# Not tilting a gyroscope

by bobie
Tags: gyroscope, tilting
PF Gold
P: 574
 Quote by Nugatory You can check this result against the formula that you've been ignoring... equal to the torque times the angle over which it is applied.
I am confused, Nugatory: in the video the girl is pushing the wheel to her left, and the platform moves to her right, wiki says that the wheel has inertia and resists any change of direction, so, I thought that the platform is moving as a reaction to the force applied to the left (3rd law).
I started the new thread to learn how to calculate this force of inertia and consequently the force required to win it. You said:
 Quote by Nugatory If you actually change the orientation of the spin, you're doing work because you're applying torque across an angle, and that does work just as applying a force across a distance does. Indeed, it's fairly easy to derive the formula for the work done by a torque from the the old classic ##W=Fd## ] (and I strongly suggest that you try doing that as an exercise).
but everybdy said that the work done is (next to) zero

I'd really be grateful if you could be so kind as to help me with that exercise, as I am totally bewildered:
- the wheel has (m = 2 kg, r = 0.318 m, ω = 5 rps, v = 10 m/sec) Lw = 6.4 J*s
- the girl-platform system has (m= 50 kg, r = 0.4 m, ω = 1/2π rps, v = 0.40 m/s) Lg-p = 8 J*s
- the angle of rotation is 90° = 1.6 rad, d = 0.64 m
Is Lg-p =F*d?
is F then equal to 8/.64?
Is work done in order to rotate the wheel 2*8/.64 or zero?
 HW Helper P: 7,107 I added this comment to the video, noting that there is an external torque involved. "no external torque" - When the girl applies an internal horizontal torque to the wheel, the equal and opposing torque that would tend to rotate her body horizontally is opposed by an external torque due to a lateral force from the floor (surface of the earth) that is exerted onto the bottom of the rotating platform and the girls feet. The closed system needs to include the earth in order to be a closed system. If the girl was hovering while in free fall inside an aircraft simulating zero g, then the girl and the wheel would be a closed system (ignoring tiny effects like aerodynamic drag). In this case the girl's body would be flipping and twisting while twisting the spinning wheel. In an ideal case with no losses, then the original direction and magnitude of the angular momentum of the girl and wheel would be preserved. Total energy would also be conserved, as any internal work performed by the girl would result in a corresponding change in the kinetic energy of the system (girl and wheel).﻿ The point here is that the external torque due to the lateral force from the floor at some distance from the center of mass of the girl and wheel equals the torque being applied to the wheel. This external torque has a horiztonal axis (parallel to the floor), and the precession of the spinning wheel results in a perpendicular reaction to that torque, which in this case is a rotation reaction along a vertical axis. Using right hand rule, in the initial state, the spinning wheel has angular momentum vector pointing to the girls left. If the girl applies a clockwise torque, then the external torque is a clockwise torque, and the torque vector is horizontal, pointing away from the girl. The cross product of these two vectors results in a rotation vector pointed downwards, so the girl and wheel will rotate to the girls right. If the girl applies a counter clockwise torque, then the girl and wheel will rotate to the girls left. The angular acceleration about the vertical axis only exists while the girl applies a torque to the wheel. What's notable is that in the initial state, there is zero angular momentum about the vertical axis, but after the external torque is applied, it becomes non zero. If the girl was in a simulated zero g environment, such as an aircraft or space station, then angular momentum is conserved, the direction and magnitude of the angular momentum vector would not change no matter how the girl twisted the wheel, and as mentioned, the girl would be flipping and rotating as a result of applying an internal torque to the spinning wheel. It would be similar to the twisting techniques used by divers, gymnasts, trampoline competitors, ... .
 PF Gold P: 574 Thanks rcgldr, that is very interesting but a bit too complicated/abstract for me, could you please translate this into practical numbers and equations with the (plausible?) data in my previous post, considering a concrete exapmle? Thanks a lot for your understanding
 HW Helper P: 7,107 The torque applied by the girl to the wheel is the same as the external torque applied by the floor to the girl which is why her body does not rotate about a horizontal axis. The platform, girl, and wheel are a complex system. The math would involve using the external torque and the angular momentum of the system to determine the rate of precession. The zero g environment would change things. The initial angular momentum vector of girl and wheel points to the girls left. No matter what she does, that won't change. If she applies a torque to the the wheel, she "tilts" and and twists along with the wheel but the systems' angular momentum vector will remain the same. When divers initiate twists while flipping (with no initial twist generated before leaving the diving board so that the initial angular momentum vector is horizontal and to the side of the diver), they will get tilted until they stop the twisting motion.
 PF Gold P: 574 How do you relate the two values of L : 8 and 6.4 J*s?
 HW Helper P: 7,107 I don't understand. In the initial state, only the wheel has angular momentum, the girl and platform angular momentum is zero. You mentioned work done. Work done would be torque times angular displacement, and would result in a change in the kinetic energy of the girl + wheel + platform system if there were no losses. The issue is that the work being done transitions between positive and negative as the orientation of the wheel changes.
PF Gold
P: 574
 Quote by rcgldr I don't understand. In the initial state, only the wheel has angular momentum, .
The platform acquires L = 8 J*s, isn't this related to Lw? if Lw had been , say, 20 instead of 6.4 wouln't it be different?

