is no longer over the resting point, gravity is going to pull the top over.  \n\nNow gravity has an acceleration of 32 feet per second squared (if I remember my ancient high school course correctly).  It can only topple the top at a specific speed.  If you spin the top fast enough, any side that gravity tries to "pull down" will translate around so quickly that it literally has no time to fall.  If you were standing on a planet with a much higher gravity, you'd have to spin the top far faster to achieve stability, and if you were on the moon, it would spin for a much longer time. Eventually, however, the top starts to slow and the tip will inscribe little circles.  That action is called "precession".  And of course, eventually the thing drops to the table and the game is all up once it slows enough.\n\nAgain, that's a layman's view of how it works.  Some college physics major is probably going to come in here and make my answer look really foolish.
574012
I've alread answered two of your questiions in my last answer....\n\nIn this one, the guys above are correct - however, I'd try working it out for myself if I was you. Your tutor will soon realise something is wrong if you are giving answers to a subject you don't understand. Read the chapter, ask you teacher if you need more help. Learn Newtonian mechanics for yourself.
602725
It is knovn as "inertia". It is mathematicaly explained by the first Newton's law (Law of inertia). Both cases are closely connected to the act of the outer force - in the first case, there hasn't been any force around to make resistance to the object's motion (to change it's direction or to slow it down). In the second case, outer forces have made so big resistance to the object's motion that it has to stop because of the loss of kinetic energy, which has been lost in object's attempt to stay in constant motion.
742316
Torqu equals force applied times radius, so the equation would be T=Fr    or    T=mar    so, T=(4)(9.8)(6), or 235.2 n/m\n\nMoment of inertia is not needed to solve this problem unless it later asks for another different answer.
65267
It's me again, Dhinesh,your classmate.\nrolling frictin is the friction caused due to rolling. it opposes rolling & it acts on a single point as there is a single pt. which has contact with surface
737399
We know that the satellite swings from 10° to 58.5° in 20'.\nWhich means 48.5° swing around the 34.25 Lat. in half period of the oscillation of a sinusoidal wave or 180°. \nIn 360° it returns to the 10° Lat.\nJust apply the formula of a sinusoidal movement. \nThe radius R is half the swing, i.e. 24.25° and the period T of a full oscillation is 40'. (I.e. 360° in 40'). The speed is 360/40=9°/min.\nRemember that an oscillation can be represented by the projection of a rotating point, i.e. a sinusoid.\nThe frequency is f=1/T = 0.025 oscillations/minute. \nThe latitude is given by the projection on the Y axis of such rotating point, i.e. y=R*Sin Alpha, being Alpha the angle of the radius with the coordinate x at a given time. The value of +- y is a function of time and the rotational speed.\nDon't forget to add the first partial oscillation from 23° to 58.5° = 35.5° which occurs in 35.5/9 = 3.944 min\n\nPlease don't take this calculations for good without checking. It is a long time I don't deal with such problems. I hope I have helped understanding the approach.
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.
172617
It happens because of  gyroscopic inertia or rigidity in space. It is due to the spinning wheel on an axle. The wheel, once spinning, tends to resist changes to its orientation due to the angular momentum of the wheel. However, on static bike, there is no such resisting force and you become victim of gravity!!
1130182
Um... I think maybe pull... if something is rotating/turning it is creating it's own little gravity field (I think!) therefore making it easier to pull.  I'm no scientist but this is an educated guess.
724154
The answers describing the damped oscillations about the exact centre of mass of the planet are correct. The answer about terminal velocity meaning you don't go much beyond the centre is completely correct. You rock rock back and forth about the centre for a while, each time not getting as far away as the last. The distance would decrease proportionally to e^(at) where a is dependent on friction.\n\nBelieve me, I wish I didn't know this, been studying mechanics, including harmonic motion, for three days now for an exam!
712396
Here's a nice explanation:\nhttp://www.4to40.com/QA/index.asp?counter=382&category=science
881328
this requires way more than anyone can do here, but, i know of a cool java program that will do it for you: \n\nhttp://www.math.rutgers.edu/courses/ODE/sherod/phase-local.html\n\nyou can even set your initial conditions and print it out!
858812
Not sure about a website but the physics would be similar to what causes corrugations in a dirt road.  The vibration frequency would be related to the speed of the wheels.
653943
gravity and inertia\n\nread more below. It is really fascinating.