hi, thank you for any help. -[problem understanding angular momentum and newton (second)2nd law] -[usual/classic problem of spinning bike wheel on rope doesn't fall] (from 28min:10s to 29min:15s and also 37:48 to 39:00) http://ocw.mit.edu/courses/physics/...echanics-fall-1999/video-lectures/lecture-24/ -Free Body Diagram Attached Below how is it the tension increases IF the wheel is spinning? more amazingly, when the additional weight is placed on the axle, tension simultaneously increases? I probably have a misunderstood concept somewhere, can't seem to figure it out. thank you for your help in advance, Glenn
Newtons third law goes for torques as well. If the string puts a torque on the wheel, the wheel will put a torque on the string, wich will increase the tension. If you were to precess the wheel with your hands you would feel a torque that would raise on of your hands and lower the other.
Torque is produced on the axle of the wheel in a direction normal to the plane of the wheel. The tension in the string holds the axle off the ground, but really you have a torque acting on the rod. t=Ialpha=alpha*1/3m*L^2 this torque must be equal to the total force of gravity to allow the wheel to do its crazy thing in the air. So where is this torque coming from? The angular acceleration is the derivative of angular momentum, and as acceleration forces are pointing radially inward, so are torque forces pointing radially outward. The strongest radial torque will be parallel to the string because at this angle, there are two points of torque acting on the rod. This is why it stays up. There is also a radial torque that is normal to the radial momentum and to the axle. This torque is pointed in the same direction as the wheel rotates, and is why the movement of a circular path can be witnessed. The result of that torque is an angular moment torquing not only the axle, but also against some of the inertia of the spinning wheel.
Assuming no vertical acceleration of the center of mass of the system, the tension in the string equals the weight of the wheel (or the wheel + 2kg weight), regardless if the wheel is spinning and precession or not moving at all. If the wheel is orbiting horizontally, then there's some additional tension related to the radial acceleration of the wheels center of mass (and the 2kg weight if it's added), and there is a component of horizontal force applied by the string to whatever the string is suspended from. The actual motion of this setup is more complex than mentioned in that class. Video #9 on this web page shows the more complex motion, with the gyroscope cylcing through horiztonal orbits of varius radius, until the rate of orbit gets close to the rate of precession. I'm not sure what the idealized pattern with no losses would be and if the initial condition could be setup so the orbital rate matched the precession rate and remained that way. http://www.gyroscopes.org/1974lecture.asp The lecturer states that the main rule during the complex motion is that the center of mass of the system will move along a horizontal plane (it won't rise or fall), but that probably assumes an idealized situation where energy is not being lost to heat as the wheel slows down.
hmm... there must be mistake in my thinking: -wheel puts torque on string(length of axle,r X weight of wheel) [nth to do with wheel spinning or not] - string puts torque on wheel(newton's third law:tension in rope=weight of wheel) (so torque = length of axle X tension) since tension=weight. torque is the same my point: since weight nv increase, tension should not increase whether spinning or not. hmm...I can accept the fact that if the rod precesses(move in horizontal plane), must have had Horizontal acceleration(must have been given by the tension), which should be the increase in tension?(i might be very wrong) but again, Why will the Tension in the Rope have any Increase(whether in the horizontal plane or anywhere)....IF...the wheel is spinning.
From a practical standpoint, a torque can not be applied to a string, only a force, and generally just a tension in the string (since the string has a relatively small amount of mass, and is not rigid). The string generates a force at the end of the axle, mostly upwards. Gravity generates a downwards force at the center of mass of the wheel. Both forces result in a torque applied to the axis of the wheel. If the wheel is spinning the reaction to the this torque is precession, which in turn generates an opposing torque due to the string and gravity, keeping the wheel axis somewhat horizontal. It doesn't increase. Where did you get the idea that an increase in tension was implied by the video? If the center of mass of the rod and wheel moves in a circular path, then there is some radial acceleration, which results in a horizontal component of tension in the string in addition to the weight of the rod and wheel. The vertical component of tension will correspond to the weight of the rod and the wheel, unless there is a vertical component of acceleration due to the initial state.
so when you say, additional tension, you mean? also in the vid by prof lewin(have put in the times at the top), he did mention that the tension increases. [so that means i have misunderstood which tension he is talking abt] hmm..if let's say case 1: the wheel is not spinning, and it is suspended frm a very taut string Just about to break. IF now, i spin it and then place it on the same string, will it break? (or better still, if i use a spring gauge to measure the tension, will i get the same reading, despite the wheel spinning or not?) erm, kinda hard to visualize, i think my free body diagram is wrong. so if nw there is an new additional horizontal component of tension in the string, plus the usual vertical component of tension(which did not change since due to weight), than if you sum up the horizontal and vertical components to find resultant tension(has this increased? compare to when wheel no spinning). still didn't quite catch: how come when the wheel is spinning About it's own centre(ignoring the movement Around the rope for 1 moment), it results in a horizonal component of tension? Thank you for your patience with me, much appreciate, i am really trying my best.
What he writes on the chalk board is tension = m g = weight of the wheel and axle. I didn't see a part where he states tension increases (but maybe I missed it). The vertical component of tension = weight of wheel and axle (as long as the center of msss doesn't accelerate vertically). The point I was making is that the wheel and axle do not (normally) precess about the center of mass in this situation, so the center of mass is moving in a circular path, which means there is horizontal acceleration. The horitontal force = m a = m v^{2} / r (where r = radius of curvature of the path). The total tension in the string will be equal to the vector sum of the vertical and horizontal forces.