 Quote by rcgldr .. when the girl peforms work by exterting a torque over some angle,..
What is concretely the value of work in this particular example, and how do I find it? don't I need to relate the two values? that is what I still do not know, how to calculate it in a concrete example
 HW Helper P: 7,107 Another issue is that in addition to applying a torque that forces the wheel to rotate about a horizontal axis extended outwards from the girl, the girl also applies a torque to prevent the wheel from precessing about a left to right horizontal axis in front of the girl, by keeping her arms straight and not allowing one arm to move inwards with respect to the other. This torque would also correspond to an external torque related to lateral force applied by the floor onto the platform onto the girls feet.
PF Gold
P: 574
 Quote by rcgldr the girl also applies a torque to prevent the wheel from precessing about a left to right horizontal axis in front of the girl, . This torque would also correspond to an external torque related to lateral force applied by the floor onto the platform onto the girls feet.
If I got it right, when she pushes her right hand down she meets no resistance but at the same time the right axis starts pushing horizontally (clockwise looking from above) and she does some work pushing anti-clockwise leaning on the platform that starts moving horizontally clockwise as a reaction (3rd law).

Is that right?
If so, a couple of questions:
- can the girl modify the speed of the platform in any way (for example) tilting the axis faster? or is the outcome determined only by Lw?
- can the situation in space can be obtained hanging the girl from the ceiling?
- is Lg-p 1/2 the work done by the girl or it follows the rules of elastic collision m*v=m1*v1?

Thanks a lot for your invaluable help
 HW Helper P: 7,107 In the wheels initial position, the girl inputs a roll torque. She also inputs a yaw torque to prevent the wheel from precessing about a vertical axis with respect to the girl by keeping both arms extended. The girl will feel a resitance to any attempt to rotate the wheel along any axis other than the wheels primary axis, and this resistance will be greater if the wheel is spinning faster. To simulate the zero g situation, the girl could be harnessed into a gymbol type apparatus to allow rotation along any axis, similar to ones somtimes seen at carnivals, but it would have to be made of very light materials, perhaps one built from carbon / kevlar fiber (the carnival ones are metal and somewhat heavy, too much inertia).
PF Gold
P: 574
 Quote by rcgldr a resitance to any attempt to rotate the wheel along any axis other than the wheels primary axis, ...).
Thanks, rcgldr,
I know that after 40 years these problems were raised , some processes are still mysterious, do you know of a reason why the wheel stops dead although it's on frictionless gimbal (here https://www.youtube.com/watch?v=IaOIWXqH9Io at 1:45) and why the big one is so light at
3:20?
BTW he says at 1:50 that there is no angular momentum, isn't it inertia that is missing?

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