| New Reply |
Force applied on a rotating disc |
Share Thread | Thread Tools |
| Jun26-12, 01:23 PM | #1 |
|
|
Force applied on a rotating disc
I understand that if we apply a force on the axis of rotation of a spinning disc, the torque is in a different direction, and the axis is moved in a direction perpendicular to the direction in which the force upon the axis was applied, determined by the right-hand rule.
But what if the force was applied directly on the disc? e.g. we have a functional gyroscope, and we apply a force (effectively a torque) on the disc, say, by blowing down on one half of the spinning wheel. What happens then? Would we still get gyroscopic precession or will the axis of rotation simply move in the direction in which we are blowing? Thanks! |
| Jun26-12, 01:33 PM | #2 |
|
Recognitions:
|
In the case of helicopters, when using the cyclic (pitch and roll control), the result is equal and opposing forces between rotor and air, resulting in the air exerting a torque on the rotor. The rotor's reaction to the torque is about 90 degrees later out of phase, so a pitch torque results in a roll reaction and vice versa. The pilot controls are advanced 90 degrees to compensate for this.
|
| Jun26-12, 04:35 PM | #3 |
|
|
Regardless of whether the force is exerted on the axis or on the disk itself, either way a force is exerted upon the spinning disk. From there on it's the same case. A point of clarification: Given the way that gyroscopic precession is usually demonstrated one easily gets the impression that gyroscopic precession happens in response to a force. It's actually a sequence: a force exerted on a spinning disk causes motion of it, this motion then gives rise to gyroscopic precession. In the usual demonstrations the initial motion is imperceptably small, but without it the motion pattern of gyroscopic precesson would not arise. The initial motion follows the force; it's in the direction of the force. The direction of the motion of gyroscopic precession is at right angles to that initial motion. The reason I expand on this is to emphasize there is no lag in the response. Yes, the gyroscopic response is at a 90 degrees angle, but that 90 degrees angle is not due to any lag. The response is instantaneous. The 90 degrees angle is very counterintuitive of course, but the reason for it is known, and it's not some lag effect, there is no "later", no out-of-phase. |
| Jun27-12, 01:18 AM | #4 |
|
Recognitions:
|
Force applied on a rotating discLink to wiki article that mentions the 90 degree advance used on the swash plate for cyclic control on a helicopter: wiki_helicopter_cyclic_blade_control.htm |
| Jun27-12, 11:38 AM | #5 |
|
|
In general, about the description of helicopter cyclic blade control: As described in the wikipedia article, the way the blade pitch is controlled has conceptually two components. One is adjustment in time: the blade needs some time to arrive at the desired pitch, so an advance in time is needed. This component is strictly about when the blade control acts, so that the desired pitch occurs at the desired point. Then there is adjustment of blade pitch to take gyroscopic effect into account. That is strictly about where the blades have their maximum pitch. This is at a spatial 90 degrees angle. Of course, since the rotor angular velocity is pretty much a fixed rate controlling when and where is indistinguishable from a control point of view. I wonder, what if there is a helicopter design where depending on circumstances the rotor angular velocity is varied a lot. Freight helicopters need to adjust for flying with or without cargo. Possibly there are freight helicopters that vary the rotor angular velocity according to the load. Let's say such a variable rotor speed helicopter exists. - At every rotor speed the adjustment for gyroscopic effect is the same: a spatial angle of 90 degrees. - Depending on the current rotor speed the time adjustment must be set differently, so that at every rotor speed the maximum pitch occurs at the desired point. The two adjustments are handled by a single mechanism, but it's still two adjustments, one is a spatial adjustment, the other a timing adjustment. |
| Jun27-12, 01:15 PM | #6 |
|
Recognitions:
|
There are model helicopters without collective, but the control is not precise, and transition into and out of forward flight involves normally unwanted changes in altitude due to the momentum of the rotor. The better model hellcopters have collective, and there's a throttle versus rotor pitch curve mix on the transmitter to maintain rotor speed (the "pilot" only has a single axis for combined throttle + collective). In aerobatic mode (enabled or disabled by a toggle switch on the transmitter), the rotor speed is kept at a near constant and fairly fast rate of rotation regardless of collective pitch (positive or negative). |
| Jun27-12, 06:02 PM | #7 |
|
|
While in the motion pattern of gyroscopic precession: Velocity component perpendicular to the plane of the disk rises from zero to a peak and goes back to zero again. Acceleration component perpendicular to the disk oscillates between positive and negative. It seems to me these velocity/acceleration patterns are logical consequences of the motion pattern of gyroscopic precession, and not in themselves a causal factor. The reason I think that way is that in the convincing quantative derivation of gyroscopic precession rate (given spin rate and available torque) that I know, the velocity/acceleraton component perpendicular to the disk is a physical consequence, and not a cause. |
| Jun27-12, 09:17 PM | #8 |
|
Recognitions:
|
If torque, angular momentum, and angular velocity are stated as angular vectors, then the equation for precession can be stated in the form of a vector cross product: {torque} = {angular velocity of precession} x {angular momentum} Some of the angular momentum is in the direction of precession, but if the angular velocity of a disk is much greater than the angular velocity of precession, it's a small effect. |
| New Reply |
| Thread Tools | |
Similar Threads for: Force applied on a rotating disc
|
||||
| Thread | Forum | Replies | ||
| Force Required to stop a Rotating Disc | Introductory Physics Homework | 5 | ||
| Tangential force applied on a disc | Introductory Physics Homework | 3 | ||
| 2 people 1 rotating disc | Introductory Physics Homework | 2 | ||
| Rotating Thin Disc | Introductory Physics Homework | 6 | ||
| rotating disc | Classical Physics | 2 | ||