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Problem understanding angular momentum and newton 2nd law 
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#1
Mar1612, 12:41 PM

P: 10

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/8...es/lecture24/ 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 


#2
Mar1612, 05:41 PM

P: 1,395

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. 


#3
Mar1612, 07:38 PM

P: 10

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. 


#4
Mar1612, 09:03 PM

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P: 7,033

Problem understanding angular momentum and newton 2nd law
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. 


#5
Mar1712, 01:08 AM

P: 10

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. but again, Why will the Tension in the Rope have any Increase(whether in the horizontal plane or anywhere)....IF...the wheel is spinning. 


#6
Mar1712, 01:57 AM

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#7
Mar1712, 04:46 AM

P: 10

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. Thank you for your patience with me, much appreciate, i am really trying my best. 


#8
Mar1712, 10:58 AM

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P: 7,033

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. 


